TurboTommy

Danno,
I don't think you're giving 100 % the true picture here.
Like you said, it's definitely area under the curve that determines overall acceleration. But, look closely at your gragh of car #2; you didn't extend the shaded area under the curve all the way to the power peak like you did with car #1's power peak. If you had, that area would have been much larger.
Also, the graph of car #2 is totally unrealistic, whereas car #1 is realistic. Except for a peaky two stroke motocross bike there is no engine that has a narrow torque curve like that car #2.
When it comes to gearing, in reality, higher reving engines will always have shorter gearing, therefore multiplying the lower engine torque number.

What I'm trying to say is that large differences in torque with relatively same hp #s will not yield those big differences in real life acceleration.

In fact, I'm going to say that horsepower rules,providing that the torque curve has at least moderate width (modern four stroke engines). Properly matched gearing will always make up for short comings in engine torque.

Funny you should mention the race between the mustang and the integra type r. That's exactly the race I had ( me in the type r - 195hp; against an early 90's mustang- 225hp)
He did get a jump on me right off the line, but by the time I was in 2nd gear, I was even. Then I shifted into 3rd (he jumped ahead a bit), and then I started to slowly pull away as I was approaching redline.

I have 130 ft/lb of torque; he has almost 300. Horsepower is what counts. He would have beaten me if he weighed less.
Please, no arguements that his engine was tired(overwhelming difference in torque, remember!)

toddk911

Then you have the old saying of "horsepower sells cars, torque wins races"

Danno

"In fact, I'm going to say that horsepower rules, "

In an abstract sense, perhaps. Given the same RPM-range of operation and same gearing, the car with the higher HP will usually have higher TQ as well. In concrete real-world terms, when trying to calculate speed at any given time or any given distance from a standing start, you'll find that you have to convert HP to TQ in order to come up with those results. Remember those basic physics problems in high-school where they ask if you were to drop something from the top of a building: at X amount of time, how far as it fallen? Or how fast is it going at that point? Same thing with a car, you have to use TQ to make those calculations.

"providing that the torque curve has at least moderate width (modern four stroke engines). Properly matched gearing will always make up for short comings in engine torque."

Yup, it's the torque on the ground that counts. In the above example, Car-2 would always be in 1 gear lower than Car-1 so the differences in torque won't be as great. I was trying to equalize as many various as possible, RPM-range, gearing, weight, etc. In the real world, what really counts is weight and linear thrust on the contact patch. F=ma is all there is and you calculate that Force from torque at the crank through the gearing and tire-diameter. That's how you can make up for low torque with RPMs, because you can multiply it through the gearing... But... if you have twice the torque with the same gearing.. look out!

gmonsen

I was reluctant to post anything again on thisthread. Don't know how much anybody really cares anyway. But, as I said before, HP is a second order derivative. The torque curve should be all that you should care about. Period. Somebody compared a Honda with a Mustang earlier. If the Mustang had the same power as the Honda, but much more torque, than with weight the same, the Mustang smokes the Honda. The Hinda could have a lot more power, say, 250 hp vs the 225 of the Mustang and if the Mustang has 300 foot pounds of torque to the Honda's 130 foot pounds, the Mustang smokes the Honda.

Ah. Anyway. Torque is what matters and I only bother saying it again, because so many guys build cars with peaky power instead of broad power, thinking there is big value in the high ultimate hp number. This is compounded with a turbo, where lag is followed by a rush of power. The reason Porsche used the smaller turbos on these cars was to let them spool up very early and have a more linear, broad torque curve. Porsche was the innovator in turbocharging cars (along with BMW's 2002 Turbo) and clearly knows hoiw to make more power through a bigger turbo. Even in those days, they could have used a bigger turbo, but chose not to in order to maintain the relationship to the 911 Turbos, bt also to provide a better driving experience through a smoother power curve. Today, if you want to upgrade the turbo, you should look for one that is not overly big and spools up fast. The GT30 ball bearing turbos are probably a great choice for that, combined with a modest displacement increase. Quick spooling and more displacement equalsd much more torque and a much broader torque curve. All you miss is the kick in the pants from lag-followed-by-big-boost feeling from a bigger laggy turbo and same displacement.

