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121 Posts
Discussion Starter #1
Some things will be the same as before, I'm still 66 years old and counting, but this time around I'll try to make my posts easier to read and understand! I only have my notes to go on because as I posted on FTE...

Disaster at The Diesel Garage Forum!!!

... Ironically I thought documenting all my work here on The Diesel Garage Forum would protect it in case my computer should crash! Well back to the drawing board with my previous introduction...

This post might seem a little weird to many, so let me explain. I'm a 66 year old engineer whose retirement hobby is analyzing topics that interest me, and I enjoy sharing these results with others. I used to analyze for dough, but now it's just to know! I often find when analyzing a topic using math and equations, that many things which I thought I knew about it turn out to be only partially true or even totally false. I realize that non-engineers don't like equations and such, but you just can't adequately understand and explain some things without them. I'll try to make equations as easy to understand as possible, and here's some reasons why you might be interested in putting up with them...

#1) Dynamometers don't measure rear wheel HP or rear wheel TQ! You do get some dyno numbers for HP and TQ, but the TQ numbers lead to endless confusion, and you'd be better off not even knowing them!

#2) Having separate dyno categories for highest HP and highest TQ is like playing high low poker, and the real dyno prize should be based on wining the whole pot!

#3) It's all about HP, and TQ doesn't have to be considered or even known, to calculate how well your truck accelerates to win a drag race, tows a load up a mountain, or makes a speed run at Bonneville! However, you do have to worry about TQ when wrenching bolts to the correct tightness!

#4) Shifting at or near the RPM for peak TQ or peak HP or wherever, isn't the best strategy for getting the lowest ET to win a drag race. In fact, after coming off the line you should stop looking at the tachometer, and watch the speedometer!

I know this won't be an easy sell, and I can't do it all in a single post, so over the coming weeks, make that months, and possibly even years if I'm lucky enough, I'll be posting on this thread enough explanations using words, equations, charts, and graphs, that will hopefully convince anyone who takes the time to read it, that the above statements are true, and I'll try to answer all questions that come up along the way until the job is done!

121 Posts
Discussion Starter #2 (Edited)
The true story of how HP and TQ were actually discovered...Part 1 of 3

Well when I tried to post this I discovered that the new site is counting each character as a word, so I'll have to post this in 3 parts! For whatever reason before I was allowed to post very long stories with no problems!

This time around I think it's a good idea to first let everyone know how all this HP and TQ stuff, which is the primary interest of most diesel enthusiasts, actually got started! While it's true that Newton, Watt, Joule, and others get the credit for all of this, that's because their original hand written manuscripts survived the ravages of time to be displayed in museums, however, the concepts of HP and TQ date to prehistoric times, and were actually discovered by an ancient caveman!

Just for the fun of it assume you're the caveman in the Geico add, and you've just invented the wheel, and you carve some notches around the perimeters of three of them and name them "gears", and then you drive some "gear shafts" into the ground and fit your newly invented gears together like is shown in this picture...

...and now you're trying to figure out how these gears operate, what they're good for, and then write some equations to quantify how they operate so you can show them to your boss and get a raise! You wind up using the equations in the lower left hand corner of the picture to get your raise, and you literally get burned by the equations in the lower right hand corner, but I'm getting way ahead of myself again!

First you "measure" gear revolutions by counting them, and observe that if you turn Gear1 6 revolutions, that Gear2 turns exactly 2 revolutions, and that Gear3 turns exactly 1 revolution. Next you "measure" the diameters of the 3 gears with a stick, and divide by 2 to get their relative radii as, R3={(2)(R2)}={(6)(R1)}. It appears as though the ratios of the gear radii, such as R3/R1=6, predict the number of gear revolutions between the respective gears, but you still don't have a theory with an equation to explain this!

Now if a reader of this post actually looks at the gear picture and uses a modern day ruler, he'll see that these relative radii aren't quite exact, but since you're a caveman you haven't learned to deal with fractions yet, and besides you've already figured out that it's the number of gear teeth that determines the gear revolutions, and you're just using the gear radii as a convince instead of counting teeth!

Next you decide to "measure" gear revolutions versus TIME in units of RPM, because your sun dial only has a one-minute accuracy anyway, and that gives, RPM1=6, RPM2=2, and RPM3=1, and in relative terms that gives, RPM1={(3)(RPM2)}={(6)(RPM3)}. Combining the RPM measurements with those of the gear radii gives the equation, RPM1={(R2)/(R1)}{RPM2}={(R3)/(R1)}{RPM3}, and now you've got an equation that seems to be a universal law for gears, but there's still no theory to explain why it's true!

At this point you get an inspiration, and realize you can improve the one-minute accuracy of your sun dial by using it to "measure" the TIME for each revolution of Gear3, and then use each revolution of Gear2 to "measure" TIME in 1/2 min increments, and use each revolution of Gear1 to "measure" TIME in 1/6 min increments, which you later decide to call 10 sec! Unfortunately no one is in much of a hurry in prehistoric times, and you don't think a sun dial with a second hand will be a big seller, so you press on with your research.

It's not until much later, when you accidently get burned while discovering HP, that you realize how important it is to have a sun dial with a second hand, because you decide to use ft-lb/sec as the unit for HP so that it can be quantified very accurately, because this HP turns out to be very dangerous stuff if you try to "handle" too much of it all at once!

So you proceed with your experiments. By now you've recalibrated your measuring stick using your foot as the standard unit for DISTANCE, and you remove the gears from their shafts, and roll them along the ground, and "measure" the linear DISTANCE in ft that each gear travels for 1 complete revolution. Then you use a piece of string to "measure" the total perimeter DISTANCE around each gear, and discover that the perimeter DISTANCE=linear DISTANCE, and that this DISTANCE is proportional to the gear radius R. Then you do a few trial "calculations" and figure out that the proportional constant is (2Pi), so that the perimeter DISTANCE={(2Pi)(R)}, and you decide to call this perimeter DISTANCE the gear circumference, C={(2Pi)(R)}!

From your clock idea you realize that in units of sec, one complete gear revolution occurs in a TIME of T={60}/{RPM} sec, and it seems natural enough to call the {DISTANCE}/{TIME} the ft/sec VELOCITY, V, and that gives the tangential VELOCITY at the perimeter of a gear as, V={C}/{T}={(2Pi)(R)}/{(60)/(RPM)}={(2Pi)(R)(RPM)}/{60} ft/sec!

After many weeks of watching the gears go round and round it was clear that since the gear teeth prevent any slippage you need to have V1=V2=V3, and this leads to, {(R1)(RPM1)}={(R2)(RPM2)}={(R3)(RPM3)}, and to RPM1={(R2)/(R1)}{RPM2}={(R3)/(R1)}{RPM3}, which is the same equation that you'd come up with months ago by just counting revolutions versus TIME, but now you've got some solid Physics to help explain to your boss why this is in fact the universal gear law equation!

But you don't want your boss to think this is too simple, so you decide to define a GR=Gear Ratio as G12={R2}/{R1}, etc..., so that RPM1={(G12)(RPM2)}, etc... You figure that by using fancy terms like "Gear Ratio" you can do a snow job on your boss by making this gear stuff sound really complicated, and that might get you an even bigger raise! Since the way you've defined the GR parameter turns out to be such that it REDUCED your beer drinking buddy to tears, you later decide to call it the Gear Reduction Ratio!

121 Posts
Discussion Starter #3 (Edited)
The true story of how HP and TQ were actually discovered...Part 2 of 3

By now you've firmly attached all 3 gears to their respective shafts, and placed the shafts into deep holes in hard-packed clay ground so that the holes function as crude load bearings. When your beer drinking buddy stops by for a visit you get him to grab hold of the shaft for Gear3, and you bet him a beer that he can't stop it from rotating when you try to turn the shaft for Gear1. Well since he outweighs you by at least 100 lb, he says why not up the ante to a case of beer!

Of course you've previously discovered that there's a lot more to this "Gear Ratio" thing than it just being a fancy term to snow the boss with, and that when you twist on shaft1 you have a huge, but mysterious, mechanical advantage over anyone trying to hold onto shaft3 and stop it from turning! While you haven't quite figured out what's causing this magical effect, you figure there's no harm in cashing in on it!

Well after losing 3 cases of beer your buddy is REDUCED to tears and won't bet anymore, and he says he's really "TORQUED" off because you must've cheated him somehow since he knows he's much stronger than you. Now that you've got a name for this mysterious mechanical twisting advantage, you shorten it to TQ, and decide to analyze it further!

So you mix up some mud about the consistency of firm modeling clay, and by "measuring" the thickness of the mud on each set of gear teeth after a wad of it gets mashed between the gears, you soon realize that the FORCES between the gear teeth are equal in magnitude and opposite in direction, so F1=F2 and F2=F3, and that means F1=F2=F3! As luck would have it you forget to write this down, and some guy named Newton winds up getting all the credit for this "equal and opposite FORCE" effect!

But now it gets a little complicated, because when you do this mud squishing experiment on the teeth between Gear1 and Gear2, and compare those results with the teeth between Gear2 and Gear3, you discover that you've got to twist much harder on shaft3 than you do on shaft2 to produce the same amount of mud squishing FORCE F3=F2, and you have to twist much harder on shaft2 than you do on shaft1 to produce the same amount of mud squishing FORCE F2=F1. So the amount of twisting TQ that's required to produce a given mud FORCE between the gear teeth increases as the radius of the gear increases!

You decide to "measure" the gear radii in ft, and you use your wrist strength to estimate the twisting TQ applied to the shafts, and you use several different mud consistencies to estimate the FORCE between the gear teeth, and you discover the relationship, TQ={(FORCE)(RADIUS)}, and by the time you get to Gear3, which has the largest RADIUS, the last batch of mud has gotten quite firm, and you POUND your FOOT on the ground trying to muster up enough twisting TQ action on shaft3 to FORCE the mud through the gear teeth. So it seems only natural that the correct units for TQ should be lb-ft, and of course that makes the standard unit for the mud FORCE on the gear teeth the lb!

Since you've now shown that TQ={(F)(R)}, this means F1={TQ1}/{R1}, F2={TQ2}/{R2}, and F3={TQ3}/{R3}, and since F1=F2=F3, you have the result that, TQ1={(R1)/(R2)}{TQ2}={(R1)/(R3)}{TQ3}, so that TQ3={(2)(TQ2)}={(6)(TQ1)}. Now that you've quantified this mysterious mechanical advantage, you call it "TQ Multiplication", and you realize you've got a TQ multiplier of 6, which gives you a whopping 6 times advantage, so that you can't possibly ever lose out on your beer bet, because no caveman is 6 times stronger than you!

