Torque Explained

[05 tC]

-Originally posted by PulsaGTi-R-


With all the dialogue going on about horsepower and torque, it became
obvious there is still a lot of misunderstanding. While there have been
many attempts to sort it all out, I'll try a different approach in hopes of
making it more understandable. Instead of just figures and calculations
(included), I also used a make believe example that uses extremes to better
illustrate the points being made. To perk your interest, consider these.

What would have the higher top speed? An athlete on a 21 speed bike, or the
same bicycle with rider, but powered with a model airplane engine pumping
out 1 (one) lb. ft. of torque at 10,000 rpm, each using gearing appropriate.

A Ford Mustang GT with a big V-8 drag races a 2 liter Honda S2000
sports car. The Honda will lose because it does not have the torque of the
V8 and torque is what wins races. True or False?

What puts out more twisting force, or torque? You tightening the axle nut
on your Ninja ZX9 or the ZX9 motor itself at the torque peak?

The answers will be self evident.
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I am going to use a bicycle as the method of explanation for reasons that
will become clear.

Imagine the distance from the center pivot of the pedal to the center of the
drive sprocket on the bicycle as 12", or 1 foot. It is not exactly this
distance, but for the purposes of clarity, let's make it 12 inches. Stand
on the pedal with all your weight, say 150 lbs. You are generating 150 lb.
ft. of torque, for as long as you stay standing on the pedal. Yes, this is
about 50% more than a Hayabusa can generate at the crankshaft.

Now, try to maintain the 150 lbs of force on the pedals at 4 (four) rpm, or
one revolution every 15 seconds. You can probably do it. Now try to
maintain that 150 lbs. of force on the pedals while moving the pedals or
center sprocket at 20 rpm. That is one revolution every three seconds.
This will get tough. Now try to maintain that force at 60 rpm. You can't
do it. You simply can not maintain that 150 lbs of force at that kind of
rpm.

Horsepower is the result of torque at a given rpm. The formula is torque in
lb. ft. multiplied times rpm and the result divided by 5252. It should be
obvious then that doubling the torque at a given rpm doubles the power (hp). Or
doubling the rpm with the same torque doubles the power (hp).

Go back to when you were just standing on the pedals. You applied force,
but you never did any work. Try the formula. 150 lb. ft. multiplied times
zero, then divided by 5252. The result is zero hp. Move the pedals at 4
rpm, and the result is 4 x 150 divided by 5252, or .114 hp. Double the rpm,
and the horsepower doubles. Double the torque, and horsepower doubles.

We should all see that acceleration is simply the result of a force at the
tire contact patch attempting to drive the vehicle forward. The actual rate
of acceleration is largely dependent on the amount of that force and the
weight of the vehicle. There are other factors, but are less significant
and not necessary to address in this discussion.

We know that we want maximum force at the contact patch, but being the speed
nuts we are, we want as much force at the rear contact patch at all miles
per hours, 30, 60, 120, 180 mph. The more force, the stronger the
acceleration and the higher the top speed.

Let us see what the bicyclist can deliver. He has huge torque, more than
any motorcycle. We want him to go 60 mph. If the rear tire is 24" in
diameter (using this for ease of calculation), the rear tire will turn 840
rpm at 60 mph. Let us assume the bicyclist can pedal at a maximum of 40 rpm
and deliver 150 lbs of force at the same time (quite difficult).

Since the bicyclist is pedaling at 40 rpm and the rear wheel is spinning at
840 rpm, there obviously must be some gearing. In this case, every one
revolution of the pedals translates into 21 revolutions of the rear wheel.
This is some tall gearing! In fact let us assume that the front sprocket is
168 teeth and rear 8 teeth, which gives us the gearing of 1 to 21.

