**2**

# Torque vs. horse power

### #1

Posted 08 April 2008 - 09:14 PM

### #2

Posted 08 April 2008 - 09:15 PM

__This is a cut/past of the info, the graphs and images do no appear on this site. This info is best viewed on the origional site linked above.__

Have any of you ever watched Bill O’Rielly on the Fox news channel? You know, the guy with the huge ego, stating his opinion as fact to "save the country" from its own evils. Bill has a section of his daily show devoted ridiculous happenings in the world of politics.

If I were to do my own "Most Ridiculous" item in the world of racing it would be based on the following statement. "Horsepower sells motor cars, but torque wins motor races." This couldn’t be further from the truth.

Like it or not, everything that goes on around us is governed by the laws of physics, and these laws are non-negotiable. The good news is that we don’t have to be Einstein to apply the basic laws of physics to racing. The fact that too few do is the reason that such ridiculous statements are common in racing.

Let’s start with a few definitions. Webster’s dictionary describes torque as "A turning or twisting force." Note that the definition does not imply motion. As applied to an engine, it is simply a measure of the twisting force at the crank/eccentric shaft. Torque is normally rated in Lbs.-Ft. Since pounds feet doesn’t exactly roll off the tongue, most of us refer to it as foot pounds.

Notice that there are two terms. Force (In lbs.) and distance (In ft.). At first it may seem strange to describe a "Turning or twisting force." in terms of distance, but a more detailed description makes it clear. If I were to put a shaft in a bench vice, attach a 1-ft. long lever to the end, perpendicular to the shaft, and then hang a one pound weight off the end of the lever, I would be applying one ft.-lb. to that shaft. Notice that the shaft is not rotating even though a torque is applied to the shaft.

If I were to replace the one-foot lever with a 100-foot lever, I would now be applying 100 ft.-lbs. to the shaft with the same one lb. weight. As you can see, the amount of twisting force on the shaft will vary depending on the length of the arm, and that requires that we specify a measure of distance to properly describe the force seen at the center of the shaft.

Let’s say for instance that I pull the shaft from the vice, and ask you to hold it in your hand. If I do this with a one pound weight hanging from the end of the one foot lever, I will be applying a force of one ft.-lb. to your hand, and you will have no problem holding on to it. If I replace the lever with one that is 10 ft. long, with the same 1lb. weight on the end, (For all these scenarios, we assume that the lever itself is weightless.) You will now have a force of 10 ft.-lbs. applied to your hand, and it will be much harder to keep the shaft from rotating, even though you are still only resisting the one pound weight.

This would have exactly the same effect as setting a torque wrench to 10 ft.-lbs., attaching it to the end of the shaft, and applying force until the wrench clicks. Ten ft.-lbs. is ten ft.-lbs. whether it is applied with a one-foot lever and a ten-pound weight, or a ten-foot lever and a one-pound weight. Torque is equal to the weight, or force, times the length of the lever. It’s that simple.

If a particular engine has a peak torque rating of 200 ft.-lbs., that force is equivalent to attaching a one foot lever to the shaft, and hanging a 200 lb. Weight from the end of it. Or…any other combination of weight and lever length which has the product 200.

Notice that I can take you from easily holding on to the one pound weight, to not even having a chance of holding it just by changing the length of the lever. (Like a one-lb. weight, and a 100-ft. lever.) Of course you say, that’s just leverage! Well…you’re right! Keep that in mind, because it is that leverage that makes all the difference, and a gear is in fact just a clever way to apply leverage between two or more rotating devices.

Let’s say that the shaft used in our example is the input shaft of the transmission from a 1993 RX-7 with the following ratios.

--------------------------------------------------------------------------------

1st 3.483 to 1

2nd 2.015 to 1

3rd 1.391 to 1

4th 1.0 to 1

5th .719 to 1

--------------------------------------------------------------------------------

If the transmission is in 4th gear, one complete revolution of the input shaft will result in one complete revolution of the output shaft, just as if there were a solid shaft running all the way through. If we attach a 1-ft. lever to the input shaft with a 10-lb. weight on the end, the torque at the input shaft will equal 10 ft.-lbs. as we have already determined. Since we have a 1 to 1 ratio from input to output, we will also have 10 ft.-lbs. at the output shaft.

If we were to keep the same weight and lever on the input shaft, but switch the transmission to third gear, we would still have 10 ft.-lbs. at the input shaft, but we would now have 13.91 ft.-lbs. at the output shaft. This value is the product of the input torque and the gear ratio. (10-ft.-lbs. times 1.391 gear ratio equals 13.91 ft.-lbs.) If we were to switch the transmission into 1st gear, the result would be 34.83 ft.-lbs. at the output shaft.

