Horsepower - Part 2

In the previous blog post we discussed Mass, Acceleration, and Force, marching towards the notion of Horespower. In this post we will discuss Work, Power, and Torque.

Work

Energy in the form of Work means exerting a force over a distance. For example, lifting a 4300 pound M5 5 feet straight up expends 21,500 ft-lb of energy (5 ft * 4300 lb = 21,500 ft-lb).

See how I casually mixed mass and force? The force required to lift a 4300 Pound (mass) M5 at the surface of the earth is 4300 Pounds (force). If I was lifting that same object the same distance on the moon, I would only need 4300/6 = 717 Pounds (force), and hence I would only have done 5ft*717lb = 3585 ft-lb of work on my 4300 Pound (mass) M5 instead of 21,500 ft-lb.

Now I can hear some of you clever car folks saying, "hold on, isn't foot-pounds the unit of torque?" It is indeed, and this isn't due to the confusion of force and mass, it’s due to the fact that horsepower and torque are very closely related, as we’ll soon see, so stay tuned.

Power

This brings us to Power, as in Horsepower.

Suppose I did 66,000 ft-lb of work on my M5 by pushing it up a slight incline with a force of 100 lb over a distance of 660 ft. My energy expended was 66,000 ft-lb, whether it took me two minutes to push it that far or ten minutes.

However, if that was accomplished in ten minutes, the Power I used was 66,000 ft-lb / 10 min = 6600 ft-lb/min. If it was accomplished in two minutes, the Power would have been 66,000 ft-lb / 2 min = 33,000 ft-lb/min.

Doing the same work in one fifth the time means I need five times the power to get it done.

Horsepower

The unit "ft-lb/min" is a unit of power. So is the "Horesepower" a unit of Power.

A standard Horsepower was a unit that was introduced as a marketing term by the Scottish engineer James Watt in 1782.

He used the unit to compare the power of his new-fangled steam engines

with the power of the draft horses they were intended to replace.

He measured the power output of a few horses, and defined his unit of a Horsepower to be his best estimate of the maximum power that a draft horse could generate over a reasonably long period of time. He settled on using a standard definition of 33,000 ft-lb/min. 

So pushing my M5 up a slight incline by using 100 lb of force over a distance of 660 ft in 2 minutes means I had to use 1 Horespower to do it. In other words, that’s about how far and how fast a big draft horse could do it at a sustainable pace.

Torque

Torque is a measure of twisting force. Say I want to unfasten a wheel nut. I get out a tire iron that is 1 ft long and I step down on it applying 100 lb of force. I am applying a torque of 100 ft-lb to that nut. That’s how much twisting power I am applying.

Say the nut does not move. I can stand on the tire iron with both feet and apply 200 lb, still at 1 ft from it, and the torque is now 200 ft-lb.

Alternatively, I can go to my tools and get a longer tire iron, say one that is 3 ft long. If I now step down on that tire iron with the original 100 lb at the end of it, I am applying 300 ft-lb of torque.

The longer the tire iron, the more leverage I get, and the more torque I am applying throughout the tire iron.

So torque is a measure of "twisting force" and is a force applied using leverage a certain distance away from the centre of rotation.

In the diagram above, say the wrench is 2 feet long. Say I applied 20 lb halfway down the big wrench, or 10 lbs at the end.

• 1 ft * 20 lb = 20 ft-lb
• 2 ft * 10 lb = 20 ft-lb

In either case the wrench "sees" the same torque, all along its length (including at the nut).  In fact, if the near hand was pushing upwards with a force of 20 lb, then the far hand would need to pull down with a force of only 10 lb to resist it. One hand is putting 20 ft-lb into the wrench, and the other is exactly resisting that with his own 20 ft-lb of torque, applied in a different manner.

So Torque is really just a “leverage-independent” way of stating force, and happens to have the same units as Work for a very good reason.

Engine Torque

Car engines generate Torque. That’s their job. Recall that a piston moves because there is pressure applied to its surface as a result of the gasses super-heating when ignited. An average gas pressure in the cylinder on the pwoer stroke is 800 psi (pounds per square inch). A cylinder with a bore of 3.5in has an area of pi*(3.5/2)^2 = 9.6 square inches. That means there is 800*9.6=7700 lb of force being applied to each cylinder during the power stroke.

How far the crank moves right to left is the same as the stroke of the engine (how far it moves up and down), which is 8.83 cm. Half of that is 0.145 ft. If the piston applies that force on average 0.145/2 ft away from the crank, that means the engine’s torque is approximately 0.145/2 ft * 7700 lb, or about 560 ft-lb. That’s likely about the torque that the M5’s cylinders generate (I made up some of the number to have it come out right, but they are all in the right ballpark).

Wheel Torque

The engine torque is transferred to torque at the wheels through the gearbox

and the differential (or "final drive").

We will discuss these in considerably more detail in a later blog post. For now, suffice it to say that the engine torque is transferred and transformed by means of gears to the wheels.

Here's how gears work.

Imagine gear A is attached to the engine crankshaft, and gear B is what turns the wheels. We see that A is going twice as fast as B. This is because B has twice as many teeth as A. The force is transferred at the point where gear A meshes with gear B. If the torque on A is 500 ft-lb, and gear A is 1ft in diameter, then the force pushing on the teeth of gear B is 500 ft-lb / 0.5 ft = 1000 lb. Because gear B has twice the teeth of gear A, it must have twice the perimeter, and therefore twice the radius. So gear B is 2 ft across. The force of 1000 lb is therefore applied 1 ft from the centre of gear B, therefore the torque on gear B is 1000 ft-lb, or double the torque on gear A.

That's the general rule of gears, no matter how complicated is the gear train. If the gear slows down the spinning by a factor of X, then it multiplies the torque by that same factor of X.

In 1st gear on the M5, every 15 turns of the engine crankshaft turns the wheels once. Therefore the torque is multiplied by a factor of 15 when in first gear. That means that if the torque generated by the pistons onto the crankshaft is 500 ft-lb, the torque transmitted to the wheels is 500*15 = 7500 ft-lb, evenly divided between the two rear wheels, when in 1st gear (I am ignoring any frictional losses along the way that wind up as noise and heat for the time being, but will come back to that later).

The rear wheels and tires together have a radius of about 13.5 inches, or 1.125 feet, therefore the force exerted by the rubber on the road is 7500ft-lb/1.125ft = 6667 lb of force driving the car forward.

Given the mass of the car is about 4300 lb, then if the tires were perfectly sticky it should accelerate forwards at 6667/4300 = 1.55 g’s, which means it is speeding up by 34 mph each second, which would get it to 60 mph in under just 2s. Woo hoo!

Unfortunately, the M5 is traction limited off the line, as discussed in the previous blog post, and takes about 4s to get to that speed. Boo hoo!

A good all wheel drive with the same torque could theoretically hit sub 3s. The next iteration of the Mercedes-Bens E63 AMG is slated to have 4 wheel drive, so we’ll soon see if this will be the case!

Wasted Torque?

Is all that torque wasted? It certainly is in 1st gear when starting from a standstill. But the exact same engine torque is available in all gears.

For example, in fourth gear the ratio is about 4 to 1, so the torque at the wheels is 500*4 = 2000 ft-lb which means we are no longer traction limited, but rather engine torque limited. Still, the 2000 ft-lb should accelerate us at around 0.4g’s at highway speeds, meaning that we can increase our speed by about 10 mph in a second and zip past that Caddy!