torque vs horsepower?

i was wondering what the differece is between horsepower and torque? ideally i would want a car with tons of horsepower and torque, but alas its not an ideal world and i am hella poor. well anyways, is it better to have tons of horsepower or tons of torque?

There is a connection between torque and power. Torque is a rotating force, work. Power is how fast the work can be done.

P=Mv*w there w is the angularspeed of the crankshaft.

This means that if we have two engines with the same amount of power but in diffrent engine speeds, the engine which delivers the power at the lower speed will have more torque. But if we use a gearbox to lower the speed from the high speed engine both engines will deliver the same torque (if the gearbox has no losses).

Originally posted by captainlag
i was wondering what the differece is between horsepower and torque? ideally i would want a car with tons of horsepower and torque, but alas its not an ideal world and i am hella poor. well anyways, is it better to have tons of horsepower or tons of torque?

Here is how I understand it:

Torque is all the low rpm/low end power, Horsepower is all the high rpm/high end power.

I like torque... not having to really rev my engine that much to move is better than having to rev to 6k to get any power...

I would say lots of torque is better to have than lots of hp... *shrugs*

-Grendel

The best and easiest way I can think of to describe the two is:

Horsepower determines how fast you can go in a gear, torque determines how fast you get to that speed.

Torque is the twisting [rotational] force applied to 'something'. Horsepower is torque multiplied by the RPM of said 'something' divided by 5252.

Horsepower
a measure of forced over time, where 1 hp is equal to 550 lbs lifted one foot in one second. Expressed in terms of torque, 1 hp equals 550 ft/lbs per second. Noticed this measurement is time based.

Torque
the measure of instantaneous rotational energy an engine can produce. Notice this measurement is not time based.

hp= torque x RPM / 5252
torque= hp x 5252/ RPM


With these two basic equations, you can figure out how much hp or torque you have at a given RPM with either measurement. Notice that because of this equation, an engine can NEVER make as much hp before 5252 RPM as it can torque, and the opposite is true above 5252 RPM. Any dyno chart you ever see where the two plotted curves don't cross at 5252 RPM is a bogus dyno chart.

Here's the four most important factors in determining how fast a car can accelerate...

-effective torque at the wheels (@ a given RPM point and gear selection)
-overall vehicle weight
-available traction
-total drag (constituted mainly of aero drag at speeds above 50mph)

So, horsepower is torque over time, accurately summed up as 550 lbs/ft per second. Horsepower is an immensely important figure in vehicle performance, if you want to take only one number and guesstimate the performance of a car this is it. Hp to weight ratio is the single best figure for guesstimating a vehicle's accelerative ability, but the thing to always remember is that horsepower has no direct relationship to vehicle acceleration. That's right, I'm saying that horsepower is very important to estimating vehicle performance, but not a direct or even necessary figure to take into account when speaking of the physics behind vehicle acceleration. And if you read my previous posts, you will notice I never said horsepower wasn't important, I said it wasn't DIRECTLY important.

Now torque, that's a different matter entirely. Engine torque output multiplied by overall gearing ratio is what determines how fast a car can accelerate at any given point, yet the maximum torque output taken by itself is nearly useless in estimating vehicle performance. The way it works is that acceleration is based upon several things, but the driving force behind it is entirely effective torque to the wheels. Effective torque to the wheels is found by multiplying engine torque output (at a specific RPM point) by the overall gear ratio you're in, as shown in this example...

Engine A makes 150 lb/ft of torque @ 5000 RPM, and has the following gear ratios:
1st: 3.230
2nd: 2.105
3rd: 1.428
4th: 1.107
5th: 0.848
Final drive: 4.400

So at 5000 RPM in 1st gear, effective torque to the wheels would be found by multiplying:

150 x 3.230 x 4.400...
Which equals 2131.8 lb/ft of effective torque to the wheels. That's a good bit of thrust for a small car, but 1st gear numbers are usually impressive. In second gear at 5000 RPM, you can see how a taller gear lowers our effective torque output:

150 x 2.105 x 4.400
which equals 1389.3, or only 65% of the available energy for acceleration at the same RPM in 1st gear.

