Race Car Acceleration Rates?

1. Oct 26, 2006

Race Car

Race Car Acceleration Rates???

Hi There,

I have been on the web for two weeks now trying to find an answer to my situation. There are no calculators that I can input the numbers below in to find out what is best. The rule in road racing is he who can accelerate the hardest out of the corner and down the strait will win the race. First my question then I will supply some information. Which will accelerate my race car faster? Being at a lower RPM with a higher torque rating but less HP or being at a higher RPM less torque rating but more HP.

INFORMATION:
Car: Mazda 1.3 Liter Rotary RX-7 5 Speed
Weight with driver: 2425 lbs

Tire Diameter: Choice number one 24.5” Choice number two 25.7

Final gear ratios: 1st 17.868 2nd 10.337 3rd 7.136 4th 5.13 5th 3.909
Rear end ratio: 5.13 (Note 4th gear in transmission is direct drive 1 to 1)

Power Range: @ 5000 RPM 79.6 HP 85 Ft lbs Torque to 7750 RPM 163.8 HP 111 Ft lbs Torque to 9000 RPM 174.1 and 101.6 lbs Torque.
Note Torque on this engine @ 5750 RPM is 100 ft lbs and peeks @7750 RPM 111 ft lbs and drops back to 100 ft lbs at 9100 RPM. So for 3350 RPM the average torque is rated at 105 ft lbs (A very flat torque curve in this area)
.
Engine Dyno Ratings based on 4th gear test:
Test #1 Road speed 120 MPH 8424RPM 168.4HP 105 ft lbs of Torque. Load at rear tires.226g. Tire diameter 24.5”
Test #2 Road speed 120 MPH 8027 RPM 165HP 108 ft lbs of Torque. Load at rear tires .244g. Tire diameter 25.7”

From every thing that I could figure out I believe I have enough information to solve my problem but do not know how. Please help and send the proper formula that relates to this problem. In this way I can work things out at a later date after making changes to my car. Waiting for a positive reply thanks guys.

2. Oct 26, 2006

Danger

I think that Hypatia and Stingray are probably your best bets on this. He's a very savvy car dude, and she builds and races some hot stuff.
The one thing that I can say flat out is that if you aren't referring to some specific class of vehicles, what you have described is not by any means a race car. It's closer to an over-muscled go-cart. My Roadrunner weighs a little less than twice yours, and is putting out 650hp; the new motor that I designed for it will be somewhere in the area of 2,000 hp. As it sits now, it's a joke on the street scene. When I left the Detroit area in '78, which is where Hypatia remained, the general rule was "if you don't have at least 1,000 hp, don't bother showing up".
As an added example, which I consider embarrassing... my car has an A-833 4-speed. First gear was good for 65mph, 85 in second, 115 in 3rd, and 4th is a .73:1 overdrive. Zero to 60mph is about 3.5 seconds. Top end is about 165mph, and quarter times are in the low 12 second range. And again I stress this thing is a joke on the street.

3. Oct 26, 2006

David Laz

I've always wondered what was preferable between torque and power when wanting to accelerate. I sometimes get the idea it doesn't matter because of the way things are geared. Could be waaaaaaay off though.

I'd like to add I'm very interested in the physics and engineering principles involved in race cars. Don't suppose anyone has any good reading material?

4. Oct 26, 2006

Danger

There's some good stuff right here in various areas. Some is in Physics, some in Engineering. Try doing a forums search. In addition, www.howstuffworks.com has a few different articles such as 'how horsepower works'.

5. Oct 26, 2006

wxrocks

pure vs practical physics

There is a big factor here that is being ignored -- the atmosphere/friction.

In a pure sense, it would be easy to calculate the perfect balance of HP and torque based on the length of the track and the radii of the corners. A few minutes of calculation would get you your settings to minimize your time.

However, the friction of your tires will change based on the wear, the temp, etc. The atmosphere will provide alot of resistance -- and the turbulence will make this difficult to get a good handle on its contribution. Then the issue with HP and torque changing depending on engine temp, fuel condition, etc.

If I remember my basics, torque is what you need for acceleration, and HP is more for top speed (in a rudimentary sense). I would try a few setups on the course and with some different tire styles and find what gives the best time.

Good Luck!

6. Oct 26, 2006

NateTG

Roughly speaking
Horsepower=(Torque * RPM)/(magic number)

If the torque is in lb-ft, and you want horsepower then the magic number is 5222.

Assuming that your efficiency is roughly constant, and you can accelerate without slipping, you'll get your best acceleration wherever you get peak horsepower. Effectively, RPM and Torque are interchangable at the engine -- that's what a gearbox does.

Internal combustion engines typically produce peak power at the high end of their RPM range. That is because, at the lower end of things, power is essentially proportional to fuel consumption. As the power gets ramped up, air intake becomes an issue, and, at the higher end there are mechanical 'failures' such as valve float.

