View Full Version : Do Cars Lose HP an the Wheels w/ Lower Torque Ratings?

Chuck C
03-26-03, 12:33 AM
eg: which would have greater rwhp:

Corvette with 350hp/350lb-ft


Corvette with 350hp/200lb-ft

or does it not matter??

What is the significance of torque in a sedan or sports car?

03-26-03, 08:02 AM
Your question is about RWHP. The numbers are the numbers.
As for what is the significance of torque, the old saying goes; HP sells cars, torque wins races.
Generally, the car with the higher torque wins. As in your example above, the 350HP/350lb-ft car with win the race as long as it can get the torque to the pavement ;)

03-26-03, 01:12 PM
Drivetrain losses aren't a percentage of the engine or anything -- it's a particular number (that is usually never specified by the OEM). For instance, it may take 20 horsepower to turn the transmission and propeller shaft and axle half shafts on a given model (say the Corvette for instance). 20 horsepower to turn all those components.

Now...you put in a 100hp 4-cylinder engine. It takes 20hp to turn the drivetrain, so you have 80hp left to push the car with.

Next...you swap out the 4 banger and put in a 350hp V8. It still takes 20hp to turn the drivetrain, so you have 330hp left to push the car with.

The drivetrain losses are a static number, and are not related to the engine installed. Obviously, all of the above is in very general terms.

03-26-03, 03:12 PM
Some transmissions are worse than others when it comes to power loss to the wheels. Mine is costing me something like 27%. Not every manufacturer is that inefficient. I think Honda does it that way by design, though.

I think I remember reading that the old GM TH400 and TH350 were really solid and efficient. But I haven't driven one in probably 17 years.

03-26-03, 03:42 PM
They're not bad if you can deal with a 3-speed.

03-26-03, 04:51 PM
"One horsepower is an estimate of the power a standard workhorse can exert: 550 ft.lbs/sec. Before applying any formula, we must first identify the units of torque on you engine. Torque may be listed as foot-pounds or as Newton-meters. As you made no specification, I will assume you automobile secifications use foot-pounds.

The power exerted by a rotating object is the torque it exerts multiplied by the speed at which it rotates. In standard English units, this would be foot-pounds multiplied by radians/second. It is a special property of radians that allows this product to be foot-pounds/second: a radian is a distance around an arc divided by the length of the radius (feet per foot).

We start with 1 horsepower. We want to get to (foot-pounds)x(rpm).

1 hp = 550 ft-lbs/sec = 550 (ft-lbs)x(rad/sec)
1 rad/sec = 60 rad/min = 33,000 (ft-lbs)x(rad/min) 1 revolution=2(pi)radians
1 rpm = 2(pi) rad/min
1 hp = 5252 (ft-lbs)(rpm)

As for source of rpms, that varies from moment to moment. The number of rpms will probably be greatest in the lowest gears. When rpms get too great, a vehicle is usually shifted to a higher gear and a lower rpm for the motor. The torque tends to be greater in lower gears, when the car is trying to speed up. Once at cruising speed, all the engine needs to do is keep the car moving.

Look at the greatest rpm listed on the scale of your tachometer. Use this as a reasonable maximum. Multiply this by your engine's greatest torque. This is an estimate of your vehicle's maximum horsepower. Actual value can vary with speed, with how well oiled the car is, even with humidity."

03-27-03, 04:43 PM
Originally posted by YellowSnake
The torque tends to be greater in lower gears, when the car is trying to speed up. Once at cruising speed, all the engine needs to do is keep the car moving.

The torque of what? The engine or at the rear wheels?

The torque at the rear wheels is NECESSARILY greater in the lower gears, because a lower gear offers a speed reduction, and a torque multiplication. A 3.71:1 final drive multiplies the input torque by 3.71 and that is the output torque number. The speed is also reduced 3.71 times. It's an inverse relationship.

03-27-03, 08:10 PM
don't forget about the fact that no two engines are created equal. 2ppl can buy exact same car, dyno it and get 2 very different results

03-27-03, 09:21 PM
Yes, that's certainly true. I think we're talking about the drivetrain losses, though, and on any given vehicle, the engine makes a certain amount of power/torque, regardless of the gear the transmission is in. True, the actual power produced and power lost will be slightly different between two given vehicles.

03-28-03, 07:53 AM
Actually, the "original" question was which car had more RWHP.
And as I already pointed out to him, both examples stated have the same HP so they would be virtually the same. The only difference in the examples was the torque figures. And that won't change the HP.
He then asked what was the significance of torque in various cars. Again, the answer was that the torque diferences in his examples would certainly prove-out that the Corvette with the higher torque would out accelerate the other even though the two have identical HP ratings. Provided that the extra torque could be utilized through adequate traction.
I'm not quite sure how this thread got into parasitic drivetrain losses.

Chuck C
03-28-03, 08:52 AM
Katshot....finally! I knew about drivetrain losses thanks to some other forums, but I was unsure about trq.

03-28-03, 08:48 PM
Originally posted by Katshot
I'm not quite sure how this thread got into parasitic drivetrain losses.

Probably because he was asking about rear wheel power, and you should always include (or at least mention) the parasitic losses when you're talking about power at the wheels. I understand that the original question was the relativity between one car and another in terms of torque numbers, but other related discussion is always good to have as background information.

03-28-03, 09:35 PM
He didn't just ask about one car to another, he "specifically" asked about two "identical" cars with the same engines except for the torque ratings.
I felt Chuck was fairly specific about what he wanted to know. And when you are comparing two identical cars, the parasitic losses would be the same. Or at least for the purposes of this conversation they would be assummed to be.
Either way, I believe from Chuck's last post, we've answered his question.
Sorry it came in such a round-about way Chuck.

03-30-03, 09:07 PM