**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!