You want torque, but you want it fast, so your bike accelerate quickly.
Engine torque is a meaningless figure, wheel torque is all that matters.
Torque sucks in 6 th gear but is awesome in first, so you want to hold those gears longer.

 

I realized I made a mistake in my HP calculation above.
so to those of you wondering why formulas above don't match, here's why:

I forgot to convert revolutions to radians.
Radians are a unitless number. By definition, there are 2*pi radians in one revolution.

So if you multiply the above calculation by 2*pi, you get

HP= torque *(RPM/60)/550*2*pi
or
HP = torque * RPM/5252

Sorry for the confusion. I hope I haven't added more!

 

I need to read and think more about horsepower and torque. This thread is fun, but I seem to be doing a less than stellar job of making myself understood.

I recognize the real world, rubber meets the road advantage of higher rpm/same torque engines. The author of the above linked article claims that advantage is just gearing.

So I'm wondering out loud, does the higher rpm engine have an inherent advantage or just advantage (as the author seems to claim) due to technology limited variables like a transmission with different gear ratios that must be shifted?

The example was two possible in the real world internal combustion engines, same max torque, one at 8k rpm and one at 16k rpm. The transmission/power control technology I imagined would hold each engine's rpm at the torque peak, 8k and 16k respectively, period. Power would be modulated (like reverse slipper clutch?) and max power applied (without breaking traction) to the rear wheel by an infinitely variable auto transmission.

In this example there would be no shifting down of the 16k engine because 1) there is nothing to shift. 2) the rpm is held at 16k constant and 3) 16k rpm IS max torque for this engine.

The author of the linked article claims that the advantage for the engine with same torque but higher rpm is just gearing. Is he correct? Seems counter intuitive to me but I don't know, I'm asking. "Cause if he's correct, and with the "gearing" advantaged removed, wouldn't both bikes accelerate at the same rate?

Is the author wrong? Am I missing something here?

 

I disagree. Torque and RPM combine to produce power.
If the two engines produce the same HP. One at 8k rpm and the other at 16k rpm, then the engine at 16k rpm is outputting half the torque.

If they are both spinning the rear wheel at the same speed, the rear wheel torque and HP output are identicle. That's why a 600 can have the same peak HP as a liter bike if the liter bike engine can't spin as fast. But you have to keep the 600 high in the rev range.

On the other hand, if both engines are outputting the same torque, then the 16k rpm is outputting twice as much power.

The HP at the rear wheel is equal to the HP at the crankshaft minus the power lost through the gear train (heat due to friction).

 

Absolutely, take the tranny off by a direct drive and both bikes accelerate at the same rate, but the 16000 rpm bike will keep accelerating while the 8k guy is maxed out to half speed.

The engine produce energy and the tranny make it usable across the speed range, multiplying or dividing torque and RPM

Totally, but quite an advantage if you ask me a free torque multiplication
Not free since it cost RPM

See it this way the engine give torque and useless RPM, trade this RPM in the tranny for more torque.

 

You would like to stay in that first gear all the way trough 120 MPH, don't you. But you have to shift, the engine don't have that power.

 

OK, but for the example I gave... both engines same max torque, both engines held at max torque rpm, both infinitely variable trans putting maximum force (minus traction control bleed off) for acceleration to the rear wheels while holding both engines at their respective torque peak rpms.

Seems to me the only two limiting factors are max torque (identical for each bike) and rear tire traction, again identical. I understand the advantage with real world transmissions on the street, but in this hypothetical, who is multiplying what to give the high rpm bike an advantage?

 

That infinitely variable tranny has a mutilplication ratio as well to divide the RPM by 2 to hold the engine at his peak since it's twice higher. That same multiplication ratio double the torque output to the rear wheel.

 

gboezio;

Maybe you're right, you seem sure of what you're saying. I admit I don't get it.


Both bikes dyno 100 ft. lbs. torque at the rear wheel, one at 8k rpm, the other at 16k rpm.

When the 8k bike reaches sufficient road speed that full force can is being transmitted to the rear tire, it is making 100 ft. lbs. of torque at the rear wheel, yes?

But somehow, the 16k bike, which also dynoed max 100 ft. lbs. at the rear wheel is now making 200 ft. lbs. at the rear wheel?

 

We can look at it another way. If I want to go 165mph to the bank it's going to take 3 to 5 miles of unabstructed asphault to get to that speed. I'll take the Interceptor. Also going to take 3-5 to slow it down.

But the bank is only two miles away. I think, very seriously, this VTR will do a 100mph in a hundred yards or close to it. It also makes a lot of noise and they're all used to Hardlys but not one that is THAT red or that fast.

Wanna take it one step further, Rand? Why is the v-pattern cylinder arrangement better than the inlines? In none-track conditions - road pitch and yaw, traffic, debris, the v-design engines are the smoothest both on and off the throttle.

 

Well, this is all fascinating **** that I want to study and learn more about.

I'm real clear about the advantage that same torque but higher horsepower/rpm engines have in the real world on the street.

Here's what I still don't know;

Is that advantage inherent and absolute? Or is "horsepower" a calculated mechanical advantage for making the most out of the limitations of available/practical transmission technology?

 

Yeah I think our application is a tad complicated
I made this image to clarify to make sure we talk about the same thing, I oversimplified the tranny by using a chain since we just care about the gear multiplication we are in tu run at peak hp.
You got to love my beautiful V-twin

I might add the energy (hp) is produced by the engine and can't be altered by the driveline (except friction losses), it's components (torque and RPM) can be traded one for the other in that said driveline.

 

gboezio;

That is indeed a very handsome looking V-twin(s)! And I appreciate you having taken the time to draw it.

But since the starting point for my "shootout" between the 8k and 16k bikes was that they each make 100 ft. lbs. max torque at the REAR WHEEL, I still don't get it!!!

