Need more power

[TheIronWarrior]

The _power_ output of an engine is the only thing that matters to performance (acceleration as well as top speed).

I will begin by assuming by "power output" you are referring to the entire power curve and not just the reported peak power figure so as to not get in to the finer details of engine performance. We will also ignore the effects of wind resistance, friction, weight, etc. for simplicity's sake, even though these have a huge impact on performance (acceleration as well as top speed).

Your statement above is not accurate. Take two identical motorcycles and change the final drive gear ratios on one of them and one will undoubtedly accelerate faster than the other. This comes as a trade-off as you will lose out on top speed. Identical engines with different gear ratios WILL have different acceleration and top speeds. This is a fact. Many other things can have an impact on performance as we are talking not about the performance of the engine as a separate unit, but of the entire system of a motorcycle.

Providing his determination that he "needs more power" was made based on lack of acceleration and he is willing to lower the top speed of the machine proportionally, changing the final-drive gearing could provide the answer he is looking without increasing the performance characteristics of the engine. From a strictly technical point of view, this changes the relationship between the motorcycle's speed and the power output of the engine, as you are moving the desired speed range up into the higher power band of the engine. With this in mind, you could argue (semantically) that this is an increase in 'power' but that's more a language-based argument than physics-based.

 

[Jos]

If you shorten transmission you get higher torque (to the wheel), lower rev, and same power. Power = torque x rpm and thus juste follows the torque (engine) curve as rpm increases.

The NC 750 has decent torque, it's power is limited by the low max rpm. In road use, it is a far better choice than small 4 inline that may produce more than 100 bhp but at very high rpm due to limited torque.

To me torque is the real thing. I own two other more powerful bikes but love the engine response of the NC. Of course for track use I use the triple.

 

[670cc]

Your statement above is not accurate. Take two identical motorcycles and change the final drive gear ratios on one of them and one will undoubtedly accelerate faster than the other. This comes as a trade-off as you will lose out on top speed. Identical engines with different gear ratios WILL have different acceleration and top speeds. This is a fact. Many other things can have an impact on performance as we are talking not about the performance of the engine as a separate unit, but of the entire system of a motorcycle.

Oh, I love these power debates!

So in the statements above, changing final drive ratios resulted in a trade off between acceleration and top speed. But, what if the manufacturer widened the ratios, or lowered the final drive ratio and added another gear on top, such that the acceleration was there at the low gear ratio, but the top speed was still there at the high gear ratio? Do you get the best of both worlds with no trade off?

 

[Jos]

You can't have your cake and eat it. If you shorten you will hit the rev limiter before you get to top speed. If you lengthen you will not get to the top speed as you will not have enough torque (at the wheel) to overcome air resistance.

If you have short first gear(s) only you will have a large drop of rpm when switching that will put you out of the torque curve and make the bike sluggish.

Factory ratios usually are best for road use. I played with this on one bike, and could not decide if it was better or worse, just different. If really you need more torque at the wheel then you may shorten secondary transmission, top speed not being an issue IMHO.

 

[TheIronWarrior]

So in the statements above, changing final drive ratios resulted in a trade off between acceleration and top speed. But, what if the manufacturer widened the ratios, or lowered the final drive ratio and added another gear on top, such that the acceleration was there at the low gear ratio, but the top speed was still there at the high gear ratio? Do you get the best of both worlds with no trade off, or something for nothing?

Consider this a 'thought-experiment':
Let's take Bike A to be a completely stock machine and Bike B will have a modified transmission.
For simplicity's sake, let's assume final drive ratios, as well as top and bottom gear ratios remain the same. This way, the top speed of the bike will remain unchanged, as will the speed/torque in 1st gear.
Let's then add a few more gears to the trans (arbitrarily say 3 more gears) and adjust the ratios accordingly so they're relatively evenly distributed. We now have a completely stock Bike A and a Bike B that has a 9-speed transmission.
Bike B will now have less of an RPM drop between gear changes, keeping the engine operating higher in the power band. Theoretically this will increase acceleration. I say theoretically because now you have more gear changes, increasing the time that there is no power to the wheel at all (clutch in). I would suspect that someone somewhere determined 5 or 6 gears in the transmission of a typical motorcycle hits the balance well, as these seem to be the most common numbers of gears.
So assuming a shift of the gears takes exactly zero time (impossible) you could consider that getting "something for nothing" in a perfectly theoretical world however this is not possible in real life.

 

[TheIronWarrior]

You can't have your cake and eat it. If you shorten you will hit the rev limiter before you get to top speed. If you lengthen you will not get to the top speed as you will not have enough torque (at the wheel) to overcome air resistance.

As an addition to the above, I had read somewhere, and experimented with, top speed on my NC700 (closed course in Mexico, of course...) and have found that you can actually hit a higher speed in 5th than in 6th for this exact reason.