As you can see, a gearbox gives us a simple way to vary the torque through leverage, and it is equivalent to changing the length of the lever. Thanks to gears, we can have any amount of torque that we want! In fact, a bone stock 12A making only 100 ft.-lbs. of torque could be geared to pull an 18-wheeler up a steep hill, as long as we are not in any big hurry to get the job done.

Let’s say that it takes 10,000 lbs of force to pull a heavy weight up a hill. No problem! We could even do it with our stock 1980-GS in 4th gear if we are willing to build a custom ring and pinion gear with a ratio of 100 to 1. (100-ft.-lbs. times transmission gear ratio of 1:1 times ring and pinion gear ratio of 100:1 equals 10,000 ft.-lbs.)

If we are using a tire with a diameter of 24", the distance from the axle center to the ground is exactly one foot, and so the force is equal to 10,000 lbs. Remember, the torque is equal to the lever length times the force. If we re-write that formula to solve for force, force is equal to torque divided by the lever length, and so that 10,000 ft.-lbs. at the rear axle results in 10,000 lbs. of force at the tire contact patch.

With this same information, we can also calculate the acceleration rate of the vehicle, but first we need to consider Newton’s second law of motion, which states that "Acceleration is proportional to force." and "Acceleration is inversely proportional to Mass." This law is normally stated more simply as "Force equals mass times acceleration." or F=MA. If we rewrite this to solve for acceleration, we get A=F/M. To find the rate of acceleration for a vehicle, we simply divide the force (In lbs. at the tire contact patch.) by the mass (Total weight of the vehicle in lbs.)

Let’s calculate the acceleration rate of a 1st. gen. RX-7. The engine has a torque peak of 100-ft.-lbs. In fourth gear, the ratio is 1 to 1, and so the torque at the output shaft is also 100-ft.-lbs. The ring and pinion ratio is 3.909 to 1, and so the torque at the rear axle will be (100 times 3.909) 390.9-ft.-lbs. The tire diameter is 24 inches, and so the lever length (Distance from the center of the axle to the ground.) is 12 inches, or one foot. The resulting force at the tire contact patch will be (390.9-ft.-lbs. of torque divided by lever length of one foot.) 390.9-lbs. of force. The total vehicle weight with a driver is 2600 lbs., and so the acceleration rate in G’s (The force of gravity.) will be force (390.9) divided by mass (2600) which equals .15 G’s.

If we do the same calculations for first gear acceleration, we find that the force at the contact patch is 1,436 lbs., and the acceleration rate is .55 G’s. It’s clear that we have used the gears for leverage, with the result being a greater rate of acceleration in 1st gear. Of course you knew that already, but now you know why.

By now it should be clear that the acceleration rate of a vehicle is determined by the weight, and the force at the contact patch, which is the result of the torque output of the engine, and all the levers/gears between it and the ground.

You’re probably thinking that we have just determined the acceleration rate of the vehicle, and even changed it with gearing, with no mention of horsepower. So torque really is the determining factor right? Wrong! We haven’t considered speed.

We can gain acceleration by changing the gear ratios, but we can’t go very fast in first gear, so what’s the point? We have effectively changed the amount of torque available to accelerate the vehicle, but our top speed is limited to about 25 mph.

Confused yet?

Read on.

--------------------------------------------------------------------------------

OK, so we all know what torque is, now let’s get to horsepower.

Referring once again to Webster’s dictionary, horsepower is defined as "A foot/pound/second unit of power, equivalent to 550 foot/pounds per second."

Put more simply, horsepower is a measure of work done over time, or the rate at which work is done.

So now we have another term to confuse things. As if force and distance weren’t enough, we now have time involved, and the shaft must actually be spinning. Why… Well, if you are just standing there holding on to a shaft with a lever and a weight, you are doing no work. If you stand there long enough, you will feel like you are working, but in fact you are doing no such thing. Don’t believe me? Clamp the shaft back in the vice, and you can leave it there indefinitely without having to feed it, add gas, and any other means of supplying it with energy.

Torque, all by itself does nothing useful. In fact, the definition of torque does not even require that the shaft be moving. I am sure that all of you want your car to do something useful, like take you to the movies, or get you around the track before the other guy. In other words, you need your car to do some work, and you want it to do that work in the least amount of time possible.

If I give you a wagon full of cement blocks, and ask you to pull it one mile up a hill, you would agree that I am asking you to do work. If I ask you and your buddy to each pull a wagon full of cement blocks up the hill, you will both be doing the same amount of work. But…If it takes you an hour, and your buddy does it in 30 minutes we have a very different situation. You have both done the same amount of work, but your buddy, by completing the task in half the time, has proven that he can develop twice the horsepower that you can. You both traveled the same distance, and exerted the same amount of force, but the third term in the definition of horsepower, time, was different.