So, the rider puts out 150 lb. ft. of torque at the pedals, but the amount
that is applied to the contact patch is just 7.14 lbs. Why? Because the
gearing is 1 to 21. The rider puts out 150 lb. ft. or torque at the 168
tooth drive sprocket. The gearing is 1 rev of the drive sprocket to 8 of
the rear wheel. The ratio is then 1 divided by 21, or .04762. Multiply the
torque of the rider (engine) 150 lb.ft. times the gear ratio of .04762 and
you end up with 7.14 lb. ft. at the rear wheel. Pretty pitiful and the
reason why bicylists can not ride a bike in the open air at 60 mph. The
resistance to movement at that speed is higher than that, hence that speed
is not attainable. They do not have enough horsepower. More torque at the
same rpm would help the problem. More rpm and lower gearing to better
multiply the existing torque would help the problem. Or some of both.

Let us get back to the questions I posed earlier.

What would have the higher top speed? An athlete on a 21 speed bike, or the
same bicycle with rider, but powered with a model airplane engine pumping
out 1 (one) lb. ft. of torque at 10,000 rpm, each using gearing appropriate.

We already know something about the rider. The rider could deliver 150 lb.
ft. of torque at 40 rpm. That is 1.14 hp. Note that the 7.14 lb. ft at the
rear wheel and 840 rpm is guess what? 1.14 hp. The speed in this scenario
is 60 mph. The model airplane engine makes 1 lb. (16 OUNCES) of torque at
10,000 revs. We obviously have to adjust the gearing to make the engine
spin 10,000 at 60. The gearing is 10,000 divided by 840 wheel rpm, or 11.9
to 1. Quite a difference in multiplication of torque from the .04762 the
rider had to use due to his slow moving legs. The little model airplane
engine that delivered 16 OUNCE FEET of torque would deliver 11.9 lb. ft. of
torque at the rear wheel, significantly more than the rider who delivered
150 TIMES more torque.

A Ford Mustang GT with a big V-8 drag races a 2 liter Honda S2000
sports car. The Honda will lose because it does not have the torque of the
V8 and torque is what wins races. True or False?

False: The Honda S2000 makes far less torque than the Mustang, but with a
9000 rpm redline, it can afford lower gearing that permits high
multiplication of the little torque it has. High enough multiplication to
equal or exceed the torque the Mustang delivers to the rear wheels. Torque
at the rear wheel is what matters. The Honda does better because its lower
gearing MORE than offsets the greater torque of the Mustang. Weight plays a
role here too, but these too take markedly different approaches to
acceleration while attaining similar acceleration results.

What puts out more twisting force, or torque? You tightening the axle nut
on your Ninja ZX9 or the ZX9 motor itself at the torque peak?

You do. But you can't spin that nut with the resistance it offers at 9000
revs! That is the difference.

Torquey engines FEEL stronger, because the strong punch often comes at lower
rpm, giving a sense of power that often is not so evident in a high reving,
but relatively torqueless engine that has high gear multiplication.

Both can deliver the same amount of the tire contact force, but in entirely
different ways. Insofar as vehicle acceleration is concerned, HOW that
force was created is of no consequence. Low gearing multiplied by small
torque and high revs or big torque multiplied by tall gearing and low revs.
It can end up producing the same result at the tire contact patch.
Obviously big torque at high rpm is the best, and is the reason why open
class bikes (big engines, big torque) with multi cylinder engines (more rpm
capability, permitting lower gearing) put down the most thrust at the rear
wheel. Consider a Hayabusa. It can just about put down a HALF a TON (1000
lbs.) of thrust at the rear wheel in first gear at the torque peak and do it
at over 60 mph. A 1500 cc Goldwing engine may have more torque but due to
severe rpm restictions, it can not run the same low gearing at the same mph
as the Hayabusa, hence not mutliplying its torque as much. Acceleration
suffers.