So when your beer drinking buddy stops by for his next visit, and he goes right over and grabs hold of the shaft for Gear3, you're thinking to yourself that all his recent beer drinking must've cleansed his memory bank along with his bladder! So he says this time lets bet 6 cases, but let me start turning shaft3 a little before you try to stop me from turning it by twisting on shaft1. Well since you've still got that whopping 6 times advantage in your favor on shaft1, where's the harm in letting him have a little head start on shaft3?

Well the answer becomes painfully obvious as soon as you try to get a grip on shaft1, which by now is turning at about 60 RPM! After you burn all the skin off your palms trying to get a grip on shaft1, you admit defeat and say this is worse than falling in a big pile of "Horse Pucky", and that gets shortened to HP as the term to describe the heating effect on shaft1!

At this point you're convinced that HP must work just the opposite of TQ. Since applying 1 unit of TQ to shaft1 generates 6 units of TQ on shaft3, it must surely be true that applying 1 unit of HP to shaft3 generates 6 units of heat ENERGY HP on shaft1! It is also clear that whereas TQ Multiplication is a mild-mannered phenomenon for wining beer bets, this HP Multiplication can be downright dangerous.

Later that night, as you're rubbing a couple of sticks together to start a fire, this HP Multiplication concept pops into your head again, and you come up with an even better application for your gears than just using them to make a second hand on a sun dial and win beer bets. It will have to be a two caveman operation, with your buddy turning shaft3 and you starting the fires at shaft1, but perhaps sometimes you can turn shaft3 and just charge other cave people to stop by and start their fires on shaft1, and then they can take their fires on home in time to cook diner?

As you continue rubbing the sticks together you realize that if you press them together with a small FORCE and rub them together very quickly with a large VELOCITY, that you generate the same amount of heat ENERGY, as when you press them together with a much larger FORCE and rub them together with a smaller VELOCITY. After starting a few fires using different combinations of FORCE & VELOCITY it becomes clear that the heat ENERGY HP being generated is proportional to {(F)(V)}!

You realize you can't determine the HP of an actual horse, because they don't exist in prehistoric times, so you're free to choose any "HP is proportional to {(F)(V)}" constant you want for defining HP, and since (1/550) seems to be as good a choice as any, you use that and define, HP={(F)(V)}/{550} ft-lb/sec. You reason that if you've guessed wrong and (1/550) isn't quite the correct number to match up with the HP generated by an actual horse, no one is likely to catch your mistake until long after you're dead anyway!

Looking at the gear picture again it's clear that since F1=F2=F3 and V1=V2=V3, it must also be true that {(F1)(V1)}={(F2)(V2)}={(F2)(V3)}, and that means, HP1=HP2=HP3 and that HP3=HP2=HP1, and it's looking like no matter how you cut it, there's no "HP Multiplication" possible with a set of gears. Since you've already confused HP with TQ once, by thinking they both can be multiplied using gears, you decide it was a good choice to use ft-lb in the HP definition, and to reserve lb-ft for TQ. That way it's unlikely you'll ever get them mixed up again! Now if you could only remember when you first heard that expression about "Horse Pucky", because it definitely seems to be ahead of its time?

As you're patting yourself on the back it dawns on you that since your boss will try and find any excuse he can to deny your much deserved raise, he'll probably say something stupid like, "fire sticks aren't gears so what does your HP is proportional to {(F)(V)} result using sticks really prove?" Therefore you decide to do some more HP heat ENERGY investigations using the gears instead of the sticks.

So now you get two pairs of heavy duty work gloves, and with the help of your buddy you do some more experiments with the gears. You find out that once they're up to speed and spinning like before with RPM1=60 and RPM3=10, you need to grip shaft3 with 6 times more gripping FORCE than is needed on shaft1 in order to exert the same amount braking action on the respective shafts, and that using this factor of 6 different gripping FORCE generates the same amount of HP heat ENERGY production at each shaft!

Since you already named the gripping FORCE the shaft TQ, this result means that in addition to being proportional to {(F)(V)}, HP is also proportional to {(TQ)(RPM)}, and since {(TQ1)(RPM1)}={(TQ3)(RPM3)} it must also be true that HP1=HP3, so that confirms there's no HP Multiplication possible using gears!

121 Posts
Discussion Starter #4 (Edited)
The true story of how HP and TQ were actually discovered...Part 3 of 3

Now that you realize there's an alternate way to define HP, you plug your equations F={TQ}/{R} and V={(2Pi)(R)(RPM)}/{60} into the original HP definition, HP={(F)(V)}/{550}, and this gives, HP={{(TQ)/(R)}{(2Pi)(R)(RPM)}}/{(60)(550)}={(TQ)(RPM)(2Pi)}/{(60)(550)}, so that HP={(TQ)(RPM)}/{5,252.1131...}, but you don't much care for the fact that now the proportional factor of {2Pi}/{(60)(550)} in this form of the HP equation turns out to be a never ending decimal, so you only use this form of the HP equation if absolutely necessary!

It turns out that burning up several sets of expensive work gloves while gripping those gear shafts to determine their TQ and their HP generated heat ENERGY, was money well spent because you eventually get credit for inventing the worlds first Prony Brake Dynamometer similar to the one shown here...

...and note that even this more advanced version still uses a sundial to measure shaft RPM!

During those same experiments you learn that if you grip a shaft with both hands you can double the amount of TQ you apply to it, so that if you only had a third hand, you'd have the caveman equivalent of a triple-disc TQ converter lock-up clutch! You also discover that when you and your buddy both let go of the gear shafts they continue to spin for a considerable length of time, and that if you grab a shaft and try to stop it, the shaft still produces some TQ and HP! This leads to your discovery of ENERGY storage in a flywheel, and its subsequent extraction in the form of TQ and HP, but your boss Mr. Millstone winds up stealing the credit for that!

So now you're all done with your experiments, and you take your picture of the 3 gears with all the equations to brief Mr. Millstone and get your raise. Well as usual, his eyes glaze over and he starts to doze off, so before his head hits the desk you execute plan B, and invite him to dinner so he can see your gear set up close and personal. After dinner you get him to grab hold of the shaft for Gear3, and since he outweighs you by at least 200 lb, he agrees to bet a nice raise plus an extra week of vacation against a 10% cut in your pay, and of course the rest is history!

That is until a month later when Mr. Millstone calls you into his office and shows you this picture...

...and explains that it's a "top secret" gear set that our competitor is working on, and it's rumored to do everything that our gear set does, only better! He wants you to drop everything and get cracking on understanding how this contraption works, because if we can't beat the competition to market we'll likely go bankrupt, and that'll mean losing your raise along with your pension plan!

So you immediately build a prototype based on the picture, and you name it the "planetary gear configuration", because you call the small yellow Gear1 that's now at the center the "sun" gear, and you call the red midsize Gear2 that's now replaced by 3 identical midsize red gears the "planets", and all 3 of these planet gears simultaneously engage both the sun gear and the inside circumference of the large blue Gear3, which you now call the "ring" gear. You attach all three planet gears to a common plate using individual pivot shafts, and you call this the planet carrier.

Using an arrangement of coaxial tubes for the gear shafts allows you to have independent connections to the sun gear, the planet carrier plate, and the outer ring gear, so any combination of these gears can be used for an input and an output, just like before! However, unlike the previous 3 gears in series where the unused gear is allowed to freewheel, in the planetary configuration you lock in place the shaft that's not being employed as an input or an output so that its respective gear can't turn!

In order to duplicate the 6:1 gear reduction ratio that won the beer bet using the new planetary configuration, you lock the shaft to the planet carrier plate, the individual planet gears are still free to rotate about their individual pivot shafts which attach them to the common carrier plate, but the common carrier plate itself is held stationary and can't rotate. The small sun gear is again used as the input, and the 6:1 gear reduction ratio is obtained as an output from the tubular shaft that's connected to the ring gear. You immediately see a big disadvantage to this planetary gear configuration, because the mark you're winning the bet from is standing right next to you, and he can easily punch your lights out if he doesn't have a sense of humor!

Well it appears relatively easy to explain to Mr. Millstone how this planetary configuration gets the same 6:1 gear reduction ratio as before. If he can stay awake long enough to examine the planetary gear picture, he'll see that since the circumference of the ring gear is twice that of the planet gears, each planet gear must make two revolutions about their individual pivot shafts which attach them to the common carrier plate, which is locked in place, for the ring gear, and the shaft that's connected to it, to make one complete revolution about the sun gear. Then since the sun gear has to complete 3 revolutions for each revolution of a planet gear, you get a net 6:1 gear reduction ratio between the sun gear and the ring gear. Well maybe this won't be so easy to explain after all!

Then you discover that you can get a 7:1 gear reduction ratio with the planetary configuration, which might come in handy for winning beer bets when you're older and not quite as strong. By studying that planetary gear picture some more you see what happens when you lock the ring gear in place, again use the small sun gear as the input, and now take the output from the tubular shaft that's connected to the planet carrier plate.

Now if the sun gear rotates in say a clockwise direction, the 3 planet gears each rotate in a counterclockwise direction about their respective pivot shafts on the carrier plate, and in so doing they force the common carrier plate to rotate in a clockwise direction around the sun gear, which itself is also rotating in a clockwise direction. Since the planet carrier and planet gears orbit the sun gear in the same direction that the sun gear is rotating, the sun gear essentially falls behind by 1 revolution for each revolution of the planet carrier, so that the sun gear has to now complete 7 revolutions for each revolution of the planet carrier plate!

Well there's no way to explain this to Mr. Millstone, so instead you decide to come up with the universal gear law equation for the planetary configuration. After much effort and a little insight concerning relative angular rates you've got the following, {(Rs)(RPMs-RPMc)}+{(Rr)(RPMr-RPMc)}=0, and rearranging terms gives, {(Rs)(RPMs)}+{(Rr)(RPMr)}={(Rs)(RPMc)}+{(Rr)(RPMc)}, and this can be written as, {(Rs)(RPMs)}+{(Rr)(RPMr)}={Rr+Rs}{RPMc}!

So if you lock the carrier RPMc=0, and that means, RPMs={(Rr)/(Rs)}{RPMr}, and you get the correct 6:1 gear reduction ratio. If you lock the ring gear RPMr=0, and that means, RPMs={Rr+Rs}/Rs}{RPMc}, and you get the correct 7:1 gear reduction ratio. If you lock the sun gear RPMs=0, and that means, RPMr={Rr+Rs}/{Rr}{RPMc}, and you get a 7:6 gear reduction ratio!