For those of you who are sticklers for details, the force required to pull the wagon up the hill at a steady speed is equal to the weight of the wagon, times the sine of the angle of the hill. If the wagon weighed 100-lbs., and the hill was at a 45-degree angle, the required force would be (Sine 45 Times 100 lbs.) equals 70.7 lbs.

If we were interested in moving the wagon by driving the wheels rather than pulling it by the handle, we could convert the force to torque by dividing the required force by the radius of the driven wheels. Let’s say that we have 6" diameter wheels. That would give a lever length (Distance from the center of the axle to the ground.) of 3 inches, or .25 feet. The required torque would then be (70.7 lbs. times .25 ft.) which equals 17.675 ft.-lbs.

James Watt, who spent the majority of his life perfecting the steam engine, created the term horsepower. He was looking for a way to measure the rate of work done by a horse so that he could make valid comparisons between horses which did most of the work in those days, and his steam engines which he hoped would do most of the work in the future.

Watt found that on average, a horse could lift 330-lbs of coal 100-ft in one minute. He then stated that the power available from one horse was equal to (330-lbs. times 100-ft.) or 33,000-lbs./ft./min. If you divide that by 60 to convert to lbs./ft./sec. you get 550-lbs./ft./sec. Watt called this one horsepower, which leaves most of us wondering why he didn’t call it one watt. I don’t have the answer to that, but I do know that 746.6 watts equals one horsepower. If you ever see an engine rated in watts, (This is still popular in some countries.) you can divide by 746.6 to determine the horsepower. Or, you can tell you pals that your bone stock 3rd. gen. RX-7 puts out One hundred ninety thousand, three hundred and eighty three watts.

--------------------------------------------------------------------------------

--------------------------------------------------------------------------------

So if one horsepower is equal to 33,000 lbs.-ft. per minute, we can rearrange that to say that horsepower equals torque times rpm, divided by 5252. How do we get there?

In the above formula, force and distance are stated in ft.-lbs., and time is stated in RPM, so we need to convert our terms. First we need to express that 33,000 lbs. of force as 33,000 ft-lbs. As you now know, that is equivalent to a 33,000-lb. weight hanging from a 1-ft. lever. Then we need to express the one-foot per minute as RPM.

The circumference of a circle is defined as the diameter of the circle times Pi, which is 3.14159. We have a one-foot lever, so if we were to spin the shaft, the outer edge of the lever would scribe a 2-foot diameter circle. The circumference of a 2 foot circle is (3.14159 times 2) 6.282 feet. If we divide 1 foot by the distance traveled in a complete revolution (1 divided by 6.282) we get .159 revolutions per minute, which is equal to one foot per minute.

So now we have: One horsepower equals 33,000 lbs.-ft. of torque per .159 RPM.

That’s still kind of ugly dealing with just a fraction of an rpm, so we divide both terms by .159 and we get: one horsepower equals 5252 lbs.-ft. of torque per 1 rpm.

This can be rewritten a few different ways that are valuable to us.

--------------------------------------------------------------------------------

Horsepower equals torque times rpm divided by 5252. Horsepower = (Torque X RPM) / 5252

Torque equals horsepower times 5252 divided by rpm. Torque = (Horsepower X 5252) / RPM

RPM equals horsepower times 5252 divided by torque. RPM = (Horsepower X 5252) / Torque

--------------------------------------------------------------------------------

If you know any two of the terms, you can calculate the third. You might also notice that torque and horsepower will always be equal at 5,252 rpm, horsepower will be greater than torque above 5252 RPM, and torque will be greater than horsepower below 5252 RPM. ALWAYS…NO EXCEPTION! Just look at any dyno sheet, and you will see what I mean. If you see a dyno sheet where this is not true, you can be sure that someone fudged the numbers to help sell a product.

Back to the issue at hand, I’m sure that the coal tugging horse and a wagon full of cement blocks probably doesn’t seem all that relevant to your racecar. So let’s look at things another way.

The definition of horsepower includes three terms. Force, distance, AND time, where torque is simply a force applied over a distance. In the case of Watt’s experiment, the force was exerted by the weight of the coal, which was being lifted from the mine. In a car, we are interested in acceleration, not the ability to lift an object. In our case, the force is exerted by the inertia of the vehicle, which resists acceleration.

So now I need to bore you with another definition. Back to Webster’s dictionary, inertia is defined as "A property of matter that causes it to resist changes in velocity." In more simple terms, your car would rather not be accelerated from 30 to 70 mph, and so an external force is required to make this happen. This force comes from your engine.

To accurately describe the acceleration capability of your vehicle, we must consider time. If we just considered force, and distance, we wouldn’t really be saying much about the car. If I tell you that my car can pull a 3,000-lb. weight 100-ft. up a hill, would you be impressed? Certainly not, because I haven’t really told you much. If I told you that I could do it in 10 seconds, while your car needed 15 seconds to do the same job, you might be impressed.