Ever wonder why acceleration feels much stronger in the lower gears than higher ones? This is one of the primary reasons, the rest being due to increased aerodynamic drag (prolly shouldn't even have mentioned that one).

This also shows how horsepower has nothing directly to do with figuring out accelerative ability, however maximum engine RPM will have a large effect on gearing ratios (so you can infer that high RPM motors will usually have great torque multiplication factors through short gearing, and most high horsepower motors are high RPM motors). This is what you are seeing when two cars with near identical horsepower outputs and weights have similar performance, it's not the horsepower having anything directly to do with acceleration, it's the idea that horsepower outputs also allow us to infer something about the gearing the car will have. Say your motor puts out half the torque of a competitor's car but equal maximum horsepower (just a generalization), well your motor is likely spinning to about twice the RPM as the your competitor's engine, and your gearing can potentially be about twice as short as his. Meaning that your meager torque output is seriously helped by great gearing, which is why your car will be competitive with his. This is EXACTLY how cars like most Hondas keep up with equal horsepower motors making much more torque, and is a central concept for this discussion. If you don't understand this concept, you will forever be in the dark about what horsepower and torque mean to vehicle performance.


In the end, it's a strange dichotomy we create when taking engine power output and separating it out into the concepts of horsepower and torque. One is useless for mathematical representations of the physics behind acceleration, yet is the best single number for gauging overall performance. The other is the true driving force behind your car's speed, yet taken alone is about as useless as a power figure can get. The secret to making a fast car is not in making high horsepower OR high torque alone, it's in taking the largest torque output possible and multiplying it with the shortest gearing that works for the intended application. Which can be nicely summed up with horsepower ratings, and continues to be why engines are rated with both these concepts.

As one last thing, I want to say that increasing torque output over a large RPM range continues to be the most effective way to increase a vehicle's accelerative ability. You can increase horsepower output without changing maximum or even average torque levels (by moving the powerband to a higher RPM range), but this will never be as effective in cost or extra speed as simply making more torque. And that's because you have to adjust gearing to take advantage of this newfound RPM potential, which isn't cheap. This is why forced induction setups are so effective at improving acceleration with minimal cash outlays, and why making an NA car go as fast will take MUCH more time, effort and money to achieve.

wow, these post are so predictable. they start with the basic question requireing only a paragraph for an answer. then a few people respond giving about as much info that is needed. then the bg guns come out and the discussion begins. i love it. i would have described it myself but i don't think i can compete with texan's description.
oh, and ivymike, gotta love www.howstuffworks.com

Check this out (http://www.cat.com/products/engines_n_power_systems/spec_sheet_library/industrial/pdf/lehh0586.pdf)
A better link (http://www.cat.com/products/engines_n_power_systems/spec_sheet_library/truck_engines/pdf/leht9326.pdf)
If this engine ran on gas, had a shorter stroke and were a wee bit smaller... I don't know if a manual would be able to handle this.

Texan, you're saying that gears multiply torque? i cant imagine that being the case or people doing gear swaps would have massive increases in torque show up on a dynamometer. Or is this something else you're talking about? just curious as to what you meant.

A chassis dyno requires you to tell it what the effective gear ratio is. Also, this is why chassis dynos are run in the gear nearest to a 1:1 ratio, which is usually 4th gear in imports.

If my equations were not the case, how would you explain acceleration in lower gears being stronger or switching to numerically higher differential gears having such effect on accceleration times? Or beyond that, what concept do you think is at work when you are riding a bicycle and switch to a lower gear (higher numerical) to make it easier to climb hills?

there are two main reasons that one would use gearing in a mechanism:
1) to multiply torque
2) to reduce speed

There are other reasons as well (direction reversal, etc), but those are the two biggies.

for more info: http://www.howstuffworks.com/gear.htm

the only explanation i've ever seen for a numerically higher gear causing more acceleration is it makes it easier for th emotor to turn the wheel.
but if gears multiplied torque, then why dont you see massive increases when you switch from say a 2.73 to a 4.11? even if its kept into the 1:1 gear it should have some effect? also, where doe shte leverage come in? a gear twice as high numerically would have to have twice the diameter, and that wouldnt fit inside a differential housing, and without oubling the leverage how do you double the torque?? not accusing, just asking

it makes it easier for the motor to turn the wheels by multiplying the torque output of the motor.