 

[670cc]

Consider this a 'thought-experiment':
Let's take Bike A to be a completely stock machine and Bike B will have a modified transmission.
For simplicity's sake, let's assume final drive ratios, as well as top and bottom gear ratios remain the same. This way, the top speed of the bike will remain unchanged, as will the speed/torque in 1st gear.
Let's then add a few more gears to the trans (arbitrarily say 3 more gears) and adjust the ratios accordingly so they're relatively evenly distributed. We now have a completely stock Bike A and a Bike B that has a 9-speed transmission.
Bike B will now have less of an RPM drop between gear changes, keeping the engine operating higher in the power band. Theoretically this will increase acceleration. I say theoretically because now you have more gear changes, increasing the time that there is no power to the wheel at all (clutch in). I would suspect that someone somewhere determined 5 or 6 gears in the transmission of a typical motorcycle hits the balance well, as these seem to be the most common numbers of gears.
So assuming a shift of the gears takes exactly zero time (impossible) you could consider that getting "something for nothing" in a perfectly theoretical world however this is not possible in real life.

Or, we could just use a Continuously Variable Transmission to eliminate the gearing steps, and have the engine constantly operating at it's peak horsepower RPM throughout the entire road speed range. The benefit of this is, of course, ignoring any additional losses in the CVT drivetrain vs traditional gears.

 

[MZ5]

I will begin by assuming by "power output" you are referring to the entire power curve and not just the reported peak power figure so as to not get in to the finer details of engine performance. We will also ignore the effects of wind resistance, friction, weight, etc. for simplicity's sake, even though these have a huge impact on performance (acceleration as well as top speed).

Your statement above is not accurate.

Yes, it is. Your assumptions and entire comparative setup are fallacious. The correct and useful comparison is two engines (or powerplants; don't assume an engine vs. a motor, nor anything else, because it's irrelevant), one with 30% higher power output than the other, and both power 'curves' completely flat. The engine with greater power output will result in greater acceleration at any given vehicle speed. The engine speed is irrelevant since the power curves are totally flat.

Your argument is fallacious because you've assumed many things, including that more power is made at higher rpm. That is normally correct, and that is why the lower-geared vehicle accelerates more quickly; your assumptions have caused the powerplant to make more power, which allows the tranny to exchange speed for more wheel torque.

He needs more power.

 

[b_rubenstein]

Consider this a 'thought-experiment':..

We don't need though experiments; work, power, forces, HP, torque, etc., are settled defined things. Anyone who took high school physics was taught that F=MA, therefore A=F/M. There's no power in the formula, because power is a scalar quantity, and force is a vector and they are different things. There are formulas to calculate HP from Torque and vice-versa and can be hound here.

 

[TheIronWarrior]

Yes, it is. Your assumptions and entire comparative setup are fallacious. The correct and useful comparison is two engines (or powerplants; don't assume an engine vs. a motor, nor anything else, because it's irrelevant), one with 30% higher power output than the other, and both power 'curves' completely flat. The engine with greater power output will result in greater acceleration at any given vehicle speed. The engine speed is irrelevant since the power curves are totally flat.

Your argument is fallacious because you've assumed many things, including that more power is made at higher rpm. That is normally correct, and that is why the lower-geared vehicle accelerates more quickly; your assumptions have caused the powerplant to make more power, which allows the tranny to exchange speed for more wheel torque.

He needs more power.

I never said a more powerful engine would NOT accelerate faster.
I also think it's comical that my assumptions were fallacious, yet your "correct and useful" comparison uses two powerplants with completely flat power curves. This is not even theoretically possible without some severe throttling in the higher RPM ranges, which would kill any performance anyway. Your "correct and useful" comparison has zero real world applications, where as my comparison which is clearly "fallacious" has been shown in many cases to produce the results noted. (try a Google search for 'gearing and acceleration' and see what you can find, there's plenty out there).
Please tell me more about how my comparison of two identical machines with different gear ratios having different performances is wrong, I'd love to hear about it.

I will certainly concede that I may have misread your initial post. I understood you meant he needed an engine with a larger power output. If your intent was to say that he needed more power delivery to the wheels at a given time, that would be correct and this is achievable in many ways, not just with a more powerful engine.

Bottom line is that changing performance of the entire system of a motorcycle is not limited to only changing the engine.

 

[TheIronWarrior]

We don't need though experiments; work, power, forces, HP, torque, etc., are settled defined things. Anyone who took high school physics was taught that F=MA, therefore A=F/M. There's no power in the formula, because power is a scalar quantity, and force is a vector and they are different things. There are formulas to calculate HP from Torque and vice-versa and can be hound here.

The thought experiment was to illustrate the situation posed by 670cc regarding effects of different gearing. Just an example without having to build a 9 gear transmission and try and install in in my bike to show the results.