Gordon

TurboTommy

gmonsen,
I guess you didn't read the part about the integra staying at least even with the stang.
You need to have a better understanding of power/torque/gearing, and maybe read the other posts more carefully, before you post with such conviction, because mostly what you said is not correct.
Like Danno said; it's the TORQUE TO THE GROUND that counts. Maybe you can figure it out from there.

Chas

Danno....you have no idea what you are talking about

hahah PSYCH...amazing once again. Perfect graphs, perfect explanation. I am printing your reply out and showing it to all my friends that don't understand why their high-hp car still can't beat my 951




Hahaha no...there are plenty...ever hear of a civic? Buddy of mine has the liscense plate TORK LES...("torque-less"). I just call him Torkles ("torkals")

...at least he can admit it

gmonsen

See? This is where it gets tiring... tommyturbo... Apparently you missed my comment that "assuming the weight is the same the Mustang smokes the Honda" and that's what it would do. All that horsepower rules is the forum bragging rights lists. Not the streets and certainly not the tracks. If you don't know it now, you (hopefull) will know it later.

;-)

Gordon

Perry 951

Actually... I have had this little device called a Road Dyno for about the past 3 years. It counts/uses ignition pulses as samples for the accelerometer plotting.

Here's the link if you want to read up on it.
http://www.charm.net/~mchaney/homedyno/dynokit.htm

Tomas L

I don't think there is much point in posting in this thread since it seems that nobody really bothers to try to understand. Therefore I'll keep this short.

TurboTommy's (and perhaps Laust's) posts is BY FAR the most correct so far.

Acceleration is determined by HP not torque! PERIOD.
From a therotical point of view. Speed is energy and in order to increase that energy, you will neeed more energy. Power x time = energy. The more HP you have the shorter time is needed to increase the speed a certain amount.
This does not mean peak HP, but that average HP in the rpm range that your gearbox imposes on you.
Your shift point should be calculated so that HP is the same at each rpm, before and after the shift.

Tomas

TurboTommy

"it seems that nobody really bothers to try to understand'

Thanks Tomas; you took the thoughts right out of my head. Some people just ramble on, jump on the band wagon and just repeat what others have said without fully understanding what they're putting in print.

You try to set things straight and realistic; they continue their ramble even with the corrected facts right in front of them!

Danno

"Acceleration is determined by HP not torque! PERIOD.
From a therotical point of view. Speed is energy and in order to increase that energy, you will neeed more energy. Power x time = energy. The more HP you have the shorter time is needed to increase the speed a certain amount."

Then take a power figure, like xyz HP, and calculate distance and speed at regular time-slices, say every 1-second away from a standing-stop, T=0 and you'll notice something every interesting.

Tomas L

Ok.
V=speed in m/s
S=distance traveled in meters
m=car weight with driver i kilogram
P=power at the wheels in watt
t=elapsed time in seconds

V=SQRT(t)*SQRT(P/m)
S=(t^1.5)/1.5*SQRT(P/m)

If we use values for a lightly modified 951.
m=1500 kg
P=200 000 W =>272 rwhp

t V S
0 0 0
1 11.5 7.7
2 16.3 22
3 20 40
4 23.1 62
5 25.8 86
6 28.3 113
7 30.6 143
8 32.7 174
9 34.6 208
10 36.5 243
11 38.3 281
12 40 320
13 41.6 361
14 43.2 403

This car will run 14s@96.7mph in the 1/4 mile if there was no aerodynamic resistance. AFAIK this is reasonably correct.

I really don't see what's interesting about this?

Tomas

Brian Morris

This argument comes up on every car board - but what's most frustrating about this board is that one of the "gurus" - Danno - is totally off base on this subject and continues to post misleading information multiple times a year. Others then jump in and say "Yeah, torque rules" and then the argument dies.

Instead of repeating all of the correct arguments why HP is what you need to accelerate a car the fastest, people should just do a quick google search. One of the links previously posted, www.vettenet.org/torquehp.html , clearly explains why a car with more HP wins the drag race. It also explains why torque feels fast, and confuses people to think it is fast.

A car's acceleration rate *in a single gear* matches its torque curve.

At a *given vehicle speed* a car will accelerate the fastest when maximizing HP.

Some people are confused by the first but should be focusing on the second. By maximizing HP in the engine, gearing and shifting you maximize the torque to the ground, which maximizes acceleration.