After all, what we are really interested in is the ability to cover distance in a period of time. The distance from the exit of one corner to the entry of the next, or the quarter mile, or maybe even the distance from one stoplight to the next.

If we consider the rate of acceleration, AND miles per hour, we have all three terms included in the definition of horsepower. Time, distance, and force. Force is the rate of acceleration, or the force of inertia. Time is in hours, and distance is in miles.

So now instead of just considering the rate of acceleration arbitrarily, let’s include miles per hour.

And while we’re at it, let’s consider the acceleration rate of two very different motors to illustrate the importance of horsepower, and the absolute irrelevance of torque.

At one extreme we have a Honda F1 motor which revs to 18,000 rpm, makes nearly 800 horsepower, but a measly 281 ft.-lbs. of torque. At the other end of the spectrum we have the Cummins turbo diesel available in the 2003 Dodge Ram which makes a whopping 555 ft.-lbs. of torque, but only 305 horsepower. So which one do you think will accelerate faster?

Everyone that you ask will answer that the Honda F1 engine will accelerate faster. Even your neighbor with the big block who claims that torque is the key to going fast. If torque were the determining factor, the Cummins diesel would win hands down. So what gives?

--------------------------------------------------------------------------------

--------------------------------------------------------------------------------

Let’s calculate the acceleration rate for both engines in a hypothetical 2500 lb. car using the transmission from the 1995 RX-7 with a two-foot diameter tire. Since we know that an F1 car will go 200 MPH, we will gear the car for that speed with both motors.

Starting with the F1 engine which redlines at 18,000 rpm, we need to calculate the required ring and pinion ratio to achieve 200 mph at redline in 5th gear.

First we convert miles per hour to miles per minute by dividing by 60.

200/60=3.333 miles per minute

The tire diameter is rated in feet, so we must convert this 3.333 miles per minute into feet per minute. There are 5280 feet in a mile, so:

3.333 X 5280 = 17,600 feet per minute

If our tire is 2 feet in diameter, the circumference is 2 feet times Pi

2 X 3.14159 = 6.28318 feet per revolution.

Now we divide the feet per minute, by the feet per revolution and we get:

17,600 / 6.28318 = 2801.13 tire revolutions per minute to achieve 200 MPH.

The engine redlines at 18,000 RPM, and in 5th gear the transmission ratio is .719 to 1. To determine the RPM of the output shaft at redline, we take the engine RPM divided by the gear ratio to get:

18,000 / .719 = 25,034 RPM

The output shaft is spinning 25,034 RPM, and we need the wheel to spin 2801.13 RPM to go 200 MPH. To find the correct ring and pinion ratio, we divide the output shaft RPM by the required tire RPM and we get:

25,034 / 2801.13 = 8.937 to 1

Going through all the same boring math for the Cummins diesel which is only spinning 3000 RPM at redline, we get a required ring and pinion gear of 1.489 to 1 to go 200 MPH at redline in 5th gear.

(Note that the F1 engine is spinning 6 times faster than the Cummins, and so the required ring and pinion ratio is exactly 6 times higher.)

With the transmission in first gear, both vehicles will be traveling at 13.76 miles per hour at the bottom of their powerband. (1,000 RPM for the Cummins, and 6,000 rpm for the Honda.) The following chart shows the acceleration rate of both engines in our hypothetical vehicle from that point to 200 MPH.

--------------------------------------------------------------------------------

--------------------------------------------------------------------------------

Note that at any point on the chart, the percent difference in the rate of acceleration is EXACTLY the difference in horsepower. For instance, at 200 mph, the Honda F1 engine is accelerating at a rate of .572 G’s, while the Cummins diesel is accelerating at a rate of .228 G’s. If we divide .228 into .572 we get 2.5, and so the acceleration rate of the Honda is 2.5 times greater than that of the Cummins.

The Cummins, at 3,000 rpm is making 305 horsepower, while the Honda is making 763 horsepower. The Honda is making 2.5 times the power of the Cummins, which is exactly the difference in the rate of acceleration. You can work this out at any point on the chart, and you will find that this direct relationship between horsepower and rate of acceleration always holds true.

I’m sure there is someone out there that still thinks I’m off my rocker, but as I stated earlier, the laws of physics are non-negotiable. After showing this article to a few people for proofreading, one person stated that the results aren’t valid because torque motors are for low rpm grunt, and aren’t meant to run at 200 MPH. Ignoring the fact that this is a ridiculous statement, let’s consider what would happen if we geared both combinations for a top speed of 100 MPH. It should occur to you that this would be a simple matter of doubling the ring and pinion ratio, and that would be correct.

The end result is that the acceleration figures would simply double across the board for both combinations. The difference in acceleration would still match the difference in horsepower, and ultimately the difference in performance would be the same.