For the below, I am going to ignore the effects of wind resistance, friction, etc. for simplicity sake.
There is absolutely no power term in the simplistic F=ma in the strictest sense, however we must understand where the Force value comes from in the particular case we are investigating:
In the particular case of a motor vehicle, the F comes from the torque delivered to the wheels. F = tw (torque at wheel) / r (radius of wheel).
The torque delivered to the wheel is scaled by the gear ratio. tw = te (engine torque) * GR (gear ratio)
As you linked, you can relate torque to power through RPM. P (power, in HP) = te * N (RPM) / 5252
From the above equations, we can easily relate (in the case of a motor vehicle) engine power to acceleration. All the junk we ignored earlier (friction, wind resistance, etc) will scale the resulting values of course, adding in factors dependent on speed, air density, etc. but the relationship between power and acceleration cannot be ignored. Power and torque are intimately related, and torque (being a vector quantity, and the cause of the "force" driving the acceleration) can certainly be related to the acceleration of the vehicle.

 

[TheIronWarrior]

Or, we could just use a Continuously Variable Transmission to eliminate the gearing steps, and have the engine constantly operating at it's peak horsepower RPM throughout the entire road speed range. The benefit of this is, of course, ignoring any additional losses in the CVT drivetrain vs traditional gears.

I was going to bring up CVTs but thought that might further complicate my (already probably unnecessarily complex) scenario. I am glad you brought them up though because in both the theoretical (ignoring additional losses, etc.) and in real world testing (to the best of my recollection, but don't ask me to cite references..) the CVT has been shown to be superior even to the manual transmission for both performance and fuel economy. I hate to admit it because I'm a die-hard manual traditionalist (and evangelist, I have converted my wife to the cult of stick-shift), but a well-designed CVT is really the transmission of the future in my opinion. Assuming we don't all transition to some space-aged nuclear-drive automobile in the relatively near future.
I think my love for the manual stems mostly from my control-freak tendencies, I want to make the decision on the gear ratio, not let the car do it for me!

 

[MZ5]

Power at the wheel is a result of power at the engine. It is not impacted by gearing. Gearing doesn’t change power, it merely makes an exchange between force and rate. That is why engine power is all that matters to acceleration and top speed both. The CVT illustrates this well. Max acceleration is achieved when the engine is operating at max power, not max torque. That’s one reason racing engines are high-rate, low-force items: Power is king, and those engines rely on the tranny to make the exchange necessary to put the maximum possible force at the contact patch at any given vehicle speed.

I observe that people get confused when they conflate powerplant needs with wheel or contact patch needs.

 

[670cc]

Power at the wheel is a result of power at the engine. It is not impacted by gearing. Gearing doesn’t change power, it merely makes an exchange between force and rate. That is why engine power is all that matters to acceleration and top speed both. The CVT illustrates this well. Max acceleration is achieved when the engine is operating at max power, not max torque. That’s one reason racing engines are high-rate, low-force items: Power is king, and those engines rely on the tranny to make the exchange necessary to put the maximum possible force at the contact patch at any given vehicle speed.

I observe that people get confused when they conflate powerplant needs with wheel or contact patch needs.

I agree. Well said.

My blood boils when people say you need torque to accelerate and you need power for top speed, as if they are entities that operate independently. Torque * RPM / 5252 = HP

 

[Jos]

To me the issue is at which RPM you get max power (typically close to max RPM before the torque starts collapsing), and what kind of power (hence torque) curve you get in between. I prefer engines with decent power at reasonable RPM and flat torque curve (like the NC) than caracterial engines with power that spikes close to the rev limiter but that you have to keep there for good response.

Or the best of both, a 1L 4 in-line, but 190 HP is not reasonable on a bike

 

[TheIronWarrior]

That is why engine power is all that matters to acceleration and top speed both.

Are you saying that two engines, both producing the same amount of power, will have identical acceleration and top speed regardless of torque, RPM, and gearing?
Arbitrary numbers below to illustrate the point:
Engine A - 200 HP, 2000 RPM
Engine B - 200 HP, 20000 RPM
Identical acceleration?

Engine A - 200 HP, gear ratio 1
Engine B - 200 HP, gear ratio 5
Identical acceleration?

I am finding that extremely hard to believe...

 

[670cc]

Are you saying that two engines, both producing the same amount of power, will have identical acceleration and top speed regardless of torque, RPM, and gearing?
Arbitrary numbers below to illustrate the point:
Engine A - 200 HP, 2000 RPM
Engine B - 200 HP, 20000 RPM
Identical acceleration?

Engine A - 200 HP, gear ratio 1
Engine B - 200 HP, gear ratio 5
Identical acceleration?

I am finding that extremely hard to believe...