-- Brian Morris
89 951

Danno

"I really don't see what's interesting about this? "

Too many simplistic assumptions. You cannot have a car that puts out 272rwhp all the time across the entire RPM-range. If you use a peak-HP figure of... say 262rwhp at 5500rpm, then you need a CVT to make those predictions more accurate.

also take this: P=200 000 W =>272 rwhp

and differentiate to remove time so that you can accurately calculate distance and time at each time-slice and what unit are we working with? The problem with using HP in calculations is you need a time-interval and the Heisenberg principal comes into play here. By the time you've used a stop-watch to record acceleration between two time-intervals (like timing-lights at the end of a strip), you've already blurred the actual starting and ending values. You really don't know what the exact HP at any given point in time, because HP has time built into it. Thus the need for derivatives and integration. We can actually set up a differential-equation that models both HP & TQ, but I doubt anyone would understand it.

"but what's most frustrating about this board is that one of the "gurus" - Danno - is totally off base on this subject and continues to post misleading information multiple times a year. "

What's frustrating? Nothing's frustrating at all except that people have no understanding of and can't apply basic calculus, that's all. Even just mastering basic physics of v=at and d=1/2at^2 would make understanding HP & TQ much easier. Hint: 2nd order derivative...

"1. A car's acceleration rate *in a single gear* matches its torque curve."

Correct. Plug into v=at and d=1/2at^2 and it's simple. But this assumes constant acceleration. Thus the need to use t=0.1s, t=0.2s, t=0.3s using unique values for acceleration each time in order to be really accurate. Thus the need for calculus where you can integrate with respect to any time T and get the answer you want. Using the torque-curve, you can come up with the real acceleration rate in any gear at any time. Integrate with respect to time and you get the car's velocity. Integrate with time again and you get the amount of ground covered.

"2. At a *given vehicle speed* a car will accelerate the fastest when maximizing HP."

Give us the math behind this. Just calculate instanenous Acceleration at any given point in time T. Or use RPM if you wish. Calculate instantaneous acceleration at both the torque-peak and HP-peak. You'll see that in order to calculate acceleration using HP at any given point, you'll need to remove one time-variable (differentiate) and convert to torque. Obviously this 2nd statement contradicts the 1st one.

The Corvette link provided above summarizes at the bottom what TurboTommy and Gmonsen said earlier:

"The Only Thing You Really Need to Know
Repeat after me. "It is better to make torque at high rpm than at low rpm, because you can take advantage of *gearing*." :-)

Another example of this is measuring the car's speed & acceleration between 5000-6000rpm in 3rd gear vs. 5th gear. The engine puts out exactly the same HP & TQ at the crank between these RPM points in both gears. Additionally, HP on the contact patch at the ground remains the same in both gears. Yes, 5500rpm in 3rd gear puts out exactly the same RWHP as 5500rpm in 5th gear. Why then does the car accelerate so much faster in 3rd gear then?

Brush up on your calculus and check out these links:
http://www.stanford.edu/~voloshin/lhowwhy.html (real charts and formulaes)
http://www.upscale.utoronto.ca/PVB/H...ginePower.html
http://www.heumann.com/m5/hp_torque.html
http://www.lhup.edu/~dsimanek/chapman.htm

I also recommend the following reading:
"Fundamentals of Physics, 3rd edition" Halliday and Resnick
"Physics" Wolfson and Pasachoff,
"Internal Combustion Engine Fundamentals", Heywood
"Automotive Handbook, 2nd edition" Robert Bosch GmbH
SAE #2000-01-3544 : Deriving Wheel Hp and Torque From Accelerometer Data

Tomas L

First of all.
I've made an error in the equations in my last post. The factor SQRT(P/m) should be SQRT(2*P/m). This makes the results incorrect. With the same input data you reach 121 mph and 397 meters after 11 seconds. The difference compared to actual 1/4 mile is that this doesn't account for aerodynamic drag, traction and assumes a constant amount of power.
Sorry about that.

Brians statements are perfectly true.
A quick try at this gives the following equation.
A=Acceleration in m/s^2
P=Power in Watt
m=total mass in kilogram
v=speed in m/s

A=P/(m*v)
Look at the equation above, it explains it.

The rest of Danno's post is confusing to express it kind.
Reading all those recommended books and still don't understand the relationship between power and torque. Sorry Danno, you are a skilled 951 tuner, but this is obviously not your area of expertise. Mixing dynamics with Heisenberg principle of uncertainty, which is applicable for quantum mechanics, kind of gives it away.

Tomas