So there it is. Horsepower is the determining factor in the rate of acceleration of any vehicle. The next article will go into more detail, and show you how these simple calculations can be used to choose appropriate gearing for any track.

Company Info

© Copyright 2003 Yaw Power Products. All rights reserved

### #3

Posted 08 April 2008 - 09:19 PM

### #4

Posted 08 April 2008 - 11:06 PM

Well thats too much math for me lol. All I need to know is that I need horsepower and torque to go fast.

85 Monte Carlo SS LT1 Powered - Weekend racer

### #5

Posted 21 April 2008 - 03:21 AM

### #6

Posted 22 April 2008 - 05:27 PM

1987 Conquest TSI, 5 spd, B71- 3" *DM Exhaust, Hooker ACM, Hood Struts, Taylor wires, silicone vac lines, fidanza, ACT Extreme, MK1 hard pipes, mbc (13psi), MK1 Shifter, MK1 steering wheel, silicone rad hoses, shp wheels - RIP

1991 BMW 318i - RIP

### #7

Posted 23 April 2008 - 12:29 PM

Almost. Great article!

First one: Red 88 - 245whp/312wtq - HIN: Nightshift - Chicago '07 WINNER

Second one: Durban 88 - Buckskin - SHP - Auto - Sold to to capable hands

Third one: White 84 Flatty - T3/T4 - GM MAS/MAFt - VELNAS - Sold to local capable hands

"...Remember: Don't crush 'em, *restore 'em!"*

### #8

Posted 23 April 2008 - 05:27 PM

I just watched a video of a cummings diesel truck that out ran cars with more horsepower. He had a higher gear ratio (low numerically) and the torque to twist it and push a vehicle that weighs more than the others to the end first.

Also when I used to race there was a guy with a 302 65 vette who reved his engine to 9600 rpm. He only had 450hp but was outrunning guys with more hp because of the rpm's combined with a very low gear ratio (high numerically) and an extremely heavy flywheel which produced the torque to get the car moving. It probably took 10 seconds for the engine to wind down in neutral.

Math is great to impress but it needs to be applied to real world situations which in this case is a relation to drag racing. All factors HP,TQ, gear ratios, weight must be applied at the proper ratios in order to produce the best results.

**Edited by marso, 23 April 2008 - 05:31 PM.**

### #9

Posted 23 April 2008 - 08:34 PM

torque is force, horse power is how much work is being acomplished with that torque over time, with gearing being a factor as to when the power peak is reached (at what speed in each gear).

I can guarantee those cars could beat that deisel if the gearing was matched propperly like in the origional article with teh race car vs. the diesel.

You are right though to point out that often a race can be won with lots of torque with a HP disadvantage when you have an idealy suited set of gears. street cars are often mis-matched gear-wise.

the S-2000 is a perfect example, the car makes more power but less torque than a quest stock-for-stock (almost half the torque), and it's 0-60 and 1/4 times, and track times are superior. It's idealy suited gear wise to it's power peak, ours is idealy suited to our torque peak, making the quest more fun to drive. You

*feel*superior with the massive quest torque, but you are not.

If you were to "make all the cars equal" then you are skewing the results. I doubt the diesel truck has the same gear ratios as a passenger car, there is a reason for that

A motor with more torque at the wheels will win the race every time is an imcpmplete statement, It depends no how quickly that wheel is turned too (rpm's). That is why low torque motors rev so high and have such numericly high gears and small diameter wheels, to multipy thier torque to the wheels at the point where the power actualy meets the road. I can guarantee you can make mroe torque at the wheels with the higher HP motor, it's a mathematical fact.

"Moving the weight faster" is a function of power, not torque. It's one of the laws of physics.

So yes, in the real world mismatches often occur so you can't win every race with HP alone, the gearing has to be right too.

### #10

Posted 24 April 2008 - 09:40 AM

torque is force, horse power is how much work is being acomplished with that torque over time, with gearing being a factor as to when the power peak is reached (at what speed in each gear).

I can guarantee those cars could beat that deisel if the gearing was matched propperly like in the origional article with teh race car vs. the diesel.

You are right though to point out that often a race can be won with lots of torque with a HP disadvantage when you have an idealy suited set of gears. street cars are often mis-matched gear-wise.

the S-2000 is a perfect example, the car makes more power but less torque than a quest stock-for-stock (almost half the torque), and it's 0-60 and 1/4 times, and track times are superior. It's idealy suited gear wise to it's power peak, ours is idealy suited to our torque peak, making the quest more fun to drive. You

*feel*superior with the massive quest torque, but you are not.

If you were to "make all the cars equal" then you are skewing the results. I doubt the diesel truck has the same gear ratios as a passenger car, there is a reason for that

A motor with more torque at the wheels will win the race every time is an imcpmplete statement, It depends no how quickly that wheel is turned too (rpm's). That is why low torque motors rev so high and have such numericly high gears and small diameter wheels, to multipy thier torque to the wheels at the point where the power actualy meets the road. I can guarantee you can make mroe torque at the wheels with the higher HP motor, it's a mathematical fact.