So you take your planetary gear equation to Mr. Millstone, of course he winds up stealing all the glory, and he even gets hired to run FORD! Before this becomes public knowledge you take your entire retirement nest egg and sell FORD short, and you expect to make a killing, but you get killed instead! Mr. Millstone takes his private jet to Washington DC, and the DC part must be included because otherwise the cave people in the great NW will assume you're referring to the state and not to the nations capitol, actually, that's close to being true once you get west of the Mississippi!

Mr. Millstone briefs Congress on FORD's plan to build a new pollution free pickup truck using an ENERGY storage flywheel. He assures Congress that FORD's new design eliminates smelly diesel fumes forever, and allows the new truck to go 10,000 miles on a single spinning up of its proprietary flywheel because that flywheel runs at 10,000,000,000 RPM, which it turns out is the exact $ amount of taxpayer's money that Congress winds up giving to FORD!

When asked about the details of this new flywheel design at the follow-up press conference, Mr. Millstone replies..."well they haven't all been finalized yet, but not to worry because I know a bright caveman engineer back at my old company who designs anything I tell him to!". Before the ever vigilant "national media" can figure out that Mr. Millstone is nothing but a scam artist, your shorts on FORD stock wipe out all your retirement money, so that from now on you've got no $ to do anything else except write these posts to diesel truck forums!

121 Posts
Discussion Starter #5 (Edited)

The above quantities are key for analyzing the performance of diesel engines and the trucks they power, and hopefully I'll remember to use CAPS for them so they'll be easy to see in my text and equations, and to emphasize their fundamental importance. The reason that VELOCITY isn't all CAPS is this posting software won't let me capitalize everything in the title?

The "caveman story" touched on how the equal and opposite FORCE between the gear teeth multiplied by the DISTANCE from the gear teeth to the center of the gear shaft determines the TQ that exists on the gear shaft, but there's a lot more than this to the complete story of TQ! Likewise there's much more to the complete story of HP than HP just being proportional to FORCE times VELOCITY or HP being proportional to TQ times RPM! The best way to tell these stories is one chapter at a time, and here I didn't capitalize time for obvious reasons!

First a word about units. The International System of Units is abbreviated SI from the French, and it's the "modern" form of the metric system. The SI units use Newton-meters (Nm) for TQ, Joules (J) for ENERGY and WORK, and Watts (W) for HP, so there's little chance of getting these quantities confused if you use SI units!

It's now more than 200 years since the metric system began to spread throughout the world, and the United States shares the dubious distinction with Burma and Liberia as being one of only three nations that haven't converted to the "modern" metric system. The first attempt to convert us came from Thomas Jefferson who considered the "new" metric system an admirable advance over the "old" English system of feet, pounds, and pints! The British are also stubborn, but Great Britain began a transition to the metric system as a condition of entering the European Common Market in 1965, and Canada successfully went metric in the 1970s.

I avoid SI units where possible, and prefer to use English units because almost everyone has a pretty good "feel" for what a 1 lb FORCE is and for what a 1 ft DISTANCE is, so if you pull on a wrench handle with a 1 lb FORCE and your hand is a 1 ft DISTANCE from the center of a bolt you're applying a 1 lb-ft TQ to the bolt! I wonder how many stripped bolts there'd be if we tightened by feel with a 1 ft wrench, but all the TQ specs were given in Nm?

In English units ENERGY and WORK are given in ft-lb, and 1 HP is a 550 ft-lb/sec TIME rate of change in ENERGY or WORK. Do you have any "feel" for what it's like to get hit with say 1.3564 J of ENERGY or perhaps 0.001285 Btu of ENERGY? Well lift a 1 lb weight a vertical DISTANCE of 1 ft and then drop it on your foot and you've just been hit with 1 ft-lb of kinetic ENERGY which is also equal to 1.3564 J or 0.001285 Btu of ENERGY!

I'm getting way ahead of the story here, but compare the above 1 ft-lb of "foot throbbing" kinetic ENERGY with the fact that my C7 injects 4,060 ft-lb of "diesel fuel" chemical ENERGY into a single combustion chamber for a single piston's power stroke to generate the engine's overall average maximum 300 flywheel HP at 2,200 RPM. That's over 4,000 times more ENERGY to push a single piston down than the 1 ft-lb of ENERGY that hit your foot! Each 1 gallon of diesel fuel contains 101,140,000 ft-lb of chemical ENERGY, and that's more than 100 million times what hit your foot! If you just dropped 1 gallon of diesel fuel on your foot from a 1 ft vertical DISTANCE that would only be about 7 ft-lb of kinetic ENERGY compared to the over 100 million ft-lb of heat ENERGY the fuel releases when combusted in the engine!

A single piston produces a peak HP of 1,025 at 27* ATDC, and about 64.7% of the 4,060 ft-lb of chemical ENERGY in a single injection gets converted into a single piston's average power stroke HP. When the piston HP for a single power stroke is averaged over the entire 180* rotation of the crankshaft you get 350 HP for each piston. Since there's only one power stroke for every 720* or 2 rotations of the crankshaft, the 350 HP needs to be divided by 4 to get 87.5 HP average for each piston, and multiplying by 6 pistons gives a total of 525 HP for all 6 pistons.

So why do you only get 300 flywheel HP if there's a total of 525 HP being produced by all 6 pistons? Well for a 28 psi boost pressure at BDC there's a total of 195 HP required during the compression strokes to compress the charge air in the cylinders! High boost pressure makes more flywheel HP, but it also results in a lot of extra pumping loss HP which drives up the required average and peak power stroke HP for the pistons, which is why highly tuned engines tend to break rods and pistons!

So 195 HP from 525 HP would still leave 330 HP at the flywheel, except that friction dissipates about 30 HP as heat ENERGY! The 35.3% of the fuel's chemical ENERGY that doesn't get converted into mechanical piston HP during the combustion process gets rejected as heat ENERGY, and some of this heat ENERGY is absorbed by the coolant and the rest goes out the exhaust manifold to help power the turbo.

Well since I've gotten so thoroughly sidetracked again I might as well show the graph of the internal cylinder parameters here...

...and the ENERGY & HP budget here...

...and I'll finish discussing those in detail at a much later time after I've covered some more of the basics.

Well back to units. In English units TQ is historically given in lb-ft to distinguish it from ENERGY and WORK which are given in ft-lb and from HP which depends on ft-lb/sec, but now days you won't find this convention universally followed! Even though when working with equations in English units reversing the order of multiplication, such as lb x ft versus ft x lb, gives the same numerical result, I think keeping the lb and the ft in the correct order helps to correctly apply TQ, WORK, ENERGY, and HP by reinforcing the notion that they're very different physical concepts!

So here's a fool-proof way, or perhaps a fools way, to remember if the lb or the ft comes first. When you use a TQ wrench to tighten a bolt, you must first pull on the wrench with a POUND FORCE, and then when the wrench slips and you bloody your knuckles you POUND your FEET on the ground in anger! So the unit for TQ is lb-ft.

When your boss gives you a WORK assignment and you fall behind schedule, he first accuses you of FOOT dragging, then he puts his FOOT to your backside to quicken your pace, and when you finally turn the WORK in late, he then POUNDS your head with his fist! So the unit for WORK is ft-lb. If you happen to be a British soccer player you actually do your WORK with your FOOT and you get paid in POUNDS!

My next two posts won't be mathematically precise because that would require special symbols not on my keyboard to designate vectors, and to denote the "dot" product of two vectors and the "cross" product or the "curl" of two vectors, and no one would want to read such a post anyway. However I'll hopefully explain in clear terms the definitions of and the distinctions between FORCE, DISTANCE, WORK, TIME, HP, ENERGY, VELOCITY, & TQ!

Senior Member
2,126 Posts
Very good article!

121 Posts
Discussion Starter #7
Vectors, Scalars, and other "technical" stuff you need to know...

Yes it's true with the way the economy is going lately and all the forthcoming proposals to fix it from those in charge you really do need to know which quantities of interest are shaped like a stick with a point on one end, and which ones aren't, or you're likely to get shafted for sure! Vector quantities are like hunting arrows and they have a sharp point on one end which indicates their pointing direction, but even the feathered end can be somewhat unpleasant if it's mishandled! On the other hand Scalar quantities are directionless and kind of like a pile of mash potatoes with the gravy running over your plate in all directions at once!

FORCE is a vector quantity represented by an "arrow" which means FORCE has both a magnitude given in lb that corresponds to the arrow's length as well as a pointing direction which gives the orientation in which the FORCE is applied. In many cases people including sometimes me only talk about the magnitude of a FORCE in lb, and the reader has to figure out the application direction of the FORCE, which is the arrow's pointing direction, from the context of the problem.

For example if I grab a full 12 oz can of beer at its center of gravity, the FORCE vector required to slowly lift the can has a magnitude of 0.75 lb, and its application direction is along a line from my hand to the center of the earth and pointing in the upward direction. If I now wish to move the can rapidly to my mouth I have to apply a small additional FORCE vector to both accelerate the can and overcome its aerodynamic drag, and this FORCE vector is orientated along the DISTANCE vector from the can to my mouth.

As in the above example DISTANCE can in general also be treated as a vector quantity which means it has both a magnitude, which is given in ft, as well as a pointing direction which gives the orientation of the DISTANCE vector from its beginning to end points, which are separated by the given number of ft. Technically speaking DISTANCE is the magnitude of the POSITION vector R just like SPEED is the magnitude of the VELOCITY vector V, but it's common practice to use all these terms interchangeably as either vector or scalar quantities!

For example the DISTANCE vector from my easy chair to the nearest replacement beer has a magnitude of 9 ft, and as measured by my satellite finder compass the pointing direction is along a heading of 45 degrees. So just like following directions on a pirate map if I follow a DISTANCE vector that's 3 paces long in a NE direction I find the treasure!

Now lets put the FORCE and DISTANCE vectors to WORK! Imagine the picture below is a grocery cart full of beer on it's way to the checkout counter. As indicated in the picture if I apply the red lb FORCE vector in a direction that's parallel to the blue ft DISTANCE vector the amount of WORK that I do is given by the product of the magnitudes of the two vectors, so that WORK={FORCE}{DISTANCE} ft-lb.

In general WORK is given by the "dot" product of the FORCE and DISTANCE vectors, and this "dot" product is equal to the magnitude of the FORCE vector x the magnitude of the DISTANCE vector x the COSINE of the angle between the pointing directions of the FORCE vector and the DISTANCE vector. In the above picture the angle between the FORCE and DISTANCE vectors is 0 degrees so that the COSINE is 1.

When I lean on the handle of the cart to support my bad back I apply an additional FORCE vector that's pointed straight down toward the floor, but since this FORCE vector results in no downward motion, and it's perpendicular to the forward motion indicated by the DISTANCE vector, the angle between this FORCE vector and the DISTANCE vector is 90 degrees so that the COSINE is 0. Therefore the "dot" product is also 0, and there's no WORK done by this additional FORCE vector that's supporting my back!

