I may be way off on this, but from what I gather on the torque RPM horsepower equasion

Just given the hypothetical numbers Kawasaki gave back in late 2011 as the ZX14R’s torque and hp output……119.9 FT lbs of torque, and 197 HP.

As I understand it the equasion is:

Torque times RPM divided by 5252 (?) = HP

So plugging the provided numbers in, this tells us that the 14’s peak HP would come at approx 8650 RPM.

And another interesting point is that while BMW claims the S1000RR also puts out about the same peak HP as the 14, it is crushed in the torque department, as only the BMW only creates 83 ft lbs of torque.
So plugging in the given BMW specs (83.5 torque and 193 peak HP), I assume the BMW reaches its peak HP at around 12,000 PRM. ???

Peak HP, alone, is not as vital to acceleration as one would initially think, from what I gather…….?

I may be way off, and any input from those more knowledgable than I am on this (everyone) is welcome. I find this to be very interesting stuff, especially since math is also an interest of mine.

* Last updated by: ZX14MAN64 on 3/11/2013 @ 11:43 AM *

NERD!!

Just kidding. You're awesome. I actually am going to do some research on that as well. Maybe I will turn it into an algebra problem for my 1st period class.

The thing is the lighter you are the less torque you need. You could almost say the same with HP but not quite. HP and Tq don't respond to weight the same way!
All the big bikes have big Tq numbers, even Harleys but it doesn't make them fast. They need torque to move dead weight but the more you get going the less you need it, this is why you rear (at least mine) won't spin in 6th gear. This is also why the lighter 1000 can hang with the big bore in roll ons!

Sure, weight also plays a vital role in acceleration, and the 1000RR is almost 100 pounds lighter.
My KX450F is probably about able to *hang* with the 14R off the line, (at least for a while) as its about half the weight of the 14.

Torque, as I understand it, is rotating force, as the crankshaft exudes. And torque is the *raw form* of horsepower for an engine driven machine.............????
But the 14s almost 35-40 more ft lbs of torque than the BMW......... That's almost a 50 % increase over the BMW's torque. So I was thinking the 1000RR was putting out close to what the 14r does, powerwise, but it clearly does not. For sheer brute force, acceleration off the line, I would imagine the 83 pounds of the BMW would be embarrassed by the 14Rs almost 120.

Its hard to understand the difference of what torque and HP really are, but I am starting to understand them a bit better and how they relate to perceived power and actual power.

It would be interesting to see these two in various side by side comparisons to see where the extra torque and weight really stand out. I think the Hyabusa was just smashed in the torque and peak HP by the 14R. That sounds impressive at first, but Kawasaki had to increase the displacement by 90cc in order to top the hyabusa.

And how RPM plays into this really makes for some interesting, at least to me.

I remember an old saying: "Horsepower sells cars, but torque wins races"

There was also another saying that says hp. determins how fast, or hard you hit the wall. Tq. determins how far you push the wall once you hit it.
I may be wrong with my quotes, but I never want to find out either one.

I guess most of us know that torque is a measure of turning or twisting force.
Generally, larger displacement engines will always have more torque than those with fewer cc's simply because the increased fuel burn volume gives bigger bangs and hence a greater force pushing down on the crankpins. A small cc turbo engine can mimick the torque of a larger NA engine as is more fuel being forced into the cylinders (under pressure) and burned, again, bigger bangs.
A torque analogy.....Imagine you have a tight wheel nut on your car, you apply the wrench and stand on it but you weigh 150lbs and it won't move so you get your mate who weighs 300lbs to stand on it and it slackens easily. That's torque.

HP or BHP is not a measure of force but a measure of the rate at which the force can be applied. You can get a smaller displacement engine with a smaller torque output to match or even exceed the HP rating of a bigger, higher torque engine by spinning it faster.
From a 'street level' perspective, high torque means you can launch hard with fewer revs. Less torque but the same or even more BHP means you need more revs for the same launch or you'll bog the engine as it just can't hook up and drive at lower RPM's like a big cc, high torque motor.

