. Good link. This could become another good thought experiment such as some other topics here did. . http://www.cbr250.net/forum/cbr250-performance/7217-gearing-horsepowers-better-half-3.html . An engine torque curve and horsepower curve are really just two different ways of looking at the same thing. Power is just torque over time. People like to look at the torque curve to get an idea of what is going on below 5,000 rpm because it has better resolution on the graph due to the horsepower curve not yet having much multiplication from the low rpm. The power doesn't yet look like much and the torque looks much bigger but they are none the less showing the same information. And power is the real metric of how much work can be done. Torque can be changed any way you want with gearing but that never changes the power at the wheel because you are trading increased rear wheel torque for decreased wheel rpm. Increasing the rear wheel torque WILL make the bike accelerate quicker but when you get to the steady state top speed for any given load such as the wind on a flat top speed run or gravity on a big hill, the gear that will be fastest is the gear that puts the rpm at the power peak. Not the torque peak. And not the redline. Even a drag strip run which is all about acceleration and using the most torque multiplication, will be the quickest on a vehicle like the CBR250R when shifting the higher gears at an rpm band that is much lower than the redline and is centered on the power peak. As was generously explained in the other thread by DieselMaxPower . The OP's link is a little too simplistic when it comes to fuel efficiency. It is possible for an engine to take in the same amount of fuel and air (disregarding the much higher losses at high rpm) at 8,000 rpm as it is at 4,000 because of the throttle plate's restriction and resulting intake vacuum. The real reason for generally better fuel economy at lower rpm is from less frictional, reciprocating, and thermal losses than the higher rpm. And from being closer to the first torque peak. . The first torque peak usually shows where the engine is operating most efficiently as set up by the cam timing, intake runner tuning, ect versus linear piston speed losses, flame front travel and combustion gas pressure curves, ect,ect. A Brake Specific Fuel Consumption chart (Which plots 4 parameters; engine rpm along the bottom, the torque that is produced along the right, increasing engine load which can be roughly increasing throttle position up the left, and finally the topo values for how much fuel is being used versus how much power is produced.) is the only way to really know the engine's best efficiency range. They usually show the best efficiency at just off of full throttle on either side of the first torque peak. Which is the bad news for fuel economy. Most engines are way too powerful to operate at full throttle for more than a few seconds without going way too fast. So small engines with barely enough power, like our 250, are more often operated near their most efficient range than an engine with plenty of extra power. The only way to get the bigger engine into it's efficient range is to pulse with a high throttle opening and then glide with the clutch in, and then pulse again. But this an advanced hypermiling technique that is not for everyone. Better to stick to a CBR250R, PCX150 scooter, or new Ford Fiesta. . Thermal losses, pumping losses, friction losses and powerband tuning, among many other factors, all come into the design equation. Using the smallest combustion chamber and the fewest of them minimizes the area which can lose heat (wasted fuel) to the head. But too small of a bore reduces the force of any given combustion pressure because it offers less area to work on. So the tendency for engine design to have a square (equal bore/ stroke) design. Slightly longer stroke than bore will be more efficient as it will move the torque peak to a lower rpm because of the longer crank arm. Interestingly, a square single will have less combustion chamber area to pick up heat than a square twin of the same displacement. Multiply it out for yourself. It's cool. And so is the CBR250R's engine. I'm sure you have noticed that the engine and radiator don't throw off much heat compared to your other bikes. Air cooled engines can run cylinder head temps of 350F/ 175C which, compared to liquid cooled engines which are limited to 100C to prevent boiling, waste even less heat. . Heat out the exhaust is another waste. Which can be reduced with Atkinson cycle engines . http://en.wikipedia.org/wiki/Atkinson_cycle . (a fancy way of saying that the cam left the intake valve open too long.) which seek to have no pressure remaining in the cylinder by the time the piston reaches the bottom. And they also reduce pumping losses with lower intake vacuum. At the expense of power versus displacement which never looks good to bench racers reading spec sheets. . Honda motorcycles is fighting for better fuel fuel efficiency in the big bike west with the ultra long stroke CBR500 platform and ultra low rpm tuned NC700. Either one of which can almost match our already exemplary CBR250R at 68 mpgUS. BMW has always been a fuel efficiency leader. The Ninja300 is another sporty and all around good bike taking a step in the right direction with longer stroke and longer gearing. Some small 110cc air cooled bikes which are popular in India can sell for $1000 and get 150 mpgUS at 40 mph.