However this back support FORCE vector increases the loading on the cart, and this increases the rolling resistance of its wheels, and this means that I have to push on the cart handle with a little larger FORCE vector in the forward direction, and this increases the amount of WORK I do by a little bit, so when it comes to supporting my bad back there's no free lunch, but my back doesn't hurt nearly as much if I lean some of my weight on the cart and push it with a little extra FORCE!

In the general case the angle between the FORCE and DISTANCE vectors is anywhere between 0 and 90 degrees, and the "dot" product forms what's called a scalar quantity, and this means that WORK only has a magnitude, which is given in ft-lb, and there's no direction whatsoever associated with WORK.

I think most will agree with the notion that when the boss heaps a large magnitude of WORK on you that this WORK has a definite magnitude that can be measured by the sweat from your brow, but the actual WORK is really without any meaningful direction! At least that's the way it is in the world of Physics in which WORK is just a number in ft-lb which measures the ft DISTANCE a given lb FORCE moved, where the FORCE is the component that's in a parallel direction to the DISTANCE moved, or conversely the DISTANCE moved is the component that's in a parallel direction to the FORCE.

Also note that WORK is a timeless quantity so that I do the exact same amount of WORK in the case where I'm thirsty, and I push that grocery cart full of beer to the checkout counter in a TIME of 1 minute, or in the case where I'm tired, and I stop and rest along the way and I take a TIME of 1 hour! The amount of WORK done only depends on the lb FORCE moving a given ft DISTANCE, and not on how quickly the FORCE moves over the DISTANCE!

That's why the $ amount of your pay check doesn't depend on just the amount of WORK you do, but rather on how fast you do a given amount of WORK, and that's your HP output! If you can do 550 ft-lb of WORK in a TIME of 1 second you're accomplishing a 1 HP TIME rate of WORK production, and for a human that's world class productivity, unless of course you're doing the WORK by just pushing a button on some machine. In equation form this is, HP={WORK}/{(TIME)(550)} ft-lb/sec.

In my world TIME seems to have a most definite pointing direction, pointing from younger, to older, to ancient, and then to deceased, and this suggests to me that TIME should be a vector quantity, but that would imply that HP must also be a vector that's pointing in the same direction as the TIME vector. However even though the world of Physics talks about TIME as being the fourth dimension in the DISTANCE-TIME continuum, and DISTANCE is definitely a vector, it turns out that TIME isn't really a vector after all, and that means HP is also just a scalar quantity like WORK with only a magnitude and no particular direction at all.

Now it requires a given amount ENERGY input to produce a given amount of WORK output, and the two quantities are related to each other by a scalar quantity which defines the efficiency of a particular machine like for example a diesel engine for converting ENERGY into WORK. So the amount of ft-lb of WORK done by an engine is determined by multiplying its efficiency, which is a number from 0 to less than 1, by the ft-lb of input ENERGY. Anyone whose watched a 2-year old child at play knows there's no particular direction associated with that ENERGY, so ENERGY can't possibly be a vector! In equation form this is, WORK={Eff}{ENERGY} ft-lb, where Eff is a dimensionless ratio.

Since HP={WORK}/{(TIME)(550)} ft-lb/sec and WORK={Eff}{ENERGY} ft-lb, you have, HP={(Eff)(ENERGY)}/{(TIME)(550)} ft-lb/sec. So if you take the TIME rate of change of input ENERGY you get the input HP, and the TIME rate of change of output WORK gives the output HP, and these input and output HP numbers are related by the same efficiency number that relates the input ENERGY to the output WORK, but don't forget to divide those TIME rates of change by 550 to convert them to units of HP.

Now look at that grocery cart picture again. As I push the cart along at a constant VELOCITY the red FORCE vector that's "attached" to the handle is unchanged, and by that I mean its pointing direction and its length or magnitude remains constant. However that blue DISTANCE vector increases in length as I push the cart, and if I push at a given VELOCITY, the VELOCITY vector is just the sec TIME rate of change of the ft DISTANCE vector, so that VELOCITY={DISTANCE}/{TIME} ft/sec.

At this juncture I can save a whole lot of words by just writing some more equations...

HP={WORK}/{(TIME)(550)}={(FORCE)(DISTANCE)}/{(TIME)(550)}={(FORCE)(VELOCITY)}/{550} ft-lb/sec

...and solving the last equation for VELOCITY gives...

VELOCITY={(HP)(550)}/{PUSH FORCE} ft/sec

As indicated in the above equation from here on I'm calling that red FORCE vector that's "attached" to the handle my PUSH FORCE, and if my PUSH FORCE remains constant my ability to push the cart at a higher VELOCITY only depends on my ability to generate a higher HP output!

Well now that I've paid for that grocery cart full of beer I've got to push it up a hill which has a 10% grade along the entire length of the parking lot to get it back to the Freightliner, because I always park at the far end of the lot to keep my truck out of harms way as much as possible. Now my PUSH FORCE must be made large enough to overcome the GRAVITATIONAL FORCE due to the component of gravity that's trying to make the cart coast back down the hill. Lets see 4 cases of beer plus the empty weight of the cart gives a total weight of about 100 lb, so it requires a 10-lb additional PUSH FORCE for the 10% grade. If I pushed on the cart handle with a weight scale, and just held enough PUSH FORCE on the scale so that the cart remained motionless on the 10% grade hill, the scale would read 10 lb.

Since hill climbing performance is basically concerned with overcoming the effects of gravity anyway, lets for now just simplify this analysis by neglecting the additional PUSH FORCE requirements to overcome aerodynamic drag and rolling resistance, and call the total PUSH FORCE=10 lb. So now I'm pushing the cart up the hill and I'm still using that bathroom scale between my hands and the cart handle to measure my PUSH FORCE. Well as I increase my VELOCITY from 0 to 1 ft/sec=0.68 MPH, and then maintain this constant VELOCITY and look at the scale, what does it read? I then increase the VELOCITY to 2 ft/sec=1.36 MPH, what does the scale read now? How about when I'm maxed out at a VELOCITY of 4 ft/sec=2.72 MPH?

The scale always reads a PUSH FORCE=10 lb no matter how high my VELOCITY is as long as I'm holding the VELOCITY constant when I take the reading and neglect aerodynamic drag and rolling resistance. That means my hill climbing VELOCITY only depends on my ability to generate higher HP outputs as given by HP={(PUSH FORCE)(VELOCITY)}/{550}={(10)(VELOCITY)}/{550}={VELOCITY}/{55} HP. So a VELOCITY=1 ft/sec requires a HP=0.0182, a VELOCITY=2 ft/sec requires a HP=0.0364, and a VELOCITY=4 ft/sec requires a HP=0.0727.

Of course the above HP numbers don't include the additional HP required to haul my own weight up the hill, but the point is the VELOCITY, the SPEED, how FAST I push the cart up the hill, or however you want to state it, only depends on my HP output.

Another way to see this is to use the ENERGY form of the HP equation, HP={ENERGY}/{(TIME)(550)} ft-lb/sec, where now instead of being an "input ENERGY" this ENERGY is the additional "potential ENERGY" that must be added to the cart to make it gain the additional vertical DISTANCE needed to climb the hill, where potential ENERGY={WEIGHT}{vertical DISTANCE} ft-lb.

Assume it's a 1,000 ft horizontal DISTANCE back to the truck so for a 10% grade that's a 100 ft increase in the vertical DISTANCE of the cart. This means the required increase in the potential ENERGY of the 100-lb cart is, potential ENERGY={WEIGHT}{vertical DISTANCE}={100}{100}=10,000 ft-lb! If I climb the hill at a horizontal VELOCITY=1 ft/sec it takes a TIME={horizontal DISTANCE}/{horizontal VELOCITY}={1,000}/{1}= 1,000 sec, and this requires a HP={ENERGY}/{(TIME)(550)}={10,000}/{(1,000)(550)}=0.0182 HP, which is the same answer as before. Likewise if I go at twice the VELOCITY I get there in half the TIME, but I have to double my HP output to do it, and a x4 VELOCITY requires a x4 increase in HP!

So when it comes to hill climbing VELOCITY and HP, why is a human engine pushing a cart up a hill any different than a diesel engine pushing a motorhome up a hill? Since it's the human engine's HP output that determines the hill climbing VELOCITY of the cart why isn't it the diesel engine's HP output that determines the hill climbing VELOCITY of the motorhome?

Well of course there're the same problems, and they've got the same solutions, which makes me wonder why motorhome salesman give the engine TQ numbers first and barely mention the HP numbers in their sales pitch for diesel motorhomes? On the other hand when trying to sell a gasser powered motorhome they give the HP numbers first and barely mention the TQ?

Well I'll answer that question with another question. How could a salesman who's main interest is a large commission convince a customer to shell out $250K or more for a motorhome with a puny 300 HP diesel like my CAT C7, when that customer could buy a much less expensive V10 gasser with 362 HP which is 20.7% more HP than in that overly priced diesel pusher?

Well of course the salesman will say that his diesel motorhome has a whopping 860 lb-ft of TQ, while that puny V10 gasser can only generate a miserly 457 lb-ft of TQ, which means the diesel's TQ is 88.2% higher than that of the gasser, and as EVERYBODY knows, it's the TQ that gets your motorhome up that hill quickly! I'll just constrain myself by saying that since I'm currently in Vegas, the typical Penn & Teller comments apply here, and I wish they'd do a show on motorhome salesmen.

In a subsequent post I'll discuss many "legitimate" reasons for wanting to own a diesel powered vehicle compared to a gasser, but having more TQ to climb hills faster isn't one of them. For some reason this "myth" that TQ is what gets you up hills fast has grown well past the status of "urban legend" and I sure hope that anyone who's read this far and still believes in this myth either posts here that they'd like some further explanation of why it's only HP that counts for going up hills fast or just send me a PM and I'll do some more posts on this topic.

Remember the "caveman" proved that you can't multiply HP with gears, but you can multiply TQ using gears, so you can get as much RWTQ as you want by simply shifting to a lower gear or changing your diff ratio! On the other hand your RWHP isn't multiplied by gearing so you need to get the HP from your engine, which is why engine HP is the only spec that counts! Of course what you really need is the engine's HP vs RPM curve because that tells you everything there is to know!

In closing I wonder how the "TQ advocates" would explain this rocket powered scooter getting up a hill with no TQ whatsoever or how the "space shuttle" manages to climb almost straight up and slip the surly bounds of earth without any TQ? During a recent shuttle launch I remember the mission directors call of 37,000,000 HP at 4,000 MPH during its really big hill climb, and that works out to be a 3,468,750 lb PUSH FORCE!