Electric motors are exceptionally good at getting things moving from a standstill as they can produce max torque from zero RPM, hence their use in very heavy earth moving plant/machinery and locomotives etc. JMO

About torque, HP, and winning races:

There's a great history lesson in this month's "Cycle World" magazine. I definitiely do recommend reading it, a full page editorial.

It describes how, since about 1920's and on, how cycles would get bigger motors / heavier everything else. bigger /heavier, and yet again bigger / heavier... and then, eventually, they get passed, in a race, by a team that figured out how to do "more Tq / lighter".

These issues appear, historically, to go in cycles (OK, pun intended ;}

I think, maybe, we're at an end where the pendulum has swung to bigger / heavier...

I personally would like a cycle with 250 HP (more baby gimmee more!) and at ~500 lbs *mass*, and then figure out how to keep it 1) on the ground, and 2) rear tire planted / not spinning, and 3) me still in the saddle at launch.

YeeeHaaaaaaaa! Go Ninjee!

I'm pretty sure that less weight is more important a low speeds and the faster you get, aerodynamic drag becomes the biggest thing that limits performance. That's why bikes rule in the 0 - 150 range while cages with way poorer power&torque/weight ratios have higher top speeds (even after the ECU 186 mph limiter) is disabled. The ZX weighs more than the BMW but the aerodynamics (which aren't so good on any bike) are probably around the same.

*HP or BHP is not a measure of force but a measure of the rate at which the force can be applied*

Now we are getting somewhere! Or are we?

I thought it was clicking, but I admit I am still a bit fuzzy on the actual difference and the relationship between the two.

Turning force is torque, yes. Rotating a crankshaft which is then transferred to the tire, and finally the road in a forward thrust. But a bike with alot of low end torque has its torque transferred to the wheel and then the road in order to use it and be a big beefy torque powerhouse. And that wheel is the same rolling wheel that also guages high HP output at higher RPMs.

Its about as clear as which came first, the chicken or the egg? LOL

But what I am gathering is that high torque means a high increase, as in acceleration. Whereas high HP means more of a sustaining of higher speeds which are slowly attained. Which means a good bottom end, as opposed to a good top end........????

* Last updated by: ZX14MAN64 on 3/11/2013 @ 7:40 PM *

http://www.youtube.com/watch?v=fgLNO3ThGD4

* Last updated by: ZX14R on 3/11/2013 @ 9:28 PM *

Don't make it difficult by trying to attach the physics concepts to riding emotions. Its really pretty straight forward.

Work done is the product of applied force and distance moved along the line of the force applied. Or in algebraic notation: W = fs where W is work and f is force and s is distance.

Power is the time rate (speed) of doing work. Or: P = W/t where P is power and W is work done and t is the time it took to do the work.

Now by algebraic substitution: P = Fs/t

but velocity (speed of movement) is equal to distance moved divided by the time it took to move that distance.

Or: v = s/t.

Subbing again: P = fv where P is power and f is force and v is velocity. Power delivered to the pavement (or the dyno) is the force of the tire against the (road) surface times the velocity of your Ninjee; assuming no tire slip.

Finally, what if the motion is rotary? Then the force is delivered to the end of a radial arm or the surface of a "wheel" and the distance moved is the circumference of the wheel or circle turned. Work then is still equal to fs; but s is the distance moved around (and around) the circle at its outer edge. That total distance is the number of circles turned times the circumference of the circle. s = N(2 pi r), where s is total distance and r is the radius of the circle or the length of the radial arm (lever arm). (2 pi r) is therefore the circumference of the circle turned. N is the number of circles turned.

Work is now equal to the product of force and radius and 2 pi and N. Power is still equal to the work done divided by the time it took to do it or: P = W/t = fN(2 pi r)/t.

OR, by substitution and regrouping: P = fr N(2 pi)/t

Now fr is the force (applied at the circumference) times the length of the radial arm (radius of circle) or the torque produced to cause rotary motion and....