Example:

Engine speed 3000rpm
Engine brake torque 100N*m
Trans drive ratio 1:1
Diff ratio 2.73

Trans input shaft speed = engine speed = 3000rpm
Trans output shaft speed = input / trans gearing = 3000rpm
Diff input speed = trans output speed = 3000rpm
Diff output speed = trans output speed / diff gearing = 3000/2.73 = 1099 rpm
Trans torque input = 100N*m
Trans torque output = input * trans gearing = 100N*m
Diff input torque = trans output torque = 100 N*m
Diff output torque = diff input * diff gearing = 273 N*m

Notice that the power output of the diff is equal to the power input? If there are no losses due to internal friction, this will be the case. (273 * 1099 ~ 3000 * 100)

Now let's look at what happens with the lower gearing in the diff:
Engine speed 3000rpm
Engine brake torque 100N*m
Trans drive ratio 1:1
Diff ratio 2.73

Trans input shaft speed = engine speed = 3000rpm
Trans output shaft speed = input / trans gearing = 3000rpm
Diff input speed = trans output speed = 3000rpm
Diff output speed = trans output speed / diff gearing = 3000/4.11 = 730 rpm
Trans torque input = 100N*m
Trans torque output = input * trans gearing = 100N*m
Diff input torque = trans output torque = 100 N*m
Diff output torque = diff input * diff gearing = 411 N*m

Notice that the power is still conserved. (730*411 ~ 100*3000)

my question is, how does the diff multiply torque, and why doesnt swtiching from a lower geared diff to a higher one, a 2.73 to 4.56 we'll say, which is nearly double the ratio, produce nearly double the torque on a dynamometer? if the reason for the gears making your car faster is more torque, then it would have to show up. also where does the torque increase come from? i've never heard of any way of multiplying a set torque amount like that without again, nearly doubling the leverage, and theres no possible way to make a 4.56 ring gear nearly twice the size of a 2.73, so where does the leverage come from?

let's quickly review how a lever works:

A force is applied at one point on the lever, which is at a given distance from the fulcrum (x1). A force is applied at another point on the lever, which is at another distance from the fulcrum (x2). The ratio of the forces is inversely proportional to the ratio of the distances (f1/f2 = x2/x1, or f1x1 = f2x2). It doesn't matter what x1 and x2 are individually, it is only the ratio that matters. Thus if x1 = 6 and x2 = 3, the ratio is 2. If you wanted a ratio of 3, you could shrink both sides and still get there: x1 = 3 and x2 = 1 gives ratio = 3.

If you draw a gear pair as a couple of circles touching each other, with radii r1 and r2, and the tangential contact force F, then the torque on each gear is the force times the radius (t1 = r1f and t2 = r2f). Since the force acting on the gears is the same, we know that the ratio of torques is proportional to the ratio of the radii (t1/t2 = r1/r2). Thus if a large radius gear acts upon a small radius gear, torque output of the smaller gear is less than torque input to the larger gear. If the smaller gear is the input, then the output will be multiplied by the ratio. You don't need to double the size of the larger gear to double the ratio - you could halve the size of the smaller gear and accomplish the same thing. In a differential, the absolute sizes of the gears would depend on the loads that they needed to carry, but the number of teeth on each would be picked to give you the right gear ratio. Since the tooth pitch has to match, the diameters of the gears will be in the same ratio.

The torque you're measuring on the dynamometer will depend on the specifics of how you're measuring it and what you're plotting against. If you plot output torque vs engine rpm, it will remain constant. If you plot wheel torque versus wheel speed, the high-ratio graph will shrink horizontally and grow vertically compared to the low-ratio graph. On a standard acceleration dyno, if you change the gears in the diff, you'll need to change the drive ratio that you input into the software to get the right engine torque curve out. Do you have a specific output plot that you're thinking about?

