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Horsepower and Torque?

Old 11-19-2007, 11:01 PM
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Horsepower and Torque?

I'm doing a project for this stupid class and i have to explain the relationship and differences between horsepower and torque. So far I've gotten a lot of info on both but a few things that keep popping up. They have to do with gearing. I've been thinking and haven't been able to find anything on the internet that shows how much horsepower and torque you can lose or gain from gearing. (And from saying "gearing" i'm not sure if i should be talking about final drive or transmission) Any help would be awesome. I know there's some smart guys on here. Thanks in advance!

-Miles
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Old 11-19-2007, 11:03 PM
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You don't really lose or gain horsepower. What happens when you change your gearing is you are moving the rpm at which the engine is spinning.

As a crude, quick example with no computation whatsoever, if you go up two teeth in the rear, instead of spinning at 4k rpm at 65mph, you'll be spinning at 5k rpm.
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Old 11-20-2007, 05:04 AM
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It's really hard to figure out how much is lost by the XMSN and final gearing. It all depends on how much mass is being moved between the crank and the tire. Think of the weight of the internal components, friction caused, resistance of slapping oil or grease inside a differential or transfer housing. Then it's different for an automatic transmission, because of all the fluid it has to move around in order to shift smoothly.

When it comes right down to it, every vehicle will give different results on loss. But you can be sure that every one WILL lose power between the crank and tires.
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Old 11-20-2007, 05:52 AM
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Let's get down to the fundamentals: engines DO NOT produce horsepower, they produce torque. Horsepower is a mathematical manipulation of the torque output based on engine rpm. It's a simple mathematical formula really. It's the reason why all engines (yes, ALL engines!) produce the same horsepower and torque [nominal value] at 5250rpm. If the engine makes 30ft-lbs of torque at 5250rpm then it's power is 30hp at that point.

There are some interesting articles on Wikipedia on horsepower you might want to check out too.
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Old 11-20-2007, 06:37 AM
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The way I like to remember it is: Torque is how much work an engine can do. HP is how fast it can do that amount of work.
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Old 11-20-2007, 08:17 AM
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Doug, That's a good way to see it.

Horsepower is the rate of application of torque.
1 HP = 550 ft-lb/sec.

Interestingly, it is also the rate of doing work, so 1 HP will also raise 550 lbs at a rate of 1 ft/sec.

Horsepower numbers are calculated from torque measurements. A dyno measures torque, then converts to HP based on gear ratios and rpm.

So say you have a torque curve from measured data. The torque curve is fit to a bunch of measured values at known rpm numbers.

Here's a suggestion:Go to the published data on HP and torque, for your engine of choice, I suggest the VTR 1000F.
Enter the torque values at every 100 rpm from the published torque curve into a spreadsheet.
Plot these values and you should get a plotted curve that looks like the published torque/RPM curve.

Then in the next column take the torque numbers and perform the following calculation:

Torque*(RPM/60)/550. This converts RPM to revs/second and converts ft-lbs/second to HP.
You will then have a column with HP for each RPM.

Plot this and you will reproduce the published HP/RPM curve.

Don't wory about losses. Friction losses are something that would best be measured experimentally.
There are rule of thumb estimates based on historical measurements.

I hope this helps to enlighten you!

Last edited by RCVTR; 11-20-2007 at 08:18 AM. Reason: correction
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Old 11-20-2007, 08:35 AM
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Another thing I meant to mention is that you can get from crankshaft torque and/or HP to rear wheel torque and/or HP by multiplying by the various gear ratios and diameters.

So if you start at the crank, you take the primary drive ratio, the gear ratios in the transmission, the final drive ratio (sprocket teeeth) and rear wheel radius. Then you would apply an estimated (or measured) loss multiplier.

This is the method used to produce the driving force calculations that you see in the publications.
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Old 11-20-2007, 11:01 AM
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Old 11-20-2007, 11:33 AM
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Originally Posted by RCVTR View Post
Another thing I meant to mention is that you can get from crankshaft torque and/or HP to rear wheel torque and/or HP by multiplying by the various gear ratios and diameters.

So if you start at the crank, you take the primary drive ratio, the gear ratios in the transmission, the final drive ratio (sprocket teeeth) and rear wheel radius. Then you would apply an estimated (or measured) loss multiplier.