N(2 pi) is the number of revolutions through a full circle (an angle of 2 pi radians or 360 degrees) and...

t is still the time it takes to complete the movement. Therefore, N(2 pi)/t = w or rotational velocity.

VIOLA! Power is equal to torque times rotational velocity (revolutions per minute, radians per second, etc.) or
P = Tw where T is torque and w is rotary velocity and P is still power being produced at the contact surface. Your number 5252 is just a correction for unit conversions between feet, lbs, minutes, and horsepower.

In the rpm range where the torque curve is "flat" then the horsepower will be directly proportional to the rpm. Torque is relatively constant and as rpm increases so will hp. Double the rpm and the hp would double. But torque curves are not always flat are they? Where torque and rpm are both increasing, hp will be rising precipitously; but where torque is dropping off even as rpm continues to climb, hp may level off or even decline.

In short, if maximum torque doesn't occur where rpm is maximum then maximum hp will not be the product of maximum torque and maximum rpm. Hp will be, quite possibly, considerably less than could be generated under that false assumption and simplistic calculation.

The dyno will tell you where torque is max and where hp is max, as you know. The rpm at which those maximums occur may be quite different. Numerous characteristics of the engine's mechanical construction and fuel chemistry and exhaust/intake function, etc will determine where the peaks occur and their quantification.

Hope this is helpful.

* Last updated by: PaulAB on 3/11/2013 @ 10:36 PM *

Quite edifying AB. And thank you for not using pi r squared... because, as we all know, pi r round.

;)

Its really good to see that Hagrid has kept his sense of houmor, with all thats going on with his cycle...

Great day in the Mornin, Hagrid :)

Bob

trivial but interesting; below 5252 rpm, the torque will always be higher than the horsepower. Above 5252 rpm the horsepower will always be higher, and at 5252 rpm they're going to be equal.

James Watt, the guy that came up with the neat formula for defining horsepower was a pretty slick character. He went out and took a average of how much weight a typical old farm horse could lift and how fast he could lift it and came up with the 150lbs/220 feet/1 minute basis which mathematically became what we know as 1 horsepower = 550lbs/ft/second or, since we measure horsepower in revs per MINUTE it becomes 1 horsepower = 33,000 lbs/ft per minute.

So where did that 5252 constant come from? Pretty simple really: that old horse that Dr. Watt had out back lifted that 150 lbs to a height of 220 ft in a minute. Since Dr. Watt didn't have a 220 ft rake handle to demonstrate his theory he did the mathematical conversion and came up with 33,000 lbs on a 1 foot fulcrum (it's rumored that he used a old wooden cooking spoon that he stole from his wife's kitchen). That 1 foot fulcrum, when rotated around in a circle like a turning wagon wheel, would cover a distance of 6.283186 feet (that's pi times the diameter of 2 feet since a circumference is calculated by multiplying pi times the diameter). So he figured that 33,000 pounds covering a distance of 6.2832 feet in a minute would give a conversion constant of 5252 each and every time he did this test with a 1 foot fulcrum. He did all of this because he was also considered the inventor of the modern steam engine (not the old unharnessed things the Romans used) and he needed some way to tell people that his steam engine could do more work than their horses. He didn't actually invent the steam engine but he was the guy that harnessed the power through cylinders and similar hydraulic devices.

So even today, a dynomometer is only measuring the force applied to the turning wheel and through mathematical calculations figures out the horsepower output using the same old numbers that James Watt used back in 1848.
HP= torque x RPM / 5252 and Torque= HP x 5252 / RPM.

1 horsepower = 550lbs/ft/second or, since we measure horsepower in revs per MINUTE it becomes 1 horsepower = 33,000 lbs/ft per minute

This is valid stuff, but...

I'd like to meet the horse that could lift a 1 lb block 33,000 ft in one minute. Yeah. Go Ninjee? That's what I'd name that critter!