121 Posts
Discussion Starter #8
TQ is a Vector Quantity and HP, WORK, & ENERGY aren't!

In this post I'll define the TQ vector and discuss how TQ shares some similarities with WORK and ENERGY, but why TQ is actually very different from WORK and ENERGY. By the time I'm done with the next few posts I hope to dispel at least some of the commonly held myths that many diesel people have about TQ, such as it's their TQ that produces the HP in their diesel engine, their TQ is measured on a dyno, their TQ is what gets them up a hill quickly, their TQ wins them a drag race, and their TQ makes their diesel engine better than a gasser!

As indicated in the picture below...

... TQ is similar to WORK in that TQ is also defined by the product of a FORCE vector and a DISTANCE vector, however in the case of TQ it's the "cross" product or the "curl" of a FORCE vector and a DISTANCE vector, and unlike the "dot" product of a FORCE vector and a DISTANCE vector which defines a directionless scalar quantity called WORK, the "cross" product of a FORCE vector and a DISTANCE vector defines a vector quantity so that TQ is a vector and TQ has both a direction as well as a magnitude!

In general, the "cross" product gives a TQ vector whose magnitude is equal to the magnitude of the FORCE vector x the magnitude of the DISTANCE vector x the SINE of the angle between the pointing directions of the FORCE vector and the DISTANCE vector, and the pointing direction of the resulting TQ vector is given by the right-hand rule which is illustrated in the picture below...

... where the red R vector corresponds to the wrench handle in the previous picture, and as indicated the direction of the light blue TQ vector is along the direction of the shaft to which the TQ is being applied by the darker blue FORCE vector, and this right-hand rule direction of the TQ vector can be determined by touching the shaft with your right wrist while using an orientation that allows you to stretch your fingers out in the direction of the R vector, and to "curl" your fingers in the direction of the FORCE vector. In this orientation your thumb will point in the direction of the TQ vector. As shown in the picture if the applied FORCE vector is trying to rotate the shaft in a counterclockwise direction the TQ vector points straight out of the shaft.

Since the FORCE vector is perpendicular to the R vector or DISTANCE vector the angle between them is 90 degrees and the sine is equal to 1. This is just the opposite case as the geometry required to produce WORK where a 0 degree angle is required. A 90 degree angle between a FORCE vector and a DISTANCE vector produces no WORK, but it does generate a TQ vector, and a 0 degree angle between a FORCE vector and a DISTANCE vector produces WORK, but doesn't generate a TQ vector!

A most important distinction between WORK and TQ is that to produce WORK a FORCE vector must "move" through a DISTANCE that's parallel to the FORCE vector, but a TQ vector is generated by a static geometry in which neither the FORCE vector or the DISTANCE vector moves! So WORK is a dynamic motion geometry defined scalar quantity, and TQ is a static stationary geometry defined vector quantity, and in that regard they're two very different things.

Now for some similarities between TQ and WORK. The magnitude of the TQ vector is measured in lb-ft, which is technically the same units as the ft-lb measure for the scalar quantities WORK and ENERGY! Another similarity between TQ and WORK is that they're both timeless quantities that only depend on FORCE and DISTANCE. As previously mentioned the FORCE does have to "move" to produce WORK, but it's not a TIME dependent motion, the only thing that counts for producing a given amount of WORK is how far the FORCE moves, not how fast it moves. Producing a TQ vector involves "no motion" whatsoever, so time can't possibly be involved with the generation of TQ!

Well you certainly can't use timeless quantities like TQ, WORK, or ENERGY, to determine how "fast" something happens, like for example how "fast" a truck goes up a hill or how "fast" it accelerates in a straight line, but you can use ENERGY and WORK for example to determine "if" a truck can potentially make it up a given hill.

To do that, you need to calculate the total ft-lb of ENERGY content of the diesel fuel that's in the truck's tank, then multiply that number by the approximately 30% engine-driveline efficiency for converting fuel ENERGY into rear wheel WORK, and then compare that ft-lb of available WORK with the ft-lb increase in potential ENERGY that's required to get to the top of the hill. If the ft-lb of available WORK is larger than the required ft-lb increase in potential ENERGY then you can make it to the top of the hill, but how long will it take to get there? That answer depends on the MPH speed at which your truck can climb a given % grade, and that's solely determined by the truck's RWHP!

Now, look at the picture below which illustrates the difference between TQ and WORK...

... well the WORK definition on the right-hand side shows a FORCE vector being used to move an object over a straight line DISTANCE that's parallel to the FORCE vector, and sure enough that's the correct definition of WORK!

On the left-hand side the FORCE vector is shown as being perpendicular to the DISTANCE vector, and that's correct for defining the TQ vector, and although the TQ vector isn't shown in this picture if you employ the right-hand rule you'll see that the TQ vector is pointing directly into the object at its center, and with this orientation the TQ vector is trying to rotate the object in a clockwise direction about its center. But wait just a minute here, the picture also seems to indicate that the object is actually "rotating" in a clockwise direction, and "rotation" isn't part of the definition of a TQ vector, because "rotation" involves "motion", and "motion" is part of the definition for WORK in ft-lb, not for TQ in lb-ft! Later I'll be emphasizing this point with some good old fashion "horse sense"!

Now look at that same picture this way. Assume the square object is actually round, and that it corresponds to a truck's driver-side rear wheel and tire. And that a TQ vector is now pointing out of the wheel at its center, and that this TQ vector is being applied to the rear wheel by the truck's rear axel half-shaft, so from now on I'll call that TQ vector the RWTQ vector. The blue DISTANCE vector is the DISTANCE from the axel centerline to the ground, and from now on I'll call that DISTANCE Dg, and the red FORCE vector is the reaction TIRE FORCE that's applied between the tire's contact patch with the ground. The RWTQ vector is trying to make the wheel rotate in a counterclockwise direction, and this reaction TIRE FORCE is resisting this counterclockwise rotation of the tire. Using the above definitions, RWTQ={TIRE FORCE}{Dg}.

If you take your driver-side rear foot, and assuming you're not the British soccer player in the previous post, that would be your left foot, and place it on the ground and apply some weight on that foot using your left leg, and then use your left leg to slide that foot to the rear, you'll feel the SHOE FORCE equivalent of the TIRE FORCE that I'm talking about. That sliding motion is what happens when a truck burns some rubber getting off the line! If you now apply some additional weight to that foot, and do the same movement with your leg that foot remains in a stationary location on the ground, and your left leg and the body that's attached to it moves forward!

Here's another way to view this TIRE FORCE, and by TIRE FORCE I mean the total from all of the powered wheels. Since the rear axel is attached to the truck's frame by suspension members, the TIRE FORCE generates a reaction FORCE in the truck's frame which is equivalent to pushing the truck at its rear bumper with an external PUSH FORCE vector that has the same magnitude and direction as the total combined TIRE FORCE vector. Instead of worrying about what's happening at each tire contact patch with the ground you can just imagine that the truck is being propelled forward by a net PUSH FORCE vector applied to its rear bumper, just like the one applied to the handle of the grocery cart or the one coming out of the end of the "rocket scooter" shown in the previous post, or the one coming out of the end of the jet powered F350 shown next door on my new drag racing thread! My next post here will continue with "More stuff about TQ..."!

121 Posts
Discussion Starter #9
More stuff about TQ...

Now I'm going to derive 3 basic commonly used equations involving RWTQ, and then explain why these equations have issues regarding their units and/or what they seem to imply about TQ! If you substitute the equivalent PUSH FORCE in place of the TIRE FORCE in the equation, RWTQ={TIRE FORCE}{Dg}, you get RWTQ={PUSH FORCE}{Dg}, and solving for the PUSH FORCE you get the equation...

#1) PUSH FORCE={RWTQ}/{Dg} lb

Each wheel revolution corresponds to advancing a DISTANCE along the road that's equal to the rolling circumference of the tire which is given by, DISTANCE={(2)(Pi)(Dg)}. The amount of WORK done as a truck advances up a hill by one revolution of its rear wheel is given by, WORK={(PUSH FORCE)(DISTANCE)}={(RWTQ)/(Dg)}{(2)(Pi)(Dg)}, and since "Dg cancels" we get we get the equation...

#2) WORK={(2)(Pi)(RWTQ)} ft-lb for each wheel revolution.

If we let RPMw be the wheel RPM then the TIME in sec for each wheel revolution is, TIME={60}/{RPMw} sec, and if we then use the #2) equation for rear wheel WORK={(2)(Pi)(RWTQ)} ft-lb in the basic HP equation, HP={WORK}/{(TIME)(550)} ft-lb/sec, we get RWHP={WORK per wheel revolution}/{(TIME per wheel revolution)(550)}={(2)(Pi)(RWTQ)}/{[(60)(550)]/(RPMw)}={(RWTQ)(RPMw)}/{5,252}, so that we get the equation...

#3) RWHP={(RWTQ)(RPMw)}/{5,252} ft-lb/sec

So just what's so wrong with these 3 equations? Well equation #1) PUSH FORCE={RWTQ}/{Dg} lb doesn't have a problem with units because {lb-ft}/{ft}=lb FORCE, but the equation does seem to imply that it's the RWTQ that produces the PUSH FORCE, and while this is "somewhat" true if RPMw=0, as soon as the wheel begins to rotate it's the RWHP that's determines how fast the PUSH FORCE moves to propel the truck at a given MPH speed.

How fast a PUSH FORCE moves is given by the relation RWHP={(PUSH FORCE)(MPH)}/{375}, and in fact even when MPH=0 and there's no RWHP because the wheel's not rotating it's the reciprocating piston HP combined with either a slipping manual clutch or a churning auto torque converter that produces the "static" RWTQ which in turn produces the "static" PUSH FORCE just before launch!

Now equation #2) WORK={(2)(Pi)(RWTQ)} ft-lb has a problem with units because using RWTQ in lb-ft in this equation gives WORK in lb-ft, whereas the correct units for WORK are ft-lb! Also, this equation seems to imply that it's the RWTQ that produces the rear wheel WORK, but this is just an "illusion" that results from the "Dg" canceling out in the derivation of this equation. As discussed in the previous post..."rotation" isn't part of the definition of a TQ vector, because "rotation" involves "motion", and "motion" is part of the definition for WORK in ft-lb, not for TQ in lb-ft!

Since HP is defined as the TIME rate of change of WORK, if the units for WORK are incorrect, the units for HP will also be wrong! Consequently, if you use RWTQ in lb-ft in equation #3) RWHP={(RWTQ)(RPMw)}/{5,252} ft-lb/sec you get RWHP in lb-ft/sec instead of the correct ft-lb/sec! Also, this equation seems to imply that it's the RWTQ that produces the RWHP, but this isn't true for the same reasons previously cited!