5750*100=575,000 -> 110 HP
7750*111=860,250 -> 164 HP
9100*100=910,000 -> 174 HP

so, theoretically if you had three different gears that put you at the same speed for 5750, 7750, and 9100 RPM, you would see the best acceleration for the 9100 RPM gear -- about half-again as much acceleration.

In real life, things aren't nearly so simple. For example, you're going to be dumping better than four times as much power into spinning up the flywheel at 9100 RPM than at 5750 RPM.

Typically, the limiting factor for cornering is the tires' contact on the road which limits your acceleration to somewhere aroud 0.6 G's. I haven't done my math too carefully, but with your numbers that means that for speeds less than roughly 30 mph, it doesn't really matter where in that RPM range you are because you can smoke the tires. You'll have to find your speed by improving your line, and the suspension.

Last edited: Oct 26, 2006
7. Oct 26, 2006

Staff: Mentor

Though the details are complicated, there is an easy answer: higher rpm is better. The simple reason is that torque and rpm are given at the motor , not at the wheel, so they don't take into account the gears (ie, it the attached link shows motor rpm but not car speed, so you can't compare 1st and 2nd gear directly from the chart) . So, upshift and you will lose torque to your wheels, even if the torque at the engine is slightly higher at slightly lower rpm. A demonstration:

Here's a sample chart I found with a quick Google: http://www.allpar.com/eek/hp-vs-torque.html

You can see that peak engine torque is at 5500 rpm, but by ratio-ing the gear ratios, 5500 rpm in 1st gear yields 3300 rpm in 2nd gear. The car provides roughly half the torque to the wheels (~235 vs ~470) in second gear as in first gear at whatever speed that corresponds to.

Last edited: Oct 27, 2006
8. Oct 26, 2006

Staff: Mentor

If you're curious about complicating factors, here's probably the biggest: torque/hp curves are static numbers. They don't take into account the dynamic forces of acceleration. That means that things like the mass of the flywheel, which don't matter in a constant-speed driving situation (unless it affects friction) make a big difference in an accelerating car. That means at higher rpm/lower gear, since the engine is accelerating faster, you lose more horsepower/torque than in higher gears. But since the difference in engine output is so large in the simplified case, these complications won't override that.

9. Oct 26, 2006

Staff: Mentor

For questions like this - and, indeed, for a real car trying to accelerate, you must assume that the wheels are grippy enough that they are not slipping. And if they are slipping, they'll slip based on the acceleration of the car, regardless of what gear you are in (if the gear is capable of producing enough acceleration torque). So your engine power becomes irrelevant if the wheels are slipping. Atmospheric factors (both the drag and engine efficiency) effect the car roughly equally regardless of gear also, so these factors are also irrelevant.

Last edited: Oct 26, 2006
10. Oct 26, 2006

Stingray

The effects of rotational inertia are actually very simple to take into account. All they do is add an effective mass to the car (when considering straightline acceleration). That mass depends on the ratio of the rotation rate between the tires and the object in question. For a flywheel, that means that the mass changes with each gear shift. It actually scales as the square of the overall gear ratio, so it's generally significant only in the first one or two gears. If you want numerical detail, I can explain further.

Taking into account atmospheric drag is also simple for most cases (leaving out high crosswinds, high-speed drifting, or heavily-winged race cars). The power lost to pushing air out of the way scales as the cube of the speed. The power lost due to the rolling resistance in the tires is pretty much directly proportional to speed, although it increases considerably near the tires' structural limits (where one is essentially exciting a mechanical resonance).

Continuing in another post...

Last edited: Oct 26, 2006
11. Oct 26, 2006

Stingray

Now let's calculate an acceleration. I'm going into some messy detail, but assume that you always have sufficient traction. More things can be taken into account, but this will then get complicated very quickly.

Let's say that the static mass of the car is $m$, and that including rotational inertia gives you $m_{\mathrm{eff}}(g)$. g is the gear ratio (a positive integer). Now say that the power available at the wheels is $$P(\omega)[/itex]. Here, $\omega$ is the rotational speed of the engine (i.e. rpm, but possibly in nicer units). Suppose that the actual gear ratios are $\alpha(g)$ (first gear would be largest), the differential ratio is $\beta$, and the tire radius is $r$. In general, various driveline losses will change depending on gear ratio, but I'll ignore that. They also depend weakly on vehicle speed and the temperatures of the air and various oils, etc etc. On top of that, standard dyno measurements have several systematic errors in them, but I'm just going to say that the dyno measurements are precisely $P(\omega)$. Assuming that the clutch and tires aren't slipping significantly, the (rotational) speed of the engine and the (linear) speed of the car are directly related to each other: [tex] v(\omega, g) =\frac{\omega r}{\alpha(g) \beta}$$