Lol... critter.

Thx Meb!

"trivial but interesting; below 5252 rpm, the torque will always be higher than the horsepower. Above 5252 rpm the horsepower will always be higher, and at 5252 rpm they're going to be equal"

Are you talking specifically about the 14R engine here? If not I can't see how this is correct. On my Toyota LC (4.2L TD) the power and torque are equal at around 3700rpm. On the 4.5L petrol engine it's around 4000rpm. Below those RPM's the torque is indeed higher than the BHP and vice-versa but the 5252 rpm figure seems meaningless when applied to those engines.

*Work done is the product of applied force and distance moved along the line of the force applied. Or in algebraic notation: W = fs where W is work and f is force and s is distance.

Power is the time rate (speed) of doing work. Or: P = W/t where P is power and W is work done and t is the time it took to do the work.

Now by algebraic substitution: P = Fs/t

but velocity (speed of movement) is equal to distance moved divided by the time it took to move that distance.

Or: v = s/t.*

Great insight, Paul. For a CLEP college test I had to have I recently studied these very equasions, such as that of work, (force(distance)) velocity , etc, or classical mechanics. And the 14R is certianly bigger than atoms but not quite as fast as light...LOL. But seriously, its interesting how you point out that work, force, and velocity are really what engine performance calls horsepower, torque, and RPM. That makes it more understandable, at least to me.

And the fact that power is work, but also with time factored in, makes engine dynamics even more interesting given how different engines hit their peaks at differing RPMs.

Very good input. Appreciated.

mebgardner, that would certainly be a impressive NINJEE horse, haha. I tend to embellish my stories to make them a bit entertaining. To put the tale into a more factual perspective we can imagine a plow horse lifing a burlap tarp loaded with grain that weighed 550 pounds. The horse would lift that grain 1 foot in 1 second when he hears the whip crack. That would be the equivalent of 33000 pounds per foot per minute... or 1 Horsepower.

Hey Pegscraper,

Im sure this is for all engines. The graph is just an easy way for us to see what is happening but if you actually do the math at 5252 is where HP will always be greater then torque and it will never catch back up.

That Youtube video above was pretty good. I hate that kid for being so smart. But I bet he cannot build a VMware infrastructure to support over 1200 virtual machines with mulitsite failover and network it all together ;-p or get a girlfriend :-D

ZX14MAN64 writes:"Very good input. Appreciated."

Thank you, but you make the mistake of encouraging me, heh,heh.

Starting from where I previously only paused (LOL): perhaps you would like to examine the magic number of 5252.
You may note it carries no units in your application but it should carry units of lb*ft (ft*lb if you want to reference true units of torque); either will do since we are ignoring vector notation and only calculating qualitative magnitudes.

5252 is required as a factor in the conversion to horsepower by reason of the units used in the calculation of the values of hp, torque and time:

Starting from Power = Tw where P is power in ft*lb/sec and T is torque in lb*ft and t is time in seconds; then:

P = Tw = T (RPM*2pi) because a revolution is 2 pi radians or one circle of rotation, so w = RPM*2pi

But that equation returns power in units of ft*lb/minute and must be divided by (60 seconds / minute) to return units of ft*lb /sec. We need the value in ft*lbs/second because a hp is defined (as has been eloquently explicated) as 550 ft*lb/second. So as an intermediate we get:

P = (T(RPM*2pi)ft*lb/min)/ 60 sec/min = (T(RPM*2pi)/60) ft*lb/sec
This gives us power in ft*lb/sec. Now we must divide by (550 ft*lb/sec)/hp to convert to hp.

So, proceeding: P = (T(RPM*2pi)/(60)) ft*lb/sec / (550ft*lb/sec)/hp = (T(RPM*2pi)/(60*550)) hp. Finally we are in hp!

Simplifying: Power measured in hp = (T(RPM) *2pi) / 33,000
Where is the magic number? 33,000 = 2PI* 5252! Viola!!