Out of curiousity, did you have an alternative explanation for how changing gearing affects vehicle acceleration? I know you said earlier that it "makes it easier for the engine to turn the wheels." Did you have any thoughts on how that might happen physically, or what that means in regard to torque?

the explanation i had was, comparing a 2:! ratio to a 4:1, the 2:1's rear wheel only gets 2 engine RPM, or every 1/2 turn of the rear wheel recieves 1 engine RPM, meaning if the engine puts out 200 ft lbs, the wheel recieves 200 ft lbs every 1/2 RPM, whereas in the 4:1 ratio, the wheel recieves 200 ft lbs every 1/4 RPM. and i dont quite follow your post about leverage, the only way i'm aware of, to increase leverage, is to increase distance from the center, or to move the fulcrum point closer to the center.
as to the chart, no, i've not got a dyno chart in mind, just from looking at ones i've seen, i've never seen one put out the kind of torque ratings that they should be, several were from very tall geared drag cars, as to the dyno settings, i dont quite follow how the gearing being inputted would be the reason, unless they didnt input the actual gearing the car has.

okay, well see if you can answer this:

if you apply 200 lbf to one end of a lever, and get 500 lbf from the other end of the lever, how long is the lever?

It sounds like you could use a little more info about how levers work.

First, reading something like this might help: http://cosmos.colorado.edu/~urquhart/play/lever.html Remember to read the "teacher's guide."
or maybe this: http://www.atlans.org/elements/machine/pg02.html
brief, but helpful: http://www.ceeo.tufts.edu/curriculum/classroom/lever.html

gears in brief: http://www.ceeo.tufts.edu/curriculum/classroom/gears.html

gears as they apply to this question:

http://www.btc-bci.com/~billben/lincoln.htm

Because the maximum torque developed by the engine is puny compared to the weight of the car, mechanical advantage must be obtained through leverage to move the car and this is accomplished with gears. By multiplying the torque available at the flywheel through speed reducing gears, 300 pounds feet of torque can become as much as 3000 (or more) pounds feet at the driving wheels.


http://www.miata.net/sport/Physics/03-Basic-Calcs.html

A transmission is nothing but a set of circular, rotating levers, and the gear ratio is the leverage, multiplying the torque of the engine. So, at the output of the transmission, we have of torque. The differential is a further lever-multiplier, in the case of the Corvette by a factor of 3.07, yielding 3100 foot pounds at the center of the rear wheels (this is a lot of torque!).

Originally posted by ivymike1031
okay, well see if you can answer this:

if you apply 200 lbf to one end of a lever, and get 500 lbf from the other end of the lever, how long is the lever?

well, i'd guess 2.5' long, but i'm not positive if the length of the lever after the fulcrum point is taken into acocunt, or how exactly to overcome that, i have a guess but thats for another discussion.

but, i've cleared up the gearing issue, had a discussion with a friend last night, was forgetting the fact torque is work over time and you need to add each revolution of the gear into the one of the wheel to get your torque numbers, i was believing the gear was sending that much torque to the wheel each revolution of its own, however many thousands of lb-ft. that may be. and thanks for the links, i'll have to read them when i get home.

It's actually not possible to answer the question as I posed it. The only thing you know about the lever from what I told you is that one side is 2.5 times as long as the other side. (each side is measured from the fulcrum to where the force is applied)

Possible solutions:

side1 is 2.5 inches long, side2 is 1 inch long (total = 3.5 inches)
side1 is 25 feet long, side2 is 10 feet long

etc...

torque is work over time
No.

Horsepower is basically power = energy x time, torque is energy = force x distance.

doh, not time, distance traveled, thanks texan.

this is pretty much why 4 cylinder cars need to be at about 3000 rpm to take off comfortably, and in a 350 big block you can take off at about 1200... just about idle:)