"Moving the weight faster" is a function of power, not torque. It's one of the laws of physics.

So yes, in the real world mismatches often occur so you can't win every race with HP alone, the gearing has to be right too.

You are actually proving my point with your statements. This whole thread is about racing not mathematical comparison. Theoretically if you have the two vehicles at the same weight, aerodynamics and optimum combinations of of torque, hp and gearing one with more torque less hp the other with the opposite they should arrive at the end of equal distance at the same time but one leaving before the other and the other catching the first at the end. The one with more hp is turning at a higher rpm to get the axles turning the same speed as the one with more torque pushing a gear ratio that turns the wheels at the same speed as the other car. There really is no way to say which is best. There must be the right combination of the two to move the car at the quickest time over a certain distance. If the vehicle weighs more you must have more torque to compensate to get the car moving and keep it moving but you need hp to increase the rate of acceleration.

### #11

Posted 24 April 2008 - 12:12 PM

A lot ov people mistake speed (acceleration) for being faster, not accounting for time and distance.

### #12

Posted 24 April 2008 - 12:18 PM

Take a 1 HP mootr, a 50 pound weight, a series of pullies, a stop watch, and 20' of rope.

You can gear that assembly (compound the pulies) with the pulies to create a high torque situation, say a 3:1 multiplication. That assembly will move the 50# weight with ease, and the rope will very quickly be spooled up, but that weight will not be moved very far, it will move exactly 1/3 of the 20 feet of rope, the weight will move 6.667 feet. Lets imagine it takes 10 seconds to do this.

Now take the same setup and remove the compounding of the pullies, now it will move the weight the full 20', exaclty 3 times further than the other setup. but it will be slower before the rope is fully spooled up, exactly 3 times slower, so we can call that 30 seconds. Note now every time, exactly 1 HP of work was beign done (minus friction).

Take the same assemblies and use a 2 HP motor, now in both scenerios the weight will move twice as fast, 5 seconds and 15 seconds repsectivly (distance will be unchanged), twice the work was done.

Now take the 1 HP high torque assembly and race it against the low torque 2 HP assembly, the 1 HP will move the rope faster (not the weight), but the weight travel less distance. the 1 HP setup has a 3:1 advantage in torque, but the 2 HP motor has a 2:1 advantage in power. the 2 HP setup will always do twice the work, even if the 1 HP setup with it's gearing is able to wind it rope more quickly.

If you raced these setups to move the 50 pound weight 60', the high torque 1 hp setup would need to wind up

__9 times__the origional 20' of rope, or 180' > 1/(6.667/60) =

__9__. the 2 HP setup would only need to wind

__3 times__the origional rope, or 60' of rope > 60/20=

__3__.

If it takes the high torque 1 HP setup 10 seconds to move a 50# weight 6.667', it will take

**90**seconds to move it 60'

It takes the low torque 2 HP motor 15 seconds to move a 50# weight 20', it will take

**45**seconds to move it 60'

Notice the time/distance ratio is exactly 2:1 in favor of the high HP motor (which is also 2:1)? In a car race you are doing the same thing, measuring time over distance, not how much torque you have to move your weight. At any point in that race, the 1 HP motor always has more torque, but it is always doing less work.

Cars in the real world have a lot of varribles, multiple transmission gears, different axle ratios, different diameter tires, and different RPM power peaks. Lets not forget, different drivers too . this all "muddies the watters so to speak" and can create a situation where torque can win-out over power. This is usualy this is when gearing is misapplied to off-peak power, and distances are almost always short. In the long run, when propperly appied, higher HP will

__always__win.

### #13

Posted 24 April 2008 - 12:30 PM

Take a 1 HP mootr, a 50 pound weight, a series of pullies, a stop watch, and 20' of rope.

You can gear that assembly (compound the pulies) with the pulies to create a high torque situation, say a 3:1 multiplication. That assembly will move the 50# weight with ease, and the rope will very quickly be spooled up, but that weight will not be moved very far, it will move exactly 1/3 of the 20 feet of rope, the weight will move 6.667 feet. Lets imagine it takes 10 seconds to do this.

Now take the same setup and remove the compounding of the pullies, now it will move the weight the full 20', exaclty 3 times further than the other setup. but it will be slower before the rope is fully spooled up, exactly 3 times slower, so we can call that 30 seconds. Note now every time, exactly 1 HP of work was beign done (minus friction).

Take the same assemblies and use a 2 HP motor, now in both scenerios the weight will move twice as fast, 5 seconds and 15 seconds repsectivly (distance will be unchanged), twice the work was done.