Now if you just use the "numerical" value of the RWTQ in equations #2) and #3), but change the units from lb-ft to ft-lb everything seems to "come out in the wash" as they say, however all 3 equations still reinforce the misconceptions that #1) it's the RWTQ that produces the PUSH FORCE, #2) it's the RWTQ that produces the rear wheel WORK, and #3) it's the RWTQ that produces the RWHP, and the reasons why I call these "misconceptions" have been previously stated.

Now for the good old fashion "horse sense" that I mentioned in the previous post! If you study this diagram carefully you'll hopefully see why there's issues with the units, and what's actually producing the PUSH FORCE, the rear wheel WORK, and the RWHP, and I'll give you a hint, it's not the RWTQ!

As shown below you can use the RWTQ to help calculate the PUSH FORCE, the rear wheel WORK, and the RWHP, or you can use the alternative and more fundamental ways shown that don't use the RWTQ for specifying the rear wheel PUSH FORCE, rear wheel WORK, and RWHP. Unless you're worried about breaking that driveshaft you don't even need to consider RWTQ at all in order to do any performance assessment of interest.

The WORK done by that poor horse as he completes each revolution of his endless boring journey is equal to his PULL FORCE F1 in lb times his DISTANCE traveled in ft, which is the circumference of his loop C={(2)(Pi)(D1)} ft, and this gives W1={FORCE}{DISTANCE}={F1}{{(2)(Pi)(D1)} in ft-lb. Now the problem arises when we substitute RWTQ={F1}{D1} in lb-ft into this equation in place of the product {F1}{D1} in ft-lb.

When you do this substitution, you're replacing a dynamic motion geometry defined scalar quantity, {F1}{D1} product in ft-lb, in which the lb FORCE F1 moves over a ft DISTANCE of C={(2)(Pi)(D1)} ft that's parallel to the lb FORCE F1, with a static stationary geometry defined vector quantity, {F1}{D1} product in lb-ft, in which the lb FORCE F1 does not move relative to the ft DISTANCE D1, because in the definition for TQ the lb FORCE F1 maintains a fixed perpendicular relationship to the ft DISTANCE D1.

Even though the "parallel {F1}{D1} product in ft-lb" gives the same "numerical answer" as does the "perpendicular {F1}{D1} product in lb-ft" it does pose some interesting, at least to me, questions regarding the relationship of the units to the underlying Physics that's implied by the equations! If you Google "torque and horsepower formula" the first site that's listed is...

... and it deals with the issue of units by using lb-ft for both WORK and TQ... "...Torque, like work, is measured is pound-feet (lb-ft). However, torque, unlike work, may exist even though no movement occurs..." The second site that's listed is...

... and it deals with the issue of units by using ft-lb for both WORK and TQ... " foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower..." The third site that's listed is...

... and it gives TQ as lb-ft and says... "...In the metric system, force is calculated in newtons, and distance is in meters, so the standard torque unit is Newton-Meters or N-M. In the Standard/English system, force is calculated in pounds and distance in feet. So the torque unit is lb-ft..." ...but this site doesn't even bother to define their unit for WORK, and just lists some formulas as if units aren't really very important anyway so just calculate a "number result" and be done with it!

Now this ft-lb versus lb-ft discrepancy and what the equations imply with regard to the underlying Physics is totally obscured if you're using SI units, where TQ is in Newton-meters (Nm), ENERGY and WORK are in Joules (J), and HP is in Watts (W) or kilowatts (kW), because now when you substitute TQ in Nm into equations #2) and #3) you need to change the units of the result from Nm to J for the WORK equation, and from Nm to kW for the HP equation!

Of course you just can't use SI numbers in my English equations without converting because JPL tried that out on one of their earlier "Mars Lander missions" and it didn't work out so well! The fourth site that's listed is the infamous "Wikipedia"...

... and I cant copy and paste this formula into a post text box because it's an "object", but if you look about half way down the page to the section title... "...Relationship with torque...", you'll see a HP formula that has TQ in Nm as an input, and it gives HP in kW as the resulting output, but there's never a mention of the apparent discrepancy with the units shown in the formula or what underlying Physics the formula is based on!

The fifth site that's listed is "RV.Net Open Roads Forum", and I'm not giving the link to that one because I don't want to make it too easy for you to read more myths and misconceptions about TQ! The final site I'll give uses lb-ft for TQ and ft-lb/min for HP, which implies that WORK is in ft-lb, but it just ignores the unit discrepancy as if it didn't even exist...

... however the main reason I'm giving the above link is so that you can read this... "...You may be surprised, I know I was, but horsepower is simply a calculation based on rpm and torque. Dynos and rolling roads only ever measure torque. Horsepower is calculated after. Here's the formula..." ...because this is one of the most commonly held misconceptions about TQ that I've seen posted many times, and this is where I'll pick up in my next post where among other topics I'll discuss what dynos actually "measure"!

I guess while I'm in town I should stop by and see if Penn & Teller will do one of their BS specials on TQ to help put an end to all this nonsense! In the meantime, if you have an occasion to use equation #2) or the more popular equation #3) just remember to change the TQ units from lb-ft to ft-lb! I suppose this is just one more example where for the sake of expediency we ignore what appears to be an inconvenient complication!

121 Posts
Discussion Starter #10
Confused about HP vs TQ, and what a dyno measures?

Well judging from some of the statements I've read on various diesel forums and on various web sites over the past several years if your answer is "yes" you're in good company! In the previous post I gave this link... ...which has the following statement..."...You may be surprised, I know I was, but horsepower is simply a calculation based on rpm and torque. Dynos and rolling roads only ever measure torque. Horsepower is calculated after. Here's the formula..."

This link... ...includes the statement... "...How they rate hoarse power is a joke I have been a diesel mechanic for 30 years. I have rebuilt engines 25HP to 2000HP. The old cat D8H low HP had only 285 HP the bore was 5.75" and high idle was only 1200 RPM that old cat with only 285 HP wood out pull a 750 HP race car..." ...and this one... "...More hp is possible in any engine configuration -- what you really WANT is more torque..."

This link... ...says... "...Horsepower Vs Torque, What’s The Difference, Why It Matters, And How To Get It?... Engines don’t make horsepower; they convert fuel into torque..."

Well that's enough examples to make my point that the internet is loaded with incorrect statements, but the thing I find discouraging is that the above statements mostly come from "diesel people"! It must be a my TQ is bigger than your TQ thing like the sales pitch given by diesel motorhome salesmen because most of the"motorcycle sites" seem to have the "HP vs TQ thing" and the "what a dyno measures thing" figured out pretty well!

Here's a couple of examples from "motorcycle people"... ... "...Engine Torque only determines the load on the Primary Drive's drive gear, that's all. It's the most misused property of an engine ever ... Engine Torque is about as useless an attribute of a bike as can be ... If you over-torque the head of a bolt, you’ll break the head off. If you built an engine with too much torque you’d break the crankshaft or sheer the teeth off of the primary’s drive gear (or some weaker part farther down in the drive train) ... Magazine articles that you've read that say they measured the torque at the rear wheel are not telling you the truth ... Rear wheel dynos measure Horsepower not Torque..."

This link... ...says... "...The word "torque" is often used incorrectly to describe low RPM horsepower. In the real world, horsepower is all that matters, because torque involves no motion whatsoever. Torque is simply the static measurement of twisting force..."

So why do motorcycle types seem to get it while diesel heads don't? Well one reason I started this thread was as a hobby to help keep my brain from aging any faster than is absolutely necessary, but as I got into this it's become somewhat of a mission to disseminate the true story regarding HP vs TQ, and what a dyno measures! However anyone reading this thread shouldn't believe anything I say just because I claim it's correct, but only believe the things you understand to be correct! That's why I go to great lengths to try and explain the Physics involved as to why things are the way they are! Please feel free to keep asking questions until you understand for yourself why things are true or perhaps make me realize that I'm in fact in error.

Since I try to use "technically precise" language let me explain just what I think it means to say you've "measured" something. If you take a ruler and hold it up next to an object you can "measure" its length. However, if you "measure" the three sides of a box can you claim that you just "measured" its volume? I'd say that you'd "measured" some parameters that could be plugged into an equation to "calculate" an estimate of the box's volume, but you most definitely didn't "measure" it.

Well the same thing is true for HP and TQ, as I don't see how it's physically possible to "measure" either HP or TQ! The definition of HP is the TIME rate of doing WORK which is, HP={WORK}/{(TIME)(550)} ft-lb/sec, where WORK is in ft-lb, and TIME is in sec. The WORK is defined by, WORK={(FORCE)(DISTANCE)} ft-lb, where the lb FORCE moves over a ft DISTANCE that's parallel to the lb FORCE. The TQ is defined by, TQ={(FORCE)(DISTANCE)} lb-ft, where the lb FORCE acts perpendicular to the ft DISTANCE, but the lb FORCE doesn't move relative to the perpendicular ft DISTANCE. You can measure DISTANCE with a ruler and TIME with a stopwatch, and a static FORCE with a spring scale, but measuring a moving FORCE is somewhat problematic!

For example, you could drive a truck to the top floor of a tall parking garage, hang one rear wheel over the side, wrap a length of rope around the tire to form a crude winch, and use the rope to lift a weight from the ground at a constant speed, and then measure the TIME required to lift the weight a measured vertical DISTANCE, and then calculate the WORK by substituting the value of the weight in lb for the FORCE in the WORK equation, and then use the WORK in the HP equation to calculate the truck's RWHP at a specific engine RPM. Of course RPM is actually a calculation based on either a measurement of crankshaft or camshaft angular position versus TIME, or a counting measurement of how many revolutions occur in a given TIME. Then you could calculate the RWTQ from, RWTQ={(RWHP)(5,252)}/{RPMw} lb-ft, where the rear wheel RPMw could be calculated from the engine RPM and the overall gear ratio. Of course many trial runs would be needed to figure out the correct amount of weight to fully load the engine at a each RPM, and the weights used would need to be measured on an accurate scale!

In the above configuration the rope is applying a constant tangential lb TIRE FORCE to the tire tread which is equal to the weight in lb, and the ft DISTANCE from the tire thread to the center of the wheel, which I previously called Dg is the RWTQ moment arm, so you could calculate the truck's RWTQ at a specific engine RPM using, RWTQ={(TIRE FORCE)(Dg)} lb-ft. Then you could calculate the RWHP from, RWHP={(RWTQ)(RPMw)}/{5,252} ft-lb/sec.

