# Finding acceleration for a car after finding torque

I am trying to calculate the acceleration of a vehicle after finding the torque $\tau$. Assuming my Horse Power is 130 and my RPM is 1000, I calculated torque as:

$$\tau = \frac{130 \times 33,000}{2\pi \times 1000} \approx 682.77$$

Assuming that the effective radius of my car tire is 12 inches, I calculated the force $F$:

$$F = \frac{\tau}{12} \approx 56.89$$

Now I will "use the Force" to find the acceleration, using the formula $F = ma$. Rewriting in terms of $a$, and assuming the mass of my vehicle is 2712 lbs I get:

$$a = \frac{F}{2712} \approx 0.0209$$

(I am not sure what the units should be here - the above torque equation I believe is using US customary units and the tire radius is in inches... so maybe the acceleration is in $ft/s^2$?)

However, for larger RPMs, the acceleration seems to decrease, according to these operations. This seems to be the exact opposite of what I expected and what should happen (right?).

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Hi Dylan to Physics.SE! You have to keep in mind that engine RPM is not wheel RPM. Other than that your conclusion is correct, higher RPM means less acceleration. What happens in a real engine though is that the power output also increases with RPM. About the units: sticking to SI units can reduce a lot of conversion problems, I would not recommend using HP, lbs, ft and inch in the same equation. – Alexander Jun 8 '12 at 7:59
Does the above torque equation refer to engine RPM or wheel RPM? Also, if power output increases with RPM, does this account for increase in instantaneous speed? How would one measure instantaneous speed at a time interval, with kinematic equations? (e.g. $s = s_{0} + at$) – Dylan Jun 8 '12 at 16:21
1) both, depends what you do with it. 2) The increase of output power with RPM is a general gasoline engine parameter, independent of any actual car speed. 3) $v = \Delta s/\Delta t$. – Alexander Jun 8 '12 at 16:30
With respect to 3), I am trying to use acceleration to determine speed. It seems you have given me an equation to find velocity using the change in speed over the change in time... Should I find the velocity with the equation $v = v_{0} + at$ and then use the equation you have and solve with respect to $s$? – Dylan Jun 8 '12 at 16:35
Not really. The equation I gave is for "measure instantaneous speed" as you asked for. To calculate it for a given initial speed $v_0$ and time dependent acceleration $a(t)$ the speed at time $t$ is $v(t) = v_0 + \int_0^t a(t) dt$. This site is not really made for discussions though, so please open a new question if it is still not clear how to approach the problem. – Alexander Jun 8 '12 at 22:34

## 2 Answers

I'd go along with Alexander's comment that working in SI units makes life a lot easier. However, assuming you have a good reason for sticking to US units ...

The torque you've calculated is in foot pounds. The easy way to see this is that the 33,000 conversion factor you've used converts horsepower to foot pounds min$^{-1}$, and the rpm is in units of min$^{-1}$. The min$^{-1}$ on the top and bottom of the fraction cancel leaving the units as foot pounds.

In the second equation you've put in the distance as inches, which makes life harder than it need be. If you take the wheel radius to be one foot rather than 12 inches you get the force equal to 682.77 pounds.

For the last step you need to be aware that the "pound" is being used as a unit of force here i.e. it's the force exerted by an object weighing one pound in Earth's gravity. In units of feet per second the acceleration due to gravity is about 32.18 feel/sec$^2$, so the acceleration of an object weighing 2712 pounds will be:

$$a = \frac{682.77}{2712} \times 32.18 = 8.09$$

and that's in units of feet/sec$^2$.

Maybe I'm just used to SI units, but I repeated the calculation using SI units and got the same result a lot quicker! Apart from anything else it makes the distinction between mass and force a lot clearer so you wouldn't have forgotten you need to multiply by the acceleration due to gravity to get the acceleration.

You're quite correct that if power is constant then torque is inversely proportional to rpm. However for most engines the power is roughly proportional to engine speed over a reasonable range, so the torque is approximately constant. If you look at the torque curves for family cars you'll see the torque is roughly constant in the middle of the engine speed range but tails off at high engine speeds. Very highly tuned engines have a torque that peask at higher engine speeds because they're tuned to develop a lot of power at high engine speeds.

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How can I use this information to calculate the speed in MPH? Using an online unit conversion calculator, converting this to $mph$ yields a very large value... – Dylan Jun 8 '12 at 16:32
You've calculated the acceleration not the speed. The acceleration is 8.09 feet/sec$^2$, so after one second you're doing 8.09 feet/sec, after two seconds you're doing 16.18 feet/sec and so on. There are 5280 feet in a mile, so 8.09 feet/sec is 5.52 mph. So after one second you're doing 5.52 mph, after two seconds 11.04 mph and so on. – John Rennie Jun 8 '12 at 17:03
I am sorry, i meant to say that calculating $mph^2$ yields a large number. I thought I needed to convert acceleration to $mph^2$ units – Dylan Jun 8 '12 at 17:37
If you use mph$^2$ the number you'd get is the speed the car would be doing after accelerating steadily for an hour and this would indeed be a large number. I make it 19,872 miles per hour, which is probably faster than most cars could manage :-) I doubt anyone would find it useful to give acceleration in mph$^2$. Feet per sec$^2$ is more likely to be useful, or for us Europeans meters per sec$^2$. – John Rennie Jun 8 '12 at 17:45
Ok. Thank you. The math is easy, but the relationships are kind of hard to envision for a non-Physicist like me. :S – Dylan Jun 8 '12 at 17:52
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You should really include units. I believe the conversion factor horsepower is 1 horsepower $=33,000$ ft*lbf/min (not seconds) It is also in feet not inches.

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