This is the method used to produce the driving force calculations that you see in the publications.
I knew there would be some good info on here. Thanks to all of you by the way. What you [RCVTR] said about calculating the HP loss by multiplying the gear ratios from the primary drive, transmission, and final drive is more what i was trying to find.
Yeah i've read a lot of articles and everyone seems to be an expert but they all say different things. I was amazed when i read all those misleading marketing ploys about HP when it's really torque that you want, whether it be down low or up high.
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Old 11-20-2007, 12:51 PM
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Old 11-20-2007, 03:46 PM
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Originally Posted by lazn View Post
That was one of the sites that i've been citing a lot of my info from. Seems like he knows some stuff. Thanks!
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Old 11-20-2007, 04:05 PM
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It's not really HP or torque loss, it's really a matter of conversion.
The torque is much higher at the rear wheel than it is at the crankshaft, because it spins slower. And first gear has the most torque for the same reason.

If two engines produce the same torque, but one can produce that torque at a higher RPM, it will produce more HP. That is the primary reason that V-twins can't compete in Superbike (at least not for a reasonable cost). They are RPM limited because the pistons are heavier. WSB Ducatis are extremely expensive and exotic, to even get close. That's why they want more displacement. More torque, so they don't need so much RPM.
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Old 11-20-2007, 07:20 PM
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RCVTR;

"If two engines produce the same torque..."

What if bike "A" has a substantial torque advantage but bike "B" makes more H.P.?

Example. The VTR makes more torque from 4500-8500 rpm than a gsxr-750 makes anywhere from idle to redline.

According to the author of the linked article, if you equalized these two bikes for weight and gearing, the VTR would accelerate harder at 4500 rpm than the gsxr could at any engine speed, despite the 30 horsepower advantage, right? As long as you kept the VTR over 4500 rpm the gsxr would never catch up.

OK. Given the 750's redline, equalizing the VTR for gearing might require something like a quick shifting 9 speed gearbox.

I'm not saying this isn't true, but I guess I'd have to see it to believe it 100%.
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Old 11-20-2007, 07:28 PM
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Scrap the 9 speed gearbox 'cause too be equal they'd both get one and the 750 retains the advantage.

Give them each a variable auto trans which keeps each engine at torque peak + or - 500rpm. The 130 HP Suzuki would never catch the 100 HP VTR.
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Old 11-21-2007, 08:52 AM
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That's what I used to think until I saw a comparison about 3 years ago of the GSXR 1000 and the RC51.

The GSXR motor made more torque everywhere and maintained it to about 12k RPM. No way the RC51 in stock form can keep up.

The twins do have an advantage because the torque pulses from the engine are widely spaced, so they can pull harder without spinning the tires.

The closer spacing of pulses allow the GSXR to make more torque with the same displacement. At least I think that's how it works.
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Old 11-21-2007, 09:28 AM
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So, would I be able to infer from this ^^^^^^^^, that because of the V2 pulse, the likely hood of spinning up the rear on corner exit is less than an I4 with equivalent torque and HP? So, we should be able to hammer on the throttle coming out of turns before the I4's, right? Since I came from an aircooled 600 I4 (where you could twist the throttle to WFO without any concern anywhere), I have been afraid of whacking her open on the SH. But I don't believe I have ever felt the rear break loose.
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Old 11-21-2007, 10:16 AM
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I was at a track day a few once and just blew by a 600F4i on the front straight, which apparently pissed him off.

He was trying to get me back in the corners, but I kept pulling away on corner exits, so he whacked the throttle open, to get me coming out of turn 2 (I was unaware of what was going on, he was behind me).

when I came around the next time there were little F4i parts all over the track. He had highsided and tumbled everything.

So yes, it's true, but you always want to roll the throttle on. The VTR will spin the rear tire. Don't just whack it open when heeled over. Smashed my nuts on the tank once trying that.
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Old 11-21-2007, 02:30 PM
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Hell of a lot more educational than an oil thread!

Last edited by nuhawk; 11-24-2007 at 09:32 AM.
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Old 11-21-2007, 04:18 PM
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RCVTR

I think you might have misunderstood my last 2 posts. I was comparing the VTR to the GSXR 750 (not the 1000) precisely because while the 750 is quicker and makes a bunch more H.P. than the VTR, the VTR makes more torque.

According to the linked article, if the VTR's torque advantage could be transfered to the rear wheel more efficiently (gearing!), it would out accelerate the gsxr-750.
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Old 11-21-2007, 04:35 PM
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Yeah RK1, I completely missed your first post.
the VTR is about as quick as anything to about 100 MPH.