Now take the 1 HP high torque assembly and race it against the low torque 2 HP assembly, the 1 HP will move the rope faster (not the weight), but the weight travel less distance. the 1 HP setup has a 3:1 advantage in torque, but the 2 HP motor has a 2:1 advantage in power. the 2 HP setup will always do twice the work, even if the 1 HP setup with it's gearing is able to wind it rope more quickly.

If you raced these setups to move the 50 pound weight 60', the high torque 1 hp setup would need to wind up

__9 times__the origional 20' of rope, or 180' > 1/(6.667/60) =

__9__. the 2 HP setup would only need to wind

__3 times__the origional rope, or 60' of rope > 60/20=

__3__.

If it takes the high torque 1 HP setup 10 seconds to move a 50# weight 6.667', it will take

**90**seconds to move it 60'

It takes the low torque 2 HP motor 15 seconds to move a 50# weight 20', it will take

**45**seconds to move it 60'

Notice the time/distance ratio is exactly 2:1 in favor of the high HP motor (which is also 2:1)? In a car race you are doing the same thing, measuring time over distance, not how much torque you have to move your weight. At any point in that race, the 1 HP motor always has more torque, but it is always doing less work.

Cars in the real world have a lot of varribles, multiple transmission gears, different axle ratios, different diameter tires, and different RPM power peaks. Lets not forget, different drivers too . this all "muddies the watters so to speak" and can create a situation where torque can win-out over power. This is usualy this is when gearing is misapplied to off-peak power, and distances are almost always short. In the long run, when propperly appied, higher HP will

__always__win.

ah the power of logically applied data. it's so soothing

**'88 TSi, 06H-20G, HAWK, Roller cam, Water inj, etc^10 320whp/370wtq**

### #14

Posted 24 April 2008 - 01:18 PM

- Inertia
- wind drag
- friction
- rolling resistance
- and a few other, smaller factors...

As you try to double the rate of acceleration, you need to quadrouple force (power, AKA HP). To go from a 14 second 1/4 mile car at 200 RW/HP, you need 800 RW/HP to get to a 7 second car.

Now add to that the same principal of wind drag. Wind drag also increases at a square rate, wind drag is 4 times as much at 200 MPH as it is at 100 HPH. Now that the 7 second 800 RW/HP car is going much faster, it has a lot more wind drag to work against. It's now more like a 8 1/2 second car (depending on it's coeficeint of drag).

Also, friction increases as you increase the forces applied to the rotating and reciproating components, and rolling resistance of the tires is going to have some similar square-rate resistainve properties. This is very difficult to measure but it is absolutely a factor. You could assume it would take say .5 seconds off a 800 HP car. you now have a 9 second car.

same "square-rate resistance" is true of wind drag, it only takes about 125 HP to reach 100 HP in an average sedan. Take the same car and try run 200 MPH, you'll need about 500 HP.

These factors can be seen in many

*real world*situations, it takes about 800 HP to move a 3000# car into the 9 second 1/4 mile range. it also only takes about 200 HP to make that same car run a 14 second 1/4 mile. Also, How many 200+ MPH cars do you now of with less than 500 HP ?

In that 1 HP vs 2 HP pulley comparison, there will be inertia, friction, wind drag etc. The percieved difference will not be 2:1, but the actual work done will be 2:1, it's just that some of that work will be "wasted" on other resistive forces.

### #15

Posted 24 April 2008 - 04:37 PM

From a physics aspect alone torque is more important. Without it hp would not exist in relation to moving a vehicle.

You need force to move an object. In our case: explosion/ piston/rod/crank and so on through the chain of events to pavement.

It is the torque that puts the car in motion.

It is motion that allows work to be done.

It is that amount of work being done over time that relates to horsepower.

No force, no torque

No torque, no motion

No motion, no work

no work, no HP. (Or if you're magic and could remove time you could have torque and motion but no HP).

Going back to Mr. Watt the one thing he does not touch on in his horse drawn formula is the horse is providing the torque through muscle contraction pulling on a tendon rotating a bone that's pushing on the ground that moves the horse and whatever is attached to it.

If he would have used a weaker horse 1 horsepower would actually do less work. Our cars would be slower with the same amount of HP from before.

### #16

Posted 24 April 2008 - 04:53 PM

**'88 TSi, 06H-20G, HAWK, Roller cam, Water inj, etc^10 320whp/370wtq**

### #17

Posted 24 April 2008 - 10:19 PM

I understand the formula for HP. I only pointed out the "no tq, no hp" rational because the Law of Physics was a baseline reference point for the answer to the question. So if "without TQ there would be no HP" goes without saying than the original question is a moot point

but the original question was which is more important in relation to moving a vehicle a certain distance in less time. There is no definative answer. They are both equally important. As I stated before you have to find the right mix of all factors to get the maximum potential out of the car.