 

I'm done thinking about horsepower and torque for the evening. I'm going to take my VTR out for a ride and experience some of it. I'm smarter than the average bear, but might have a mental block on this stuff. I once asked a close relative to explain it to me.

When he got done, I still didn't understand it to my satisfaction. And he was a professor/chairman of the mechanical engineering dept. of a major university at the time.

 

Ah, equal torque, equal speed = equal horsepower.
But something have to give, the torque have to be reduced to 50 lb/ft on the engine reving 16000 to be equal at the wheel, or go twice as fast, if not, the engergy can't be lost and the clutch in the tranny is taking 50 % of the power output as friction and heat.


No problem at all it was fun

 

This is a great thread.
I want to talk a little more about torque.
I talked about a flat torque curve. The definition of a flat torque curve is that torque does not vary with RPM.
so if we had an engine that had a completely flat torque curve, then the acceleration of the motorcylce would be constant across the RPM range at wide-open throttle. Rear wheel torque is constant driving force is constant and F=ma. This constant torque makes the bike easy to ride hard. If the torque increases with RPM, you get the top end rush. If you have as much torque as you have traction, this can be scary. It torque decreases with RPM, you have a Harley. Stupid motorcycles... anyway. Relatively flat torque is what makes almost all modern sportbikes so much better than they used to be.
The twins have relatively low redlines, They can produce lots of torque at low RPM, so they are always ready to jump, when you open the throttle.

A continuously variable transmission, that keeps the engine at maximum horsepower, seems like it would be pretty scary on a high-output motorcycle. I'd rather have constant torque, and then gear it for the best drive off of corners, without spinning the tires over a wide range of engine RPM.




psssst...
I'm buying an RC51.
Two ends of time will be neatly tied...

 

Bottom line, the lower in the RPM range your torque comes on the quicker you'll get moving (or pulling a wheelie). big cylinders like the superhawk can make more torque down low because the cylinders are twice the size of thier 4-banger counterparts. That equals more fuel for firing. Once you are making the torque the only thing that matters is how fast you can get through the rev range and on to the next gear. Big cylinders don't spool up as fast as small ones, that's why the 4-bangers eventually pull away (and they are firing at twice the rate).

 

This is true, lol.

 

I have often wondered the same things.

The one thing I have not seen on this thread is the practical definintion of torque and HP.

Think of torque as the ammount of force that a rotating shaft can produce. Since it is rotating force, you have to torque is measured in ft-lbs and after taking into account for the gearing of the bike (and assuming neglagable losses) can be converted to a "linear" force. Since F = ma and we can consider the bike/rider to be a constant mass, more force = more acceleration. The lower the gear the more linear force can be applied to acceleration!!!

HP is a completely different matter. HP is a measure of how much foce can be applied in a given time or ft-lbs/sec. I like to think of pumping water out of a well. Any motor can lift a gallon of water(torque), but when selecting a pump for a well you want to know how many gallons of water can be pumped every minute (HP). The same can be said for motorcycle engines. At high speeds (100+mph) a motorcycle is not acclelerating very much and force the motor is producing is making a hole in the wind. Since this is like pumping water from a well the ammount of torque is not as important as the ammount of HP.

As we all know high torque motors are the best when the light turns green or at the exit of a corner at reletivly low speeds, but as speeds increace high HP is needed to "beat the wind" and this is why the I4 bikes pass on the straights and those of us with twins pass on the exit of the corners.

Another way of thinging about it...

If you compared 2 bikes that were both displaced 1Liter, one a V-twin and another that was in I4. The twin has a 500cc chamber that will produce a bigger bang then the 250cc chamber of the I4. This bigger bang will produce more torque and linear force assuming same transimssion. This leads to more accleration. Since the I4 has smaller moving parts, it can produce more bangs per minute (RPM) then the twin and therefore more overall HP. This will give the I4 the advantage over the wind and more "top end" even with less torque.


The way I think of it, both types of engines are good, this provides the scentific justification for 2 bikes!!!

 

Most Dyno test are done with just the engine HP and Torque and then released to the public by the factory .Just dont add the gear box into the picture and make it easier for you!

 

The way it was explained to me, the reason that you would accelerate past the peak torque curve is to allow the next gear to begin closer, or within the power band to maintain acceleration. There is a point in the power curve where it provides no value to keep accelerating - this is typically just past the peak torque and as the torque curve starts to drop off more sharply.

Additionally, (just a fun fact) I read somewhere, sometime ago that the SAE, in measuring resitance in drivetrains or other energy transfer mechanisms determined that the roller chain is considered to be the most efficient in commonly used applications (as opposed to gears, hydrostatic drive, belts, air, etc.). Another reason why a higher quality chain and sprockets in good condition can actually make a difference in the acceleration and performance of a bike.

 

This makes a lot more sense, thank you. HP has more to do with time rather than performance. Where torque is how much force can be exerted, HP is how fast it can be done. Is that right?

 

Do I remember my physics properly? 1 hp is enough power to move 1 ton - 1 inch - in 1 second. Or is it 1 foot?

 

I think it's one foot.

 

OK thanks for that. Put that into perspective with the Superhawk with rider.

If you alot for the bike and the rider, generously, the grand total is .375 tons.

If you have 105 of them available, this thing almost becomes a time machine up to about 130 mph. Especially for a rider that is 155 to 165. I'm not talking twisties along the riverbeds but big long high speed curves on the long roads that we have in Texas.

It will do the riverbeds with some of the best but a well tuned Superhawk on roads that you can reliably run 75-80 mph averages (with intermittent bursts that will put you and the bike away) are going to really show you what the bike is worth.

 

There's only one thing to say about Horsepower and Torque. You can never have enough of either....