http://www.corral.net/projects/subzero/gtech.html is an interesting website, for it supplies formulas that can be used with an accelerometer to compute torque and horsepower.† However, it plays a little loose with definitions, and leaves unsaid how the details of some calculations work.† So, I played a little bit.† For purposes of this discussion, I am using a Valk and rider that weigh 1000 lbs, accelerating from a complete stop at the rate of 15 ft/sec^2 (about 1/2G).
* = multiply
/ = divide
^ = raise to power
= = equals†† :D
The first formula from the website
V = A * T
computes instantaneous velocity at any given time, not average velocity.† Thus, the bike velocity is:
after one second - V = 15 ft/sec^2 * 1 sec = 15 ft/sec
after two seconds - V = 15 ft/sec^2 * 2 sec = 30 ft/sec
after three seconds - V = 15 ft/sec^2 * 3 sec = 45 ft/sec
Incidentally, 30 ft/sec is 20.45 mph.
The second formula
D = V * T
can compute the distance only if the velocity is constant, and thus, we cannot use the velocities as computed above (which one would you use?Ö after all, the velocity is not constant because you are accelerating).† The proper way to compute distance is this formula
D = 0.5 * A * T^2
So, in the three second example above, we get a distance of
D = 0.5 * 15 ft/sec^2 * (3 sec)^2 = 67.5 ft.†
I sure like it when the units work out!† In any event, distance isnít actually needed for any other computations, but if you want to know how far youíve gone, this is the formula to use.
The next formula
HP = V * A * M
must compute the instantaneous horsepower at an instantaneous velocity since acceleration is part of the equation.† This is OK since horsepower must be increasing as velocity increases.† Before we proceed with the computations, we need to discuss the relation between force, weight, and mass.
Weight is the force that holds things on the ground due to the acceleration of gravity.† If there is no gravity, like in space, then our Valk still has mass, but it no longer weighs anything.† But, we are on the ground and the Valk weighs 1000 lbs.† Our task then, is to convert that weight into mass so that we can use the equation above.
In the English system, if you push a mass of one slug so that it accelerates at 1 foot/sec^2, then you are exerting a force of one lb, and thus 1 lb = 1 slug ft/sec^2.† But, what the hell is a slug?† Well, letís turn the definition around.† The definition of 1 lb is 1 slug ft / sec^2, and thus by solving for slugs we get 1 slug = 1 lb sec^2 / ft.† In other words, it is the amount of material that requires you to push with a force of one pound to move it one foot in one second from a dead stop.† When we say how much something weighs we are actually specifying the force exerted on a scale due to the acceleration of gravity.† Thus, weight/force and mass are related by acceleration, as in
F (force/weight) = M (mass) * A (acceleration)
If we have a Valk and rider that weigh 1000 lbs on the earth, and we want to compute its mass, then to do this we must know that gravity creates an acceleration of 32 ft/sec^2.† Then, the Valk has a mass of
M = F / A†† (solve for mass)
M = 1000 lbs / 32 ft/sec^2 = 31.25 lbs sec^2/ft.†
M = 1000lbs / 32 ft/sec^2 = 31.25 slugs† (express in terms of slugs)†
So, a Valk with a weight of 1000 lbs has a mass of 31.25 slugs.† Now we can proceed.
Now to go back to our example, letís say we want to know the horsepower after 2 seconds.† But, the formula
HP = V * A * M
is not quite accurate because the answer will be given in ft lbs of power.† So, using our velocity at the two second point (30 ft/sec), it should look like
FT LBS = V * A * M
FT LBS = 30ft/sec * 15ft/sec^2 * 31.25 slugs
(now express slugs in terms of pounds)
FT LBS = 30 ft/sec * 15 ft/sec^2 * 31.25 lbs sec^2 / ft = 14,062.5 ft lbs/sec
It is so cool that the units cancel out!† Must be doing it right.† Now one HP is equal to 550 ft lbs/sec, so
HP = 14,062.5 ft lbs/sec / 550 ft lbs/sec/HP = 25.57 HP.
This is interesting, but Iíd like to know how this works out so well.† How come we donít need to look at RPMís or torques to get the HP?† I decided to look at this from a perspective of torques.† We know that
F (force) = M (mass) * A (acceleration), that is, the rear wheel must push against the ground with a certain amount of force to accelerate the bike at the specified rate.† In the example case that would be:
F = 31.25 slugs * 15 ft / sec^2 = 468.75 lbs
(express slugs in terms of pounds)
F = 31.25 lbs sec^2/ft * 15 ft / sec^2 = 468.75 lbs
of force needed to accelerate the bike at the rate in the example.† Now, torque is force exerted over a moment arm of some length (L) as in
Tq = F * L.†
The Valk rear wheel is 79.5 inches in circumference by measurement.† Taking the formula and solving for the radius, we get
Circumference (C) = 2 * pi * R (radius)
R = c / (2 * pi)
R = 79.5 in / (2 * 3.14159)
R = 12.65 inches or 1.05 ft.†
Multiplying the two together yields the torque
Tq = F * L.†
Tq = 468.75 lbs * 1.05 ft = 494.25 lbs ft of torque.
Now in the example, our velocity at 2 seconds is 30 ft/sec.† Using the wheel circumference, we compute what the RPMís of the wheel must be to achieve that velocity.
30 ft/sec * 12 in/ft = 360 in/sec
revs/sec = 360 in/sec / 79.5 in =† 4.53 revs/sec
RPM = revs/sec * 60 sec/min (convert to minutes)
RPM = 4.53 revs/sec * 60 sec/min = 271.7 RPM.
Now we can take the HP/torque formula
HP = (torque * RPM) / 5252 =
HP = (494.25 * 271.7) / 5252 = 25.57 HP Ė the same number!†
So, back to the question.† How does the formula provide horsepower without using RPM or torque?† The answer comes in two parts.† First, the force in the torque is derived from the acceleration.† Second, RPM is only needed to compute the distance that the force was exerted (otherwise known as work).† But, we already know that the force was exerted over 30 feet in 1 second (from the velocity), and we know that the force necessary is already implicitly defined in the acceleration quantity.† Thus, the interim steps are not necessary.
Lastly, you can take the torque and HP figures back to the motor by using the final drive ratio for the gear you are in.† For example, clearly the Valk does not put out 494.25 lbs ft of torque.† However from my tables (at the URL below), we discover that in second gear, the ratio of motor rotation to wheel rotation is 7.54, so the engine is outputting about 494.25 lbs ft / 7.54 = 65.55 lbs ft of torque at 271.7 RPM† * 7.54 = 2040 RPM.† This jives pretty well with the computations I made using only gear ratios and rear wheel circumference.†
Valk RPMís and gear ratios available at: