Brian Morris

The funny part is that the formula:

a = P / (v * m)

is derived in the first link that Danno gave:

http://www.stanford.edu/~voloshin/lhowwhy.html (real charts and formulaes)

I've read this page before and I think it gives a good explanation and examples of the relationship between Power and acceleration.

--Brian Morris
89 951

Tomas L

Cool, then I got it right.
I started with the relationship between speed and energy and ended up with that formula. At first I got it wrong by a factor of 4, but it did not seem realistic that a slightly tuned 951 would be able to accelerate at such a rate at 120 mph that it would loose traction...

Tomas

JustinL

Still no-one has a graph of power, torque, and acceleration? I think this argument could be settled with some imperical evidence. I for one am getting a little confused. If what Brian and Thomas is saying is true, then the road dyno that Perry used is WAY off in it's calculations as torque and acceleration looked very similar.

Intuitively it makes more sense with instantaneous torque to the wheels being the driving "force" in instantaneous acceleration. F=Ma I can understand this as it is easy to visualize. Is this wrong?

Tomas L

They should look similar.
The whole secret is in understanding Brians two statements.
When you accelerate past peak HP in a gear you will come to a rpm where HP is the same as it is at the rpm you will drop to if you shift up. At this point acceleration at the higher rpm in the lower gear will be the same as acceleration at the lower rpm in the higher gear will be.
F=m*a is correct but can only be applied at the wheels, not at the engine torque.

Once you understand this you will be on the way...

Tomas

Danno

Justin, two charts on the previous pages showing acceleration in each gear. You provided one when you graphed Perry951's data.

"If what Brian and Thomas is saying is true, then the road dyno that Perry used is WAY off in it's calculations as torque and acceleration looked very similar."

No Brian and Thomas are saying the same thing Perry is. If you calculate using HP as a starting value, you convert it into torque, then linear thrust along the way with removing a time and distance variable with each step. So either way you calculate it, the results will be the same. It's like this progression when thinking of distance, velocity and acceleration:

D = m
V = m/s
A = m/s/s

The progression is just 1st & 2nd order integration and derivatives of each other.

It's funny how we're rehashing a 1800's debate here. Newton resolved all that with calculus. One thing that people don't understand is the concept of time. You have to accelerate over a period of time in order to get power (integration & 2nd order derivative). So you can simplify the following equation:

"a = P / (v * m)"

Note that this is not a fixed equation, you have to integrate it with time and speed in order to make it accurate. That's because at each time-slice and each distance away from a standing-start, the acceleration value changes. At a certain speed, the power results in zero acceleration.

So let's simplify it with:

P = Fd/t ; (power = Force * distance / time)
v = d/t ; (velocity = distance / time)

a = P / (v * m)
a = Fd/t /' (d/t * m)
a = F/m

In the simplest terms, acceleration is proportional to the force available, divided by the mass it has to drive, that's it. Add time and distance and you have power. The thing with a=F/m is that it's an instantaneous value. You have to change the linear-thrust Force at the contact patch as the engine-revs through its torque-curve and the resultant acceleration value will be different as well; thus the need for calculus.

Going back to the 3rd vs. 5th gear acceleration, with the same power delivered to the wheels in both cases, why does 3rd gear accelerate quicker than 5th? That's because that same power is delivered over a shorter distance in less time. Going back to P=Fd/t, with the same power over a shorter distance and time, you have more Force to plug into your F=ma equation.

Or you could've calculated it easier by just multiplying the engine-torque through the gearing and calculating the torque on the ground at the wheels. Simple.

JustinL

Let me sum up what I've got here and you guys tell me if I'm on the right track. I have a good understanding of basic calculus, but not much beyond that.

1. In a frictionless world acceleration is determined by F=ma; in our case m is constant.

2. F= torque at wheels x wheel radius

3. a= (torque at wheels x wheel radius) / mass

4. as the only variable in 3. is torque, it would stand to reason that a is a function of torque to the wheels.

5. Power = work/time= Fd/t ( this is where I'm starting to get lost)

6. Power = a x v x m = Fd/t

7. a= (Fd/t)/vm ok what have I got here?



Thanks for humouring me on this guys... it's been a long time since first year university.


Justin

8. sonofabitch I figured out what I had it's a=F/M BRILLIANT! I've done a complete circle!

2+2

It has been way too long since I have had physics and dynamics so I wont comment.

Danno

Justin, looks good! Basically HP & TQ are to each other as Velocity and Acceleration. Just add a component of time to it. Good job!

One the things that's been implicit in my posts here is you have to model physics through the actual phenomenae you're trying to model. That is, you have to caculate forces from the start, that is BMEP in teh combustion chamber, pushing on a piston of X-square-inches, pushing on a rod of Y-length, rotating a crank of Y-stroke, etc. We've been saved this complicated step by the manufacturer who've provided us with torque & hp figures at the crank, so we can leave out all the engine-internals.

The only thing we have to do then, is to model what happens to that torque & HP in order to push the car down the road. Thus you have to account for the gear-reductions & multiplications. So a good mathematical model has to account for each step along the way. That's why I posted that gear-chart previously. If you change your gearing, or tire-diameter, then the accleration figures will change as well.

So the one component that has not been accounted for is the polar moment of inertia of the tires and wheels (to a lesser extent, the crank, driveshaft, gears and half-axles). This angular acceleration should also be incorporated into a=F/m with the m being split up into linear and rotational inertia.

TurboTommy

"Or, you could've calculated it easier by just multiplying the engine-torque through the gearing and calculated the torque on the ground at the wheels".

Exactly. But that's why horse power is what counts. A high hp, low torque engine will have higher torque multiplication factor through the gears; therefore increasing that low engine torque by the time it gets to the pavement. Acceleration will be the same than a high torque engine that doesn't rev very high. It all comes out in the wash in the hp figure.
It's probably hard to understand complicated formulas but you absolutely have to understand how gearing works.
A little OT here, but when I'm in the market for a new sports car, I actually go to the extent of finding out what the gear ratios are (dealers always look at me stupid). If I don't like the gear ratios, I'll find out if I can modify them before I buy the car. The fastest car of given weight and hp will always be if top gear just makes it to redline, first gear is not to short, and the other gears are nicely spaced in between.

I always shake my head when car makes advertise, "yeh, but we have this x amount of torque".
It totally doesn't mean anything. However, what does count is the RPM spread from peak torque too peak hp. The wider this gap, the more feeling of grunt down low, and the more forgiving your engine will be to a lazy operator. This is the same thing as "area under the curve", and since you can't always be in exactly the right gear, and shifting does take some small amount of time, a wide power band tends to produce somewhat of a faster vehicle.

Tomas L

As this is an important area to understand, if you wish to improve the performance of your car, I will try to describe this in a way that is less theorethical.

Imagine that you are painting a wall. The width of the brush is engine bore and the length of the brush strokes is engine stroke. Then the area you paint with one stroke (width x stroke length) is equivalent to engine torque. The number of brush strokes you make in a minute is rpm.
Neither of these are interesting by itself. No matter how big area (torque) you can paint in one stroke, if you have zero strokes per minute (rpm), no paint will be applied to the wall. The same goes for rpm, if the brush is so small and the stroke is so short that the area painted each stroke is zero, the it doesn't matter how many strokes you will make per minute, no paint will be on the wall.
It's when you multiply the area/stroke with the number of strokes/min and get area per minute you will get something interesting. The power of an engine is in this case equivalent to the area painted per minute.
You can reach the same painted area/min either by having a large area/stroke (torque) and a low count of strokes/min, or you can have a smaller area/min and do many strokes/min. The result is the same. The best thing of course is if you can have a big area/stroke and many strokes/min

The work of the gearbox can also partly be explained with this reasoning.
If you have a gear ratio of 1:2, then your rpm (strokes/min) will be half of what it was, and the torque (area/stroke) will be twice of what it was. Since 0.5 x 2 = 1, you can se that the power (area/stroke) will be the same after the gearbox (except for frictional losses).

The first of Brians statements should be reasonably easy to understand.
This is an example to show the truth in Brians second statement.
Lets assume a car with the following torque and power.
At 4000 rpm: torque 350 lbft and 267 hp (Torque peak)
At 6500 rpm: torque 283 lbft and 350 hp (HP peak)
Consider a car that's moving at 50 mph.
The tires have a circumference of 6.6 feet.
The tires will then rotate at 667 rpm.
The gearing needed to connect the engine to the wheels for each rpm are:
4000 / 667 => 6:1
6500 / 667 => 9.7:1
The torque at the wheels will the be:
@4000 rpm: 350 x 6 = 2100 lbft
@6000 rpm: 283 x 9.7 = 2758 lbft.
This clearly shows that you should shift gears so that you will kepp the engine near the max power rpm.
You can also see that the ratio of the wheel torque figures (2758 / 2100 = 1.31) is the same as the ratio of the power figures for each rpm (350 / 267 = 1.31). This means that if you, at a given speed, can shift gear so that you end up at an rpm with higher hp, the increase in torque at the wheels, will be proportional to the increase in HP between the to rpms.

I hope this will make things a little clearer. This is not that easy to grasp but nevertheless very important.

Tomas

Chris Prack

This is one I did on a Superflow.