Here is a sample BSFC map from another vehicle and a chart of the optimum shift rpm's of the CBR250R. The shift points were charted by DieselMaxPower from a dyno printout to show when the engine torque as multiplied by the drive train at the wheel would be greater in the next gear. Notice that the top three shifts are giving the most average rear wheel torque and power when shifting at about 9,300 rpm. Redline is 10,500. . . . . Notice that the top three shifts on the CBR250R are giving the most average rear wheel torque and power when shifting at about 9,300 rpm. Redline is 10,500. . . . .

Here are a couple more BSFC graphs again showing the best efficiency for any engine is just below the first torque peak and near full load. While cruising with small throttle openings, a high powered engine is wasting half it's gas. . . . . . .

Thats intresting. Yes it's true that RPM isn't necessarly related to MPG. An engine should operate within it's comfortable RPM range for best MPG. For example, some engines will run at 4,000 RPM and get what some get at 2,000 RPM. Even though the fuel is injected more frequently at a higher RPM, a less amount of gas is injected due to a lower workload on the engine [as in not lugging it out where theres no power]. BUT, some engines that like low RPMs need to stay at low RPMs for best MPG. If all the powers down low, it's wasteful to rap it out and it works it without no power and more frequent gas injections. But yea depends on the engine. I personally don't care a lot about HP. Torque is raw force, HP is just the force combined with speed. Think about it like this. A guy doing a bench press. Lets say he can bench 200 pounds. He can push 200 pounds up in say 4 seconds. Now say another guy can do 100 pounds but in 2 seconds. They's exerting the same amount of force essentially. Or 2 football players. One guy is 300 pounds, one is 150 pounds. The 300 pound guy can run a top speed of 6 MPH. The 150 pound guy can run 12 MPH. If they run they fastest into each other, it's the same force. They would take equal hits force wise. Thats why a little guy can knock a big guy down with the right speed and technique. Or more relevent to motorcycles. Lets use bicycles. Ok think of 2 guys. A toned, racing bicyclist, and a big strong powerlifter. Both has the same exact bicycle and the same exact gearing. In a race, the pro bicyclist will probly win. The strong guy can probly exert 400-800 pounds to the pedals. He may even win off the line. But the pro will win performance wise for the most part. Now, put a 1 horse trailer with a horse inside behind the bike. I bet the pro will barley be able to pull it, the strong guy much easier. Also, lest say they bicycles is tied to a wall. The strong guy can stand and put all his force to the pedals, as well as the racer. He is still putting down torque. It don't need any RPM. Torque is raw force. HP is RPMs. The racer may make more HP by pedaling faster even with less force. Related, but not the same. I have a Dodge Ram 3500 Cummins diesel. A BMW sportscar will smoke it in a race. It has more HP. But lets see it pull 25,000 pounds of round hay bales at 70 MPH up a steep grade! My diesel has a redline of just 3,375 RPM or so. It makes all it's power from only 1,600 RPM and up. It only has 230 or so HP, but it has gobs or torque, something like 460 ft/lbs@1600. Its not just torque, its RPM. A Harley is pretty slow vs even a small Japanese inline four. But going up a huge steep highway grade, with a big fat passenger, saddle bags and rear trunk full to the max, on a super windy day at 70+ MPH, the torque is what makes it barley even necessary to downshift. That four will struggle and need to get closer to the redline, the H-D will hardly brake a sweat. But in a drag race the Harley is laughable, a Ninja 250 will compete with it, but that Ninja won't run up that hill like that all relaxed at 2,800 RPM. A H-D will get 46 MPG or so at 2,900 RPM just like a Japanese will get the same at 4,000 RPM. Its all about what its designed to do.