Put another way with algebraic expression:

P = (T * RPM * 2PI)/(2PI * 5252) = (T*RPM)/5252; power valuation returned in hp. This is, of course, your opening statement. "Torque times RPM divided by 5252 (?) = HP" Torque must be in units of ft*lb of course (or lb*ft, if you prefer and ignore vector notation.)

Now it is an artifact of the units of conversion in use in this equation that any time RPM exceeds 5252 then the HP valuation will exceed the given value for the amount of torque. This is simply because:

Power in hp equal any torque value times RPM divided by 5252. HP = T * (RPM/5252).

When RPM is equal to 5252, then, in the equation, the fraction that is multiplied by T has the value 1; and P valued in hp would equal the value of the torque measured in lb*ft. Obviously a lower value for RPM than 5252 returns a fraction valued at less than 1 and a higher value of RPM than 5252 returns a fraction greater than 1. Once RPM exceeds the value of the unit conversion factor, values of hp will always exceed values of ft*lb of torque since the fraction will always be greater than 1. So what? They are two different quantities measuring two different concepts. Power is "how fast you can do work" and torque is simply "how hard you can twist/turn a body" rotationally. Torque is the dynamic equivalent of force in their roles of causing accelerations respectively either in rotation (torque) or in translation (force).
Linearly, either greater force or greater velocity or both in combination will generate more power. Rotationally, either greater torque or greater RPM or both in combination will generate more power.

Once again, hope this is helpful!

Now where's that dead horse I left around here somewhere......sorry folks......

* Last updated by: PaulAB on 3/13/2013 @ 12:09 PM *

*Linearly, either greater force or greater velocity or both in combination will generate more power. Rotationally, either greater torque or greater RPM or both in combination will generate more power. *

That summarizes it well. Thanks.

The fact that when RPM exceeds 5,252 the engine is enhancing its the *seed of horsepower (torque) by combining it with velocity (rpm), and HP output will exceed the torque.

But when the RPM is below 5,252, the available torque is then not yet used as efficiently, by not being combined with enough velocity (rpm) to generate more HP than torque.

To me, the term synergy comes to mind.

* Last updated by: ZX14MAN64 on 3/13/2013 @ 7:54 PM *

"The fact that when RPM exceeds 5,252 the engine is enhancing its the *seed of horsepower (torque) by combining it with velocity (rpm), and HP output will exceed the torque.
But when the RPM is below 5,252, the available torque is then not yet used as efficiently, by not being combined with enough velocity (rpm) to generate more HP than torque."

Sorry, but I have to take issue with "efficiently" in the above. 5252 is just the conversion factor for the units used in the measure of hp and Torque and time. Its not a matter of efficiency; as a matter of fact the physics equations do not consider efficiency at all, nor any parameters of the engine's operation. These figures assume 100% efficiency, if you will, and merely state the mathematical relationship between the concepts; because of the way the units are chosen (defined), the ratio of RPM to 5252 is necessary to reconcile the independent choice of units for torque and horsepower (hp based on the average power output of a workhorse in the pre-industrial days). There is no significance to the absolute value (quantization) of either value with respect to the other. If hp was defined in terms of the work done by shetland ponies or clydesdales instead, then the number 550 would be smaller or larger respectively and the magic number would be smaller or larger than 5252 and finally, the rpm at which the absolute value of hp being produced always exceeded the absolute value of the torque would also be smaller or larger respectively. It is "trivial" as someone has already pointed out. I only wanted to explain its origin.

Best wishes,
PaulAB

* Last updated by: PaulAB on 3/13/2013 @ 7:55 PM *

Very interesting and educational. Thanks for the clarification.

So the term efficiency only comes into play with the engine's other variables (intake, exhaust, A/F ratio, etc) alter the torque/ Horsepower correlation as displayed on a dyno, correct?

Such as to what rate HP begins to fall or rise at a certain point?

* Last updated by: ZX14MAN64 on 3/13/2013 @ 8:01 PM *