This can also be inverted, so $\omega = \alpha \beta v/r$. Now the mechanical power supplied by the engine goes into accelerating the car and overcoming aerodynamic and other losses. Group the latter conditions into $P_{\mathrm{drag}}(v)$, so one ends up with an acceleration

$$m_{\mathrm{eff}}(g) a = \left[ P(\alpha(g) \beta v/r) - P_{\mathrm{drag}}(v) \right]/v$$

This equation can be solved directly to give the time required to accelerate between two given speeds in a single gear:

$$t = \int_{v_{i}}^{v_{f}} \left(\frac{ m_{\mathrm{eff}} v }{ P - P_{\mathrm{drag}} } \right) dv$$

Generalizing this to include shifting is straightforward, but an important point is that torque never enters this equation. You could write it in terms of the torque the engine is producing, but it would be more complicated. The interesting thing to note that is that the acceleration at a given speed is always proportional to the power delivered.

Also, if the drag is negligible, the above equation can be written a little differently:

$$t = m_{\mathrm{eff}} \left( r/\alpha \beta \right)^{2} \int_{\omega_{i}}^{\omega_{f}} d\omega / \tau(\omega)$$

The two limits in the integral are the different rpm's corresponding to different velocities, and $\tau(\omega)$ is the torque "at the engine." This equation is nice because the integral depends only on properties of the engine. The driveline is irrelevant. This is as close as you can get to the handwaving idea that improving "average torque" over the rev range makes a faster car.

Edit: I've edited this a bunch of times to fix and add things.

Last edited: Oct 26, 2006
12. Oct 26, 2006

marcusl

Wait a minute, Danger, he's trying to accelerate out of a corner on a track. Try to take your Roadrunner through a corner at speed! (Or even slow...) I used to drive a big GTO--great fun unless I had to corner or try to stop:surprised

13. Oct 26, 2006

Danger

I readily admit that I'd never keep up to a Porsche on a winding course, but it's really not all that bad. It has that excellent torsion bar front end, with a rear stabilizer bar and Strider shocks set on 'extra firm'. I'd downshift going into a turn, and could start standing on it 2/3 or 3/4 of the way through. One of my plans for the rebuild is to incorporate the Citroen style adaptive hydraulic suspension.

14. Oct 26, 2006

rcgldr

The key factors are peak horsepower and the shape of the torque curve (and the weight and aerodynamic drag). Gearing will compensate for the difference between low and high revving engines.

Assuming your car is geared efficiently, you want to exit a corner at a speed a bit faster than an ideal shift point so you use all of a gear before shifting again, which costs time, assuming you haven't installed an XTRAC no lift sequential shifter with 50ms down to 30ms shift times. Most cars shift slow enough, and forwards acceleration is limited when traction is used up for cornering, that it's often faster to take and exit corners in the next higher gear, like taking a corner in 3rd gear instead of 2nd.

So which tire to use depends on the exit speeds of the corners at a specific track. The difference in this case is pretty small, and in the case of slightly differing torques, you could adjust your line and apex speed which will determine how soon you can apply the throttle for corner exit speed.

You can ignore power for calculating acceleration. You'd be integrating the rear wheel torque versus time from one shift point to the next while in a specific gear. Instead of using calculus, you could setup a spread sheet with the rear wheel torque curve values and then do numerical integration with a large number of rows with a small time step from row to row.

Last edited: Oct 26, 2006
15. Oct 26, 2006

Race Car

To russ_watters

Thank you that link is what I was looking for. Now I can select the proper rear end gear ratio and tire sizes to best match my horse power curve and transmission gears. Thanks for the strait heads up answer. I race in the Pacific North West area of USA. If you’re around in the area free track entrance and dinner for you and a friend contact me if you would like to come out for a day.

Best Regards,
Race Car

16. Oct 28, 2006

rcgldr

A race track is different than a drag race. You're dealing with various speeds at corner exits and top speed achieved before the next braking point. The time it takes to shift also plays a factor, as a car is slowing down, especially if at high speeds, while you are shifting gears. Acceleration just after corner exit is more important than acceleration just before a braking point. So the trade off is do you take a small straigth section in just one gear, giving up a bit of initial acceleration, but eliminating a shift in the middle of the straight, or do you use two gears on the same small straight section, with more initial acceleration, but up to 1/2 second or so of time spent slowing down during a gear shift? Also, when exiting a corner, will the lower rear wheel torque from a taller gear on corner exit allow a driver more "touch" and/or a cleaner line when trying to keep the car accelerating and turning at maximum grip?

The acceleration applet below can help you figure out setups and when to shift:

http://www.cartestsoftware.com

Last edited: Oct 28, 2006