Well my point with all those convoluted word examples above is to illustrate that you can't directly measure either HP or TQ! Now it's true that if you do a Google search with "torque horsepower measurement" you'll come up with many sites like this one... ...that claim to build sensing devices that "measure" HP & TQ... "...Sensing Systems performs accurate torque and horsepower measurements using bonded strain gages and wireless telemetry systems..." ...but if you scratch beneath the surface you'll see that what's actually being measured are parameters that allow the calculation of HP and TQ! For example a strain gauge is just a modern day version of a spring scale for measuring FORCE! So called in-line TQ transducers or load cells employ strain gauges or other techniques to determine shaft TQ, but in the final analysis they're just measuring a FORCE on the shaft's surface and using the TQ moment arm DISTANCE to the shaft's centerline to calculate the TQ!

Speaking of new versus old, lets use the picture diagram below to talk about an old-time Prony Brake Dynamometer similar to the one I first used in 1960 when I was an undergraduate EE student studying electric motors.

The two spring scales apply tension to the leather belt, which in turn applies a frictional load to the pulley on the motor's driveshaft. The tension can be increased by raising the beam to which the scales are attached, and this increases the frictional load on the motor. Since the frictional FORCE between the pulley and the belt is similar to the TIRE FORCE I discussed previously, I used the same parameters, TIRE FORCE, Dg, RPMw, RWTQ, and RWHP in the diagram. Even though the intended use of this type of dyno is to "measure", oops... I mean "determine" shaft TQ and HP of an engine,you could in principle strap a belt around a rear tire and determine the RWTQ and RWHP of a truck!

For this dyno the measured parameters are Fa and Fb using the two spring scales, RPMw using a tachometer to count the number of revolutions in a one-minute TIME, and Dg using a ruler. Next the TIRE FORCE is calculated from, TIRE FORCE={Fa-Fb} lb, and then the RWTQ is calculated from RWTQ={(TIRE FORCE)(Dg)} lb-ft, and then the RWHP is calculated from, RWHP={(RWTQ)(RPMw)}/{5,252} ft-lb/sec.

I also used a larger version of a Prony Brake Dynamometer that employed a wooden clamp around the pulley to provide the frictional load on the motor, and the clamp had a long moment arm that was restrained by a single spring scale. Well both of these frictional loads were in fact very similar to the TIRE FORCE when burning some rubber because they also caused a lot of smoke, and if they weren't doused with water on a regular basis they'd catch fire!

Since HP can also be defined as the TIME rate of ENERGY production, and since all of the shaft HP is being dissipated as heat ENERGY in the prony, brake you could in principle measure the thermal mass and the temperature versus TIME and calculate the ft-lb heat ENERGY output versus TIME, and then calculate the HP from, HP={heat ENERGY}/{(TIME)(550)}. However just measuring the FORCE with spring scales and the moment arm DISTANCE with a ruler, and then calculating the TQ, and then measuring the RPM and calculating the HP is much easier and more accurate!

Modern day engine dynos typically use water brakes to load the engine, then they measure a moment arm DISTANCE and a FORCE to calculate the TQ, then measure the RPM and calculate the HP. You could design an engine dyno that first determined HP by having the test engine drive an electrical generator hooked to a resistive load, and then measure the output voltage and current and calculate the kW power output from the generator, then convert that to HP using 746 Watts per HP, and then measure the RPM and calculate the TQ!

So there're many ways to skin the HP and TQ cat by measuring various parameters that then allow calculations of HP and TQ. Whether you measure parameters that allow TQ to be calculated first, and then calculate HP from TQ using RPM, or you do it in the reverse order, depends on the relative practicality and accuracy of making the required parameter measurements! Therein lies the reason why chassis type dynos in general, and inertia type chassis dynos in particular, first determine the RWHP, and then calculate an "adjusted" RWTQ, and I'll explain later what "adjusted" RWTQ means!

288 Posts
i agree, damm your nowledge is crazy! keep it coming A+

121 Posts
Discussion Starter #13 (Edited)
My "unified explanation" for "all types" of dynos...

For now don't think about "dynos" as being either "engine dynos" or "chassis dynos", but rather consider that "dynos" in general can be classified according to the types of devices they employ to load the output shaft under test and to absorb that test shaft's HP generated ENERGY. As I'll explain later this "output shaft under test" can be a test shaft that's "directly connected" to an engine's flywheel or it can be a test shaft that's connected to a chassis dyno's roller drum, and this roller drum "test shaft" only has a very "indirect connection" to an engine's flywheel that involves gear ratios and possibly even tire slippage and or clutch slippage. It would be an advantage if you had a test shaft with a "direct connection" to a truck's rear axel shaft, but it already takes far too long to switch from one truck to the next during a "dyno contest"!

The prony brake dyno that I discussed previously employs a friction brake to load the output shaft being tested and this prony brake employs friction to absorb the test shaft's HP generated ENERGY and it dissipates this absorbed ENERGY as heat ENERGY, and this involves dousing the friction brake with water to keep it from catching fire. The friction brake is restrained from rotating with the shaft under test by an arm that's connected to a FORCE measuring device at a known DISTANCE from the test shaft's centerline. Using "measurements" of FORCE and DISTANCE the test shaft's TQ can be "calculated", and then a "measurement" of the test shaft's RPM allows the test shaft's HP to be "calculated".

Other types of dynos use water brakes or eddy current brakes to load the shaft under test and to absorb that test shaft's HP generated ENERGY and dissipate it as heat ENERGY, but other than using a different kind of ENERGY absorber brake to load the test shaft these types of dynos operate just like the prony brake dyno in that they employ a TQ moment arm to restrain the rotation of their ENERGY absorber brake, and they use measurements of FORCE and DISTANCE that are associated with their ENERGY absorber brake's TQ moment arm to calculate the test shaft's TQ, and then they use measurements of the test shaft's RPM to calculate the test shaft's HP.

Water brake dynos, eddy current brake dynos, and prony brake dynos are all called "load dynos", and all three of these types of "load dynos" can apply a constant RPM steady state load to the shaft under test, so that the test shaft's TQ and HP can be calculated at a constant steady state test shaft RPM.

An "inertial dyno" absorbs the HP generated ENERGY from the shaft under test by storing this ENERGY as kinetic ENERGY in a flywheel type rotating mass that rotates at the same RPM as the shaft under test. In order for an inertial dyno to maintain a load on a test shaft the test shaft's RPM must be continuously increasing so an inertial dyno can't apply a constant RPM steady state load to the shaft under test like a load dyno can. At the end of a test run the stored kinetic ENERGY in the rotating mass is dissipated as heat ENERGY as the rotating mass is brought to a stop using a friction brake.

With an inertial dyno one has the option to directly calculate the test shaft's HP from "measurements" of the "angular position" of the rotating mass versus TIME, or one can use these angular position versus TIME measurements to first calculate the test shaft's TQ, and then use the test shaft's "measured" RPM to calculate the test shaft's HP. The reason the test shaft's HP can be directly calculated using an inertial dyno and why this isn't practical with a dyno employing a prony brake, water brake, or an eddy current brake is that with an inertial dyno the ENERGY generated by the applied HP is being stored in the rotating mass, and this stored ENERGY can be accurately calculated as a function of TIME from angular position versus TIME measurements to give the test shaft's HP directly from HP={ENERGY}/{(TIME)(550)} ft-lb/sec. For the other types of dynos this absorbed ENERGY is being continuously dissipated as heat ENERGY and it can't be accurately calculated versus TIME, so that instead the test shaft's TQ is first calculated from FORCE and DISTANCE measurements, and then the test shaft's measured RPM is used to calculate the test shaft's HP!

Note that in the above discussions I used the terms "the test shaft" or "the shaft under test", and this "shaft under test" can equally well refer to an engine's output shaft that's connected to an engine dyno, or to the roller drum shaft that's part of a chassis dyno. Said another way the devices used in an engine dyno to absorb a test shaft's HP generated ENERGY from a test engine's flywheel are exactly the same devices that are used in a chassis dyno to absorb the HP generated ENERGY from a test truck's rear wheels. However an engine dyno connects these ENERGY absorbing devices directly to the output shaft from an engine to determine the engine's flywheel HP and TQ, whereas a chassis dyno connects these ENERGY absorbing devices to the output shaft from the chassis dyno's roller drum to determine the HP and TQ that exists on the roller drum's shaft, and then the chassis dyno's computer uses that roller drum shaft HP and TQ information to calculate the truck's rear axel shaft HP and TQ, which is the truck's RWHP and RWTQ! Of course in the case of an inertial type chassis dyno the roller drum itself is the HP absorbing device!

Independent of which type of ENERGY absorbing device is employed in a chassis dyno the conservation of ENERGY says that the chassis dyno's roller drum shaft HP "is equal to" the truck's rear axel shaft HP which is the truck's RWHP, but the chassis dyno's roller drum shaft TQ is in general "not equal to" the truck's rear axel shaft TQ which is the truck's RWTQ! The truck's rear axel shaft is separated from the chassis dyno's roller drum shaft by the truck's rear wheels and flexible tires along with the chassis dyno's roller drum, and that separation between the two shaft's constitutes an effective gear ratio that changes as a function of tire deformation under load, and that's what complicates relating the dyno's roller drum shaft TQ to the truck's rear axel shaft TQ or its RWTQ! As the "caveman" finally figured out, TQ is "multiplied" or "divided" by gear ratios, but HP isn't!

For an "inertial type" chassis dyno the roller drum's shaft HP can be directly calculated, and that roller drum shaft HP is "equal" to the truck's RWHP! For the other "load type" chassis dynos the roller drum's shaft TQ is first calculated then the roller drum's RPM is measured and used to calculate the roller drum's shaft HP, and that roller drum shaft HP is "equal" to the truck's RWHP! For any type of chassis dyno the chassis dyno's roller drum shaft HP is always "equal to" the truck's RWHP!

After you get the truck's RWHP by from the chassis dyno's roller drum shaft HP you could then calculate the truck's RWTQ from the truck's RWHP and the truck's rear wheel RPMw using, RWTQ={(RWHP)(5,252)}/{RPMw} lb-ft. In principle you could also use the calculated value of the roller drum's shaft TQ to calculate the TIRE FORCE that's being applied to the roller drum's surface, and then calculate the RWTQ from the TIRE FORCE and the DISTANCE Dg from the truck's rear axel to the roller drum's surface using RWTQ={(TIRE FORCE)(Dg)} but this approach is less accurate than simply calculating the RWTQ from the RWHP using the rear wheel RPMw!