[ExtremeCiv]

my head hurts

[jDmacCordh22a]

I really enjoyed this write up, thank you, you made everything about as easy to understand as it possibly could be. :clap

[CanSi2G]

Im not sure about the s2k vs. Mustang explanation. Allthough the rest wasn't too bad

[LevelSevenCivic]

whoops... i repped you and i didnt see that it was originally written up by "PulsaGTi-R"
He shall be repped accordingly. you get the rep anyways just cuz i can do that

[05 tC]

Originally Posted by LevelSevenCivic

whoops... i repped you and i didnt see that it was originally written up by "PulsaGTi-R"
He shall be repped accordingly. you get the rep anyways just cuz i can do that

His last post was on 11-17-2002 http://www.superhonda.com/forum/sear...0&pp=15&page=1


I don't think he'll mind me getting a rep. Heck, this info would have been lost in the Si/SiR forum 4 years ago if I hadn't brought it here

[RED MX5]

That's a well written article, and it has some analogies that I haven't seen before, so it was a good read.

I'm actually had very good success using a physics based approach to explain torque to other gearheads.

A mass is accelerated when a force is applied. In the case of an automobile, this force is applied at contact patches (the area where the tires contact the road). We call this force torque because it is a twisting force. It is produced by the engine and multiplied by the gears. In lower gears (first and second) we have more acceleration because the lower gears multiply the engine torque more than higher gears.

The Honda S2000 mentioned in the article produces very little engine torque, but the overall gear ratio is around 18:1 in first gear, and torque at the contact patches in first gear exceeds 2,400 ft.lbs. 2,000 ft.lbs. is enough smoke the back tires of a Viper, so the idea that the S2000 somehow lacks torque simply displays a misunderstanding of how torque is multiplied and how that multiplied torque accelerates the car.

It's funny to me that some know-it-all gear heads are willing to throw out fundamental laws of physics (F=MA) because they don't understand how it relates to performance.

[DriftKingBP]

I never really put any thought into it, that makes alot of sense, that article was definately ADHD friendly

[WVU97GS-R]

Originally Posted by RED MX5

That's a well written article, and it has some analogies that I haven't seen before, so it was a good read.

I'm actually had very good success using a physics based approach to explain torque to other gearheads.

A mass is accelerated when a force is applied. In the case of an automobile, this force is applied at contact patches (the area where the tires contact the road). We call this force torque because it is a twisting force. It is produced by the engine and multiplied by the gears. In lower gears (first and second) we have more acceleration because the lower gears multiply the engine torque more than higher gears.

The Honda S2000 mentioned in the article produces very little engine torque, but the overall gear ratio is around 18:1 in first gear, and torque at the contact patches in first gear exceeds 2,400 ft.lbs. 2,000 ft.lbs. is enough smoke the back tires of a Viper, so the idea that the S2000 somehow lacks torque simply displays a misunderstanding of how torque is multiplied and how that multiplied torque accelerates the car.

It's funny to me that some know-it-all gear heads are willing to throw out fundamental laws of physics (F=MA) because they don't understand how it relates to performance.

because they aren't gear heads they are ricers. btw nice r32 in the sig

[RED MX5]

Originally Posted by WVU97GS-R

because they aren't gear heads they are ricers. btw nice r32 in the sig

Actually it is usually drag racers who bring up torque and get it all wrong. I'm not sure exactly what a "ricer" is, but I've been accused of being one myself, and I usually smile and agree.

The R32 is for sale. We've got an R33 and Bigbird now, and three Skylines is really just too darn many.

[WVU97GS-R]

Originally Posted by RED MX5

Actually it is usually drag racers who bring up torque and get it all wrong. I'm not sure exactly what a "ricer" is, but I've been accused of being one myself, and I usually smile and agree.

The R32 is for sale. We've got an R33 and Bigbird now, and three Skylines is really just too darn many.

how much for an r33 gts-t oh if you want to know what a ricer is just look for APC stickers on their cars.

[ryans]

Torque explained.

Hondas don't have any.

that's all you gotta say

[RED MX5]

Originally Posted by WVU97GS-R

how much for an r33 gts-t oh if you want to know what a ricer is just look for APC stickers on their cars.

Only the R32 is for sale, but if you're seriously interested, drop me an e-mail or PM and I can get you pointed in the right direction.

[RED MX5]

Originally Posted by ryans

Torque explained.

Hondas don't have any.

that's all you gotta say

Only if you actually know what that means.