Thanks for the reply. I wondered if there were any gear heads here that would want to talk about this. Yours is the common view based on intuition but it isn't quite right. This would actually make an amusing episode of Top Gear. Car vs truck: TOWING! I found an interesting dyno test where the Ford 6.7 Powerstroke engine made 700 foot pounds and 350 hp. The truck will be better to tow with since it has greater torque and power down low. So when you take off from a light, it is already in the power band at the torque converter stall rpm. It is easier to access it's POWER band right off the line. The car is tuned for high power closer to redline. At the expense of not having as much torque or power at lower rpm. So even in first gear it's rpms are to low to have the torque or POWER to yank the trailer to get started. . But, if you were to race two trucks and two trailers at the drag strip, the driver who makes his shifts on either side of the engines power peak would beat the other driver who shifted on either side of the torque peak. Power is doing the work of moving the trailer. And the rear wheel torque of the winning truck will avreage out to be HIGHER than the earlier shifting truck who used his engine at the range of greater torque. The slower truck sent more torque to the trans but the faster truck stayed in a lower gear more often and used a greater POWER band to get greater rear wheel torque, hence power, at any given speed. . Here is the crazy part. . Once it finally got moving so it could use it's gears at the power peak, a 400 hp car will pull the trailer up the hill at a faster speed than a 350 hp truck. Even though the car doesn't make as much torque as the truck anywhere. Power is doing the work. A lesser engine torque can be multiplied with gears but the power at the wheels is always the same as the engine is putting out.. The trans is trading higher wheel torque for lower wheel rpm. The power at the wheel is the same as the engine at any gear. And higher power will do more work than lower power. Most towing vehicles have giant engines which are tuned for low rpm so they have tons of torque, and so, power, down low to yank with, and plenty of peak power to keep moving up the hill from the large displacement. But it is always the amount of power that is doing the work. . This is interesting. The first google search I made came up with test data on a dyno which states that the rear wheel power was the same in any gear. If anything, the power showed to be slightly higher in 5th than in 3rd. Must be the dyno is not really that linear at different speed ranges because that is the complete opposite from what most people would think where they would guess that the lower gear would give the most rear wheel power. And in reality the power would be the same discounting different losses in the trans at different gears. . http://www.dieselpowermag.com/featu...mmins_power_stroke_and_duramax_diesels_built/ . "We strapped down the 2011 ½ Ram 3500 dualie to the rollers first. The guys at ATS made three dyno pulls with the new engine, one in Third, one in Forth, and one in Fifth gear. The results were surprising. Regardless of which transmission gear we tested the trucks in, the power rankings were all the same"

After comparing my newly acquired Ninja 250, with it's massive 26hp and 13lb/ft (which equals less than zero torque) to my old crusty KLX686 with 50hp and 48lb/ft I can say without doubt that if ridden to the same level of accelerative force, the KLX would win the MPG contest. It just requires so much less time accelerating that it makes up for barn door aero! Majority of my riding is city and backroads. Now, where is that 300lb single cylinder 650 sport bike we all want?

What mpgUS did the KLX get? Most Ninja 250's get about 52. I can hit 66 mpgUS at 65mph on my 09 Ninja250 and 62 on my 07 Ninja650. My CBR250R with longer gearing can hit 80 mpgUS on the same ride.

By that argument, a Top Fuel dragster should get great fuel economy because it takes so little time to accelerate.

Here is an interesting quote by DieselMaxPower at the Honda site which explains the conundrum of the power at the rear wheel not being multiplied by using a lower gear even though the vehicle does accelerate much faster. An example would be The CBR250R with an engine power peak at 8,700. If you are rolling along at a steady speed in first at 8,700 rpm, and then crack full throttle, the bike will accelerate violently. We can feel this very obviously. Now do the same in top gear. The bike is felt to start going faster but the rate of acceleration is much, much less. Even though we used the same amount of power from the engine and at the wheel. The acceleration is much greater with the lower gearing but we have put the same amount of power to the ground over time as in the higher gear. There must be some quantity that is increasing at the same rate in either gear (disregarding the higher resistance of speed to wind drag in top gear). . Ready? .. The ENERGY of the system is increasing at the same rate in either gear. But we are very bad at perceiving energy so it seems like top gear does very little and first gear is doing much more with the same amount of power. . Example of how bad we are at perceiving differences in energy: . Cruising along at a steady speed in a car so you can't feel the wind, 10 mph feels pretty much the same as 70 mph. But 70 mph has much more energy and would be a very bad time to run into a wall. . . From DiesleMaxPower . http://www.cbr250.net/forum/102283-post75.html . "Torque doesn't do one thing while horsepower does another. Torque is a measure of what's happening to the bike. So is horsepower. It depends on what you want to look at. Torque can give you a way to look at the accelerations. This is great for most humans because we can perceive things like position, speed, and acceleration. Horsepower lets you look at energy. Humans don't have a good way to perceive energy, except for maybe temperature changes which we don't include. When we analyze the dynamics of the bike we look at the speed of the bike (thats the momentum side) and the kinetic energy (thats the energy side). In the simple analysis we're doing now the two are so simplistic that they will always be consistent with one another. It doesn't matter if you decide that torque moves the bike or horsepower. The reason I like to look at HP because the math is much easier, as we've shown with our torque charts. You don't need to transform it through gears."

I would be interested in seeing those efficiency curves for a BMW F650GS. I had a 2003