Of course I can see no reason whatsoever for ever wanting to calculate the truck's RWTQ to begin with because the RWHP is all you ever need to know unless you're worried about breaking something in the driveline! You can estimate the engine's FWHP from the estimated driveline loss and the RWHP, and then calculate the estimated FWTQ from the estimated FWHP using the engine RPM, and doing it this way doesn't require knowing anything about the tranny or diff gear ratios! Also you don't need to know the RWTQ to do any truck performance calculation because they all depend only on the RWHP! In a later post I'll show how you can use the RWTQ as part of a hill climb analysis, but the use of the RWTQ is completely unnecessary and even likely to cause confusion. I'll show two better ways to do the same analysis using only the RWHP!

The truck's "actual RWTQ" is given by RWTQ={(RWHP)(5,252)}/{RPMw} lb-ft, but the chassis dyno sheets I've seen give an "adjusted RWTQ*" that's calculated as, RWTQ*={(RWHP)(5,252)}/{RPM}, and this RWTQ* would be the actual RWTQ if there was an overall net 1:1 gear ratio from the engine flywheel to the rear axel because then RPMw=RPM. If you dyno with the tranny in a 1:1 gear ratio, then your actual RWTQ={Diff Ratio}{RWTQ*}. In general if the GR is the overall net Gear Ratio from the flywheel to the rear axel then actual RWTQ={GR}{RWTQ*}.

The DLE is the Drive Line Efficiency which is a number between 0 to less than 1 that's defined by RWHP={(FWHP)(DLE)}, so that FWHP={RWHP}/{DLE}. Also, RWTQ={(FWTQ)(DLE)(GR)}, and FWTQ={RWTQ}/{(DLE)(GR)}. If you substitute RWTQ={GR}{RWTQ*} into the equation FWTQ={RWTQ}/{(DLE)(GR)}, you get FWTQ={RWTQ*}/{(DLE)}, so that the RWTQ* you get from a dyno run, combined with an estimate of the DLE gives the estimated FWTQ, and the RWHP from a dyno run, combined with an estimate of the DLE gives the estimated FWHP.

Prior to the "hack job" I'd posted many pages showing all the details for several different ways of deriving the "performance equations" for an "inertial type" chassis dyno and it will be awhile before I get around to reconstructing all that work so for now I'll give the summary below.

The following describes how a rear wheel inertia type chassis dyno determines a truck's RWHP from "measurements" of the angular position of the dyno's drum versus TIME. The dyno's computer data acquisition system continuously "measures" the angular position of the dyno's drum versus TIME, and then it calculates "estimates" of the dyno drum's "instantaneous" RPMd at each specific TIME.

The RWHP from the truck that's required to increase the dyno drum's RPMd from RPMd1 to RPMd2 in a TIME interval of T sec is given by the "inertial dyno equation" ...

RWHP={(Wd)(Rd^2)(RPMd2^2-RPMd1^2)}/{(T)(6,454,618.5)} HP

...where Wd is the dyno drum's weight in lb, and Rd is the dyno drum's radius in ft.

The above equation says that if a given RWHP from a truck is applied to the dyno's drum at a TIME when the drum's RPMd is RPMd1, then the amount of ENERGY that's transferred from the truck's rear wheels to the dyno's drum during the following T sec TIME interval will cause the dyno drum's RPMd to increase from RPMd1 to RPMd2 in that TIME interval of T sec!

Now that's about as simple as it can get folks. Please note that the above calculation of the test truck's RWHP doesn't have anything whatsoever to do with the test truck's engine RPM, its tranny and diff gear ratios, or the diameter of its tires. There could just as well be a bunch of hamsters running on the dyno's drum and as long as the hamsters added the same amount of energy to the drum in the same amount of time as the truck's rear wheels did you'd calculate the exact same RWHP or in the case of the hamsters their paw HP!

The only "legitimate" reason why a dyno computer "needs" to "measure" the test truck's engine RPM is so that it can give a graph of RWHP vs RPM! However based on some of the stuff I've read on "inertial dyno" web sites another reason why a chassis dyno computer would need to know the engine's RPM is if the guy who programmed it didn't know what he was doing and thought that he needed the computer to first calculate the truck's RWTQ and then get the truck's RWHP from the RWTQ!

Yes these days nothing much surprises me anymore, and it's quite possible that the people who actually program inertial dyno computers fail to realize just how easy it is to calculate the truck's RWHP directly from the "measurements" of the dyno drum's angular position versus TIME using the above "inertial dyno equation", and that they needlessly complicate the problem by first calculating the TIRE FORCE that's applied to the dyno's drum, and I won't show all the details here, but that calculation shows that the TIRE FORCE is proportional to a bunch of "stuff" that turns out to be almost the RWHP in disguise, and that the TIRE FORCE is inversely proportional to {(RPMd)(Rd)}, and since Dg={(RPMd)(Rd)(GR)}/{RPM}, that means that since RWTQ={(TIRE FORCE)(Dg)}, the term (RPMd)(Rd) cancels which in effect also removes a direct dependence on Dg, and you get, RWTQ={("stuff")(GR)}/{RPM}, then you could calculate the RWHP from the RWTQ, but note that you need both the GR and the engine RPM to get the RWHP doing it this way!

When you get to this equation, RWTQ={("stuff")(GR)}/{RPM}, you can "manipulate" the "stuff" until it's of the correct form to recognize that the "stuff" is actually just the right-hand side of the "inertial dyno equation" multiplied by the factor (5,252), so that you wind up with, RWTQ={(RWHP)(5,251)(GR)}/{RPM}! Then if you solve for RWHP you get, RWHP={(RWTQ)(RPM)}/{(5,252)(GR)}, and if you substitute RPM={(GR)(RPMw)} you get the more familiar, RWHP={(RWTQ)(RPMw)}/{5,252}!

Now it turns out that in addition to the above types of calculations a dyno computer is also doing a lot of extra "behind the scenes black magic" involving various so called "correction factors" that the "dyno computer" uses to "play with" the measurements it takes before printing out a curve of your truck's RWHP vs RPM!

At some point I'll do a post on the effect of doing your inertial dyno run using different tranny gears or diff ratios. As you can see below which is from 3 consecutive runs in 3 different gears...

...doing a run on an inertial dyno with a lower transmission gear ratio or diff ratio gives a higher RWHP determination! This "effect" is unique to an inertial dyno which requires that the truck's engine RPM must be continuously increasing in order to get a RWHP determination, and this in turn causes a number of complications which I'll address later.

I'll also do a post on all the various so called "correction factors" including "atmospheric corrections" and various "smoothing corrections". The graphs below show some of the different ways the dyno computer has manipulated the exact same data run files to get different results. This first graph isn't corrected for atmospheric effects and has a #3 smoothing correction. The second graph also has a #3 smoothing correction but it employs an SAE atmospheric correction factor. The third graph has a #0 smoothing correction and it also employs an SAE atmospheric correction factor. Note the effects of theses various correction factors on the HP and TQ listed on the graphs. I'll be giving a simple road test procedure you can use to check the accuracy of your dyno results!


121 Posts
Discussion Starter #15
Road Test for your RWHP...

If you're going a constant speed on a flat road and mash the throttle to the floor you apply a RWHP=RWHPa+RWHPd to the road where the component RWHPd is required to overcome aerodynamic drag and rolling resistance, and the component RWHPa is available to accelerate the truck. Here's how to determine your truck's RWHP by making a timed acceleration measurement from MPH1 to MPH2 followed by a timed "coasting in neutral" deceleration measurement from MPH2 to MPH1.

Step #1... go to a truck stop and weigh your truck to get its weight W, of course a CAT scale is always preferred for weighing CAT powered truck's!

Step #2... find a flat stretch of road with no head wind or tail wind and select a gear that puts your RPM at the value RPM* where you want to determine a single point RWHP* at RPM* on your overall RWHP vs RPM curve. The numbers I'll use below are for my 300 FWHP CAT C7 truck towing my 40' 5er in 4th gear where RPM*=2,200 determines the RWHP* at the RPM* for the maximum 300 FWHP.

Step #3... reduce your RPM about 100 RPM below RPM* to a value RPM1 so that your speedometer reads a MPH1 value that corresponds to a convenient reference mark like 55 MPH at 2,100 RPM for my truck. Then increase your RPM to about 100 RPM above RPM* to a value RPM2 and get a higher MPH2 value with a reference mark like 60 MPH at 2,300 RPM for my truck.

Step #4... now reduce your speed well below MPH1 and settle into a constant speed about 500 RPM below RPM*, then mash the throttle to the floor and start your stop watch as you pass through MPH1 and stop it as you pass through MPH2, and measure the time interval Ta required to accelerate from MPH1 to MPH2.

Step #5... you're now traveling at a speed greater than MPH2 so shift into neutral and coast, and as you coast through MPH2 start your stop watch and then stop it as you coast through MPH1 so that you've measured Td the time interval that aerodynamic drag and rolling resistance acted on your truck to decelerate it from MPH2 to MPH1.

Step #6... plug the above measurements into these equations...






... this gives your RWHP* at RPM* which is the average value of your RWHP between RPM1 and RPM2 and your RWHPd at MPH* which is the average value of your RWHPd between MPH1 and MPH2. To get the best accuracy for these estimates repeat these measurements several times, and do them while going in each direction on the same stretch of road to average out any slight changes in grade or wind condition.

Here's an example calculation for my truck with my 19K 5er in tow where my W=32,000 lb, MPH1=55, MPH2=60, Ta=7.5 sec, and Td=10.8 sec. Plugging these numbers into the equations gives...

RWHPa={(W)(MPH2^2-MPH1^2)}/{(Ta)(16,452.6)}= {(32,000)(60^2-55^2)}/{(7.5)(16,452.6)}=149.1 HP

RWHPd={(W)(MPH2^2-MPH1^2)}/{(Td)(16,452.6)}={(32,000)(60^2-55^2)}/{(10.8)(16,452.6)}=103.6 HP

RWHP*=RWHPa+RWHPd=149.1+103.6=252.7 HP

RPM*={(0.5)(RPM1+RPM2)}={(0.5)(2,100+2,300)}=2,200 RPM

MPH*={(0.5)(MPH1+MPH2)}={(0.5)(55+60)}=57.5 MPH

So my RWHP is 252.7 HP at 2,200 RPM, and my RWHPd is 103.6 HP at 57.5 MPH.

On my CAT C7 I've got a VMS 240 CL made by SilverLeaf Electronics, Inc.

so I can get a real time display including a histogram of FWHP and RPM, and as I passed through 2,200 RPM it read FWHP=299 HP so my DLE=Drive Line Efficiency where DLE={RWHP}/{FWHP} and DLE is a number between 0 to less than 1 is...

DLE={RWHP}/{FWHP}={252.7}/{299}=0.845 or 84.5%
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