By the way Chad. You found a good way to smoke the brain.

**Edited by marso, 24 April 2008 - 10:35 PM.**

### #18

Posted 25 April 2008 - 12:59 AM

Is that a good thing?

Thanks for bringing up watts laws, I leaned what I know about physics while persuing my degree in electronics. Watt created many of the foundations of electronicis physics. This is how I was able to see HP for what it is, power. 1 HP is ~746 watts. We had the same arguments in theory class. The instructors proved how power is power, everything else is just a varrible to derive it. There is no way for any circuit to do more work than another circuit if it has less power, regardless of voltage or current arguments. Same is ture of torque and velosity vs. power.

In the electronics field, torque is like voltage, It's a force. Invack, it's turne name is "electromotive force". Current is like velosity, It's the "stuff" in motion that you can see. Power is still power. Apply all the voltage (torque/force) you want to a circuit, if the current (velosity) is low, then the power is low. I can come up with hundereds of real world annalogys on this one

Like static charges: A static shock from a typical metalic object is up to or even over 10,000 volts, but very low current (nano amps). It won't kill you, or even injure you. A 240 electrical outlet with 20 amps can easily kill you though.

Torque is indeed a nessesary object of study.

It is however the study of power you must focus on when calculating/predicting the outcomme of a race of moving an object a given distance over the least time.

Increasing torque (force) while decreasing velosity will not yield any gains in power = the ability to move the object to the end point quickest. A towfold increase in torqe applied at 1/2 the RPM (velosity) does the same work and will move that object to the finnish in the same time.

A motor that makes 1/2 the torque, but spins 3 times faster will win every time because it's doing 1/2 less, but doing it 3 times more often. Think of a bucket brigade of moving water, if you cut the bucket size in half, and ran 3 times faster, you will get more water from A to B in any given time frame, right ?

Engine horsepower works the same way, it asks "what is the torque at a given moment, and how fast is the engine turning at that moment?" The formula then determins how much work is being done.

In the bucket birgade, you could muster a 50 fold increase in bucket size, but if it took you 100 times longer to get there, you are doing 1/2 the work, right? In this case, bucket size (torque) does not dictate how much work is being done unless you also specify speed or frequency of the event (RPM's).

1000 ft/lbs at 1000 RPM wont win a power race against 100 ft/lbs at 11,000 PRM (10 times the torque) if the 100 ft/lbs can happen 11 times more often (higher RPM's).

Moving mass from A to B the fastest is a power race, not a force race. Put a 10:1 transmission gear set on the 100 ft/lb motor, it will make 1100 ft/lbs and spin at the same RPM. So yes, extera torque does matter, the low torque motor has more torque when geared to the same RPM.

Yah, I know, beating a dead horse here , but this is how I learned it, and maybe a few others will pick up some knowledge in the process.

### #19

Posted 25 April 2008 - 01:53 PM

but the original question was which is more important in relation to moving a vehicle a certain distance in less time.

I like to think of an engine's horse power number as "containing more information." Peak torque just doesn't tell you much. Like chad pointed out, you can gear up to make a high torque engine go fast, but you'd rather gear down, or gear reduce, to make a high HP engine work well in a heavy car. The most extreme example of this is sited in the vettnet link I always post. A huge waterwheel spinning at 1 RPM might make 30,000ft/lbs of torque but you wouldn't want to try to gear that force into something effective in any vehicle. A jet engine spinning at 20,000RPM on the other hand can doesn't make alot of torque, but can be gear reduced to 1500 shaft HP and pull a house off its foundation or propel an M1A1 70 ton tank to 50mph. You never learn what an engine's torque output is and think you know how the engine will perform do you? You always want to know what RPM the engine makes that torque at, and as soon as you know that, you can calculate what HP it makes and get a good idea what to expect from the engine. Therefore, HP contains more information. The single number HP represents not just how hard an engine can work for an instant, but how long it can sustain a given level of work. It incorporates time into the equation so it's more valuable as a point to evaluate by. It goes without saying that there would be no HP without TQ for the simple fact that there is no such thing as an engine that makes HP but not TQ.

I'm not trying to wire brush your nuts here man, but no they aren't, and yes there is a definitive answer for a given car.

And once you've found said "right mix" you will have also found the definitive answer you just said didn't exist.

**'88 TSi, 06H-20G, HAWK, Roller cam, Water inj, etc^10 320whp/370wtq**

### #20

Posted 26 April 2008 - 08:05 PM

89 Star ESIR SHP SOLD!!!!!

Black 89 SOLD!!!!!

89 Fiji sold

92 Dodge Spirit R/T

89 turbo voyager msd 2 coil, LSP MBC, Starquest Intercooled (o: (kids/grocery runs)

89 olds cutlass ciera stock 160 HP 185 TQ (daily driver beater)

00 grand am gt

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users