The only big bikes I've really compared it to are the 929 and the RC51.
the 929 feels lifeless unless you wring it's neck. the RC51 is way more fun than the 929, but I still liked the VTR best for street riding.
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Old 11-21-2007, 05:08 PM
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RCVTR

Yeah, that article was pretty cool. Helps explain why, for me at least, the VTR is easier and more fun to ride on the street than an I-4 600 or even a 750. Also helps explain how Ducati put the 750 Superbikes out of business dispite relative horsepower parity.
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Old 11-24-2007, 08:09 AM
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Yup. It's all about a flat torque curve.
Flat torque = linear power.
When you have displacement, you can make the torque flat and still have top end power.
Then, with a twin, you get more grunt off the corners without spinning the tires.

Guess who's probably going to build a new V-twin superbike when they change the displacement rules? ( a little rumor I heard). Already built it...
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Old 11-26-2007, 09:52 AM
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Thanks a bunch guys. Seems like there could be some engineers in here, lol. I don't know about some of you guys but i have a really hard time comprehending this kind of stuff. Unless i can get a really clear picture in my mind of what is mechanically going on, then i'm lost. But the information you all have provided will definetely get me a ways farther than what i would've had. So thanks again!

-Miles
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Old 11-26-2007, 09:56 PM
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For clarification, 2 engines with hypothetical flat torque curves engines
Engine A 100 lb/ft torque, redline 8000 RPM
Engine B 100 lb/ft torque, redline 16000 RPM

Both engine run trough equal transmissions, at equal speed, in High gear 1:1 ratio (mean that one turn at the input of the tranny makes one turn to the output)
Both engine run at 8000 RPM, in this situation the output is 100 lb/ft to the tranny output, they put down the same horsepower.

Now the guy with engine B is not running at full potential so he downshift two gears and the rev jumps to 16000 RPM and still put 100 lb/ft of torque to the input shaft. His transmission ratio is now 2:1, so it multiply the torque by 2 and divide the speed by two, so to the output shaft now see 200 lb/ft of torque. The horsepower is now doubled.

Engine A 100 lb/ft torque, redline 8000 RPM 152 hp
Engine B 100 lb/ft torque, redline 16000 RPM 304 hp

So one turn at a given torque is work, and the rate at witch this work is done is horsepower.

Your computer fan motor will develop enough torque to pull a fully loaded semi truck up a 5 miles hill. Even if it will take a wicked tranny, the torque it produce is more than enough, assuming that you have a Year to do such a thing.
Make this fan motor turn a 15 quadramegazillions RPM, it may take a minute lol. And gives 5000hp !!


And for an even more simplified way, now that this semi truck is up the hill, you have 1000 cardboard boxes to unload, you're able to pick up one at the time. One box power, now some drug dealer comes by and sell you some steroids, this multiply your force, now you can take 4 boxes, you got four box power now.
But aunt cookie knows that your shoes sucks and bring you some Nike air along with her amphetamine cookies, now you're flying and can run at double the speed, now you can haul 8 boxes per trip you got 8 box power now.

So you can increase horsepower (or boxpower, your choice) by adding work or the speed at witch the work is done.

Last edited by gboezio; 11-26-2007 at 10:21 PM.
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Old 11-26-2007, 11:11 PM
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gboezio;

I understand that other things being equal (test weight, max torque, gearing/transmission) the 16000 rpm bike will always have a 2 to 1 advantage in gearing. But is that the only advantage? Is the higher horsepower just a way of expressing the gearing advantage?

Here's what the linked article made me curious about- I hope this makes sense.

Take the two bikes in your example- 8k and 16k . Same max torque.

Now imagine both bikes have an infinitely variable automatic transmission and power control mode which keeps each bike's engine running at its torque peak while transmitting as much power as possible to the rear wheel without dropping engine rpm.

Does the 16k bike have a gearing advantage? A horsepower advantage? No advantage at all?

Will it still have a lower E.T. and higher trap speed?

Last edited by RK1; 11-26-2007 at 11:21 PM.
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Old 11-27-2007, 06:10 AM
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Talking

Totally, the infinitely variable transmission, will actually maintain the exact 2:1 ratio trying to maintain both engine to their peak horsepower, doing so cutting the high rev engine RPM in half, witch multiply the torque by 2.
Power is doubled and ET are lowered, more work done over time, since ET are all about time.
In the real world tough, the torque drops as RPM increase depending of the engine configuration, because of the Volumetric Efficiency (ability of the engine to breathe at a given RPM). So the torque is decreasing, but the energy figure is still higher than where the RPM will land shifting a gear, so we keep pulling until no more gain could be obtained from holding the said gear.

