Car Engine Problem: Acceleration Function of RPMs

[plusaf]

one more view, of course...

One RPM spin equals x tire spins. Ok that makes sense. Then you used the mechanics equations to find acceleration and such?

My goal was to solve this using power analysis.

why involve power?
to make a car accelerate, you have to get the tires, where they contact the pavement, to push backwards against the earth. then the car moves forward.

the tires have a rolling diameter, (not the wheel diameter, either, guys!), so to create Force to shove back on the Earth at the end of the lever arm known as the Radius of the Tire, it takes Torque!

Torque can be found on the graph of the rat motor (referenced below). since it's pretty flat, let's assume for now that it is a perfectly flat, constant torque curve, versus rpm.

next, you've got to take the car from zero to some max speed, which means the tires will have to go from Zero rpm up to some really nice high speed.

on the other hand, as you'll notice from your own experience with stick-shift cars as well as some of those nasty torque curves, that most AVERAGE engines we drive today really don't like to be very far from a 1500-3000 rpm range, or maybe 1500-4500 for some of the modestly zippier ones. I'm not talking Acura NS-X's here, with titanium connecting rods, gang...), so your engine wants to run in a 2:1 or 3:1 rpm range, while you want your car to cover the, say, 20-to-80 mph range very smoothly.

2:1 is not equal to 3:1 or 4:1.

something must be done.

the answer: gears. gear ratios. transmissions, automatic or manual.

the torque converter in your automatic or the clutch in your stick shift let's things stay loosy-goosy slippery so that you CAN get the car off the line from ZERO speed without forcing the engine to also run a zero rpm, which we know they don't like to do...

once the car is rolling in the first gear, you can mash the accelerator, and the torque is multiplied by the transmission and rear (front, nowadays)-axle ratio (call it Final Ratio) to the axle driving the tires.

the transmissions and clutches and such simply allow the engine to be kept in its happy 1500-4000 rpm range while the gear ratios deliver the torque to the tires to push the car forward.

yes, at high speeds, it does take power to push the air out of the way, so most cars ARE horsepower-limited on their high speed end.

at the low end, it's how much torque you can apply to the driving wheels/tires before they break loose from the pavement (recall that sliding coefficient of friction is lower than static...? here's where physics "meets the road" in real life...)

and that's a little more about "how cars really work."

see how much trouble horsepower can get you into?


+af

 

[krab]

You seem to think power and torque are unrelated measures of performance. They're not. Power is torque times rotation rate, or force times speed. So if you use Torque, you are forced to take the engine torque, divide by transmission gear ratio, divide by rearend gear ratio, divide by wheel radius, to finally get: Force to the road. With power, you simply take the engine power, divide by road speed and you have Force to the road. You see how much trouble it is dealing with torque instead of power? This is because energy is conserved: force, or torque is not.

 

[whozum]

I was going with power analysis initially, somehow I got lost in making my equations and ended up with some funky stuff. When you told me r(t)/v(t) = constant, then that simplifies things a lot, but it also simplifies my equation to $$a = \frac{k}{m}$$. Dimensional analysis fails with k constant.

 

[plusaf]

torque is not conserved?

You seem to think power and torque are unrelated measures of performance. They're not. Power is torque times rotation rate, or force times speed. So if you use Torque, you are forced to take the engine torque, divide by transmission gear ratio, divide by rearend gear ratio, divide by wheel radius, to finally get: Force to the road. With power, you simply take the engine power, divide by road speed and you have Force to the road. You see how much trouble it is dealing with torque instead of power? This is because energy is conserved: force, or torque is not.

really?
why not, other than powertrain frictional losses... and why doesn't that equally non-conserve power?

 

[whozum]

Can anyone address the question two posts above, also?

 

[krab]

My car has 82 ft-lbs torque at 7000 rpm. Let's trace that through:
At the crank:

torque= 82 ft-lbs, rotation speed= 7000 rpm, power= 109 hp

Gear ratio reduction in first gear = 3.166, so

After the transmission:

torque= 260 ft-lbs, rotation speed = 2211 rpm, power= 109 hp

Gear reduction through differential to wheel axle = 4.313, so

At axle:

torque= 1120 ft-lbs, rotation speed = 512 rpm, power= 109 hp

Notice the torque is not conserved? It is 82, then 260, then 1120. Notice the power is the same throughout? Of course there are frictional losses, so a few % of the power is lost. Maybe there is only 100hp at the rear wheel. But then there is a similar 8% reduction in torque at the rear wheel to 1027 ft-lbs.

 

[shyboy]

That's right, but some crude equations could be helpful
Let's assume that the energy is conserving, so power on the gears is conserving.
Let's imagine gears as ideal cyliders. Then power at any given moment

$$P=F\times v=F\times \omega \times r=T \times \omega$$

here F is a force and T is a torque. Now, if we increase the rotational frequency, we lose torque and vise versa.

 

[whozum]

Ok, can we fit this thread into the data on this one: https://www.physicsforums.com/showthread.php?t=71727 ?

 

[whozum]

I have more detailed numbers if you wish
Times at
60'
330'
1/8mi
1000'
1/4mi
final speed, speed at 1/8mi

 

[krab]

Does 3560 lbs include driver?

I find for avg. of 100hp: 1/4mile time=17.66s, trap speed=78.5mph, top speed=107.6mph, 0-60time=9.9s, 60ft time=3.1s, 1/8mile time=11.4s, speed=64.4mph.

 

[whozum]

Sorry, the actual car weight is 2500 lbs, and I weigh 160lbs, plus an ambient 10-15 lbs for other stuff. These calculations, can you elaborate/

 

[krab]

Sorry, the actual car weight is 2500 lbs, and I weigh 160lbs, plus an ambient 10-15 lbs for other stuff. These calculations, can you elaborate/

That changes things a lot. I'm sorry to have to tell you you are only getting an average of about 80hp to the ground. The formula I use is in a previous post in this thread. Do you know your top speed? If you do, I can make the calculations more accurate. Right now I'm just guessing on air resistance.

 

[whozum]

My trapspeed was 77mph, the highest I've ever gotten it to was 115.

I know my car sucks, but can you show the method used to get to that answer?

 

[krab]

Here is the code, written for a Matlab-like program. To understand it, you need to know that variables like x and v (in feet and feet/sec) are vectors. The formula for v comes from a straight-forward integration of the differential equation in post #28. Then the time vector t is found from a numerical integration of dx/v.

! b is the air resistance coefficient; it is about 0.013
! for cars, about 0.0055 for motorcycles
b=?4
! ?1 is power in hp. p is power in ft.-lbs./sec.
p=?1*550
vinf=(p/b)^(1/3)
! ?2 is weight in pounds of vehicle (don't forget the fat driver)
m=?2/32
! ?3 is initial acceleration g's. This may be limited by traction or
! in any other case is given by first gear ratio.
v1=p/?2/?3
x1=v1^2/64/?3
t1=v1/32/?3
x=[(x1+.5):1319.5]
v=vinf*(1-(1-(v1/vinf)^3)*exp(-3*b*(x-x1)/m))^(1/3)
iv=1/v
ttt=t1+integral(x,iv)
generate tf 0 ,, t1 100
xf=16*?3*tf^2
display `1/4 mile:'
=ttt[#] !seconds
=v[#]*30/44 !mph
display `top speed:'
=vinf*30/44 !mph
vout[1]=88
display `0-60:'
=interp(v,ttt,vout)
display `60 ft time:'
=sqrt(120/?3/32)

 

[whozum]

I meant from a physics perspective, rather than a computational. Sorry, that just looks like a bunch of garbage to me, because I don't know MATLAB syntax.