The question was not directed at me, but my answer is:

Engine A - 200 HP, 2000 RPM
Engine B - 200 HP, 20000 RPM
Identical acceleration? Yes. The RPM is irrelevant. If engine A makes the same horsepower at 2000 RPM as engine B at 20000 RPM, simple math says that the torque output of engine A is 10 times that of engine B. The final product is the same 200 HP. For a proper useful wheel speed, the transmissions will need to be geared differently by a factor of ten, and ignoring friction losses and power curves (since only one RPM point was specified), the rear wheel horsepower is the same on both bikes, and the acceleration is the same.

Engine A - 200 HP, gear ratio 1
Engine B - 200 HP, gear ratio 5
Identical acceleration? Insufficient information to fully answer the question. However, the power output at the rear wheel would be the same in both cases. Changing the gear ratio just inversely changes RPM vs torque in the power equation. If gear ratio 5 causes a higher rear wheel speed than gear ratio 1, it also reduces the torque by the same factor, thereby keeping the rear wheel power the same.

 

[TheIronWarrior]

Engine A - 200 HP, gear ratio 1
Engine B - 200 HP, gear ratio 5
Identical acceleration? Insufficient information to fully answer the question. However, the power output at the rear wheel would be the same in both cases. Changing the gear ratio just inversely changes RPM vs torque in the power equation. If gear ratio 5 causes a higher rear wheel speed than gear ratio 1, it also reduces the torque by the same factor, thereby keeping the rear wheel power the same.

Sorry, should have stated "All else being equal."
Correct, rear wheel power will be the same, as power is independent of gearing.
Also correct, rear wheel torque will differ by a factor of 5.
As the torque at the rear wheel is what drives the acceleration (or you could think of it as power if you want, but you need to also factor in the rear wheel rotational speed, which is affected by the gearing, and you end up using the rear wheel torque anyway) the accelerations will be different.

Anyone who doubts the effect of gearing on acceleration should try the following:
Take your car/motorcycle and reach a constant RPM in 1st gear. Fully open the throttle and gauge the acceleration.
Then, reach the SAME constant RPM in top gear. Fully open the throttle and gauge the acceleration.
You will notice that the acceleration in 1st gear is much greater than the acceleration in top gear because GEARING AFFECTS ACCELERATION.

This is a valid comparison because an engine operating at WOT at a given RPM will produce the same power and torque (which is why dyno charts plot T/HP vs RPM) and the effects of losses due to wind resistance, etc. cannot account for the full difference.

 

[670cc]

and the effects of losses due to wind resistance, etc. cannot account for the full difference.

You greatly underestimate the effects of rolling resistance and aerodynamic drag. The motorcycle has a terminal top speed because at that point ALL of the engine's horsepower is used to overcome the mechanical losses (internal heat), rolling resistance and aerodynamic drag, and there is no excess power left to accelerate it. As you speed up between standstill and top speed, more and more power is lost to friction and drag. Air resistance is proportional to velocity squared, if I remember correctly. Twice the speed, four times the drag. Acceleration in top gear vs 1st gear is vastly different due to friction and drag losses.

With an example motorcycle, the engine at WOT and at 6000 RPM puts 50 horsepower to the rear wheel in 1st gear. At WOT and 6000 RPM in 6th gear, it also puts 50 hp to the rear wheel. The gearing change from 1st to 6th that allows the rear wheel to spin faster relative to the engine speed also reduces the rear wheel torque by the same factor, leaving the horsepower number the same. However, in 6th gear, the vehicle is moving faster through the air and over the ground. More power is used to overcome the rolling friction and aerodynamic drag, so the acceleration is less. It's that simple.

If we could find a way to levitate above the ground and ride in a vacuum, yet somehow still propel the vehicle, we wouldn't need to have this friendly, educational discussion.

 

[TheIronWarrior]

Let's consider the following scenario:
We will have two bikes so designed that all losses are equal when Bike A is travelling twice the speed of Bike 2. This is completely possible.
Both bikes will also have equal mass by design. This is also completely possible.
We will also design these bikes such that the engines are producing identical power when Bike A is travelling twice the speed of Bike 2. This is again completely possible.
Let's put the same diameter wheel on the back, such that the wheel RPM of Bike A is twice that of Bike 2 when Bike A is twice the speed of Bike 2. Again, easy peasy.
As discussed, power relates to torque through RPM, meaning that Bike A travelling faster (and thus, having twice the RPM) is now putting half the torque to the ground as Bike B.
Tell me, which will have the greater acceleration, A or B?

My line of thought:
F = m * a
As all losses are made equal by design, the only variable in the F term is the torque.
This means that the net force on the bikes is proportional to the wheel torque only.
If we are putting down twice the torque, we are getting twice the acceleration with both bikes producing the same power.