Power formula : ( RPM X torque )/5252 = hp

So an hypothetical engine say a V-twin, that has a torque peak of 70.59 lb/ft @7000 RPM, will translate to ( 7000 X 70.59 )/5252 = 94.59 hp at his peak torque, so not worth shifting yet. So at 9000 rpm the torque dropped to 64.2 lb/ft ( 64.2 X 9000 )/5252 = 110 hp still the power is the highest.
Let's hold it a bit longer, ignoring that piston that goes to insane speeds and is aimed straight at you ***
The torque keeps dropping and now is a 40 lb/ft of torque due to choking and valve float/ bounce. ( 40 X 11000 )/5252 = 83 hp

So horsepower take torque into account, and the mechanical advantage of the RPM. As a bike accelerate the torque to the rear wheel gets lower, first gear high, lower,lower. A high HP engine will maintain this torque longer (higher speed) giving a better acceleration.
Hope this make sense

Last edited by gboezio; 11-27-2007 at 06:29 AM.
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Old 11-27-2007, 07:07 AM
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Originally Posted by gboezio View Post
Totally, the infinitely variable transmission, will actually maintain the exact 2:1 ratio trying to maintain both engine to their peak horsepower, doing so cutting the high rev engine RPM in half, witch multiply the torque by 2.
Power is doubled and ET are lowered, more work done over time, since ET are all about time.
In the real world tough, the torque drops as RPM increase depending of the engine configuration, because of the Volumetric Efficiency (ability of the engine to breathe at a given RPM). So the torque is decreasing, but the energy figure is still higher than where the RPM will land shifting a gear, so we keep pulling until no more gain could be obtained from holding the said gear.

Power formula : ( RPM X torque )/5252 = hp

So an hypothetical engine say a V-twin, that has a torque peak of 70.59 lb/ft @7000 RPM, will translate to ( 7000 X 70.59 )/5252 = 94.59 hp at his peak torque, so not worth shifting yet. So at 9000 rpm the torque dropped to 64.2 lb/ft ( 64.2 X 9000 )/5252 = 110 hp still the power is the highest.
Let's hold it a bit longer, ignoring that piston that goes to insane speeds and is aimed straight at you ***
The torque keeps dropping and now is a 40 lb/ft of torque due to choking and valve float/ bounce. ( 40 X 11000 )/5252 = 83 hp

So horsepower take torque into account, and the mechanical advantage of the RPM. As a bike accelerate the torque to the rear wheel gets lower, first gear high, lower,lower. A high HP engine will maintain this torque longer (higher speed) giving a better acceleration.
Hope this make sense
This helps to explain why 7+ speed gearboxes were banned from GP racing in the early 70's. I think it was Kawasaki or Suzuki that ran a 50cc bike with a 13 speed transmission prior to the rule change in the day when they had a 50cc and 80cc GP bikes.

Last edited by HRCA#1; 11-27-2007 at 07:11 AM. Reason: Don't understand how this thread works entirely
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Old 11-27-2007, 07:40 AM
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Exactly, since the engine is actually at peak hp only a fraction of every gear, the average horsepower during typical acceleration would be a better measurement, engine mated with close ratio gearboxes (assuming a negligible shift time) will maintain the engine longer in it's peak power band, raising the average hp, specially with race engines that are peaky since all VE is tuned to respond at a given RPM making the power band very narrow.
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Old 11-27-2007, 03:01 PM
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Originally Posted by gboezio View Post
So horsepower take torque into account, and the mechanical advantage of the RPM. As a bike accelerate the torque to the rear wheel gets lower, first gear high, lower,lower. A high HP engine will maintain this torque longer (higher speed) giving a better acceleration.
Hope this make sense
So let me get this straight. Despite the fact that your loosing insane amounts of torque after it's peak, you still remain on the gas because you get more hp? What does hp have thats so important if all your going to want is a flat torque curve? I thought that torque is what was most important. Could you put the meaning of hp in an easier way?
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Old 11-27-2007, 03:48 PM
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You now have the basics for understanding the relationship between the HP and torque curves. As the torque drops off in the higher RPM range, the slope of the HP curve decreases. If the slope of the HP curve goes to 0, you're done.

A flat torque curve provides a broad range of usable torque and linear HP over the RPM range, making the bike drive off of corners and making it easier to ride. Remember the early MotoGP bikes? Only Honda had flat torque. Everything else was practically unridable.

Last edited by RCVTR; 11-27-2007 at 03:57 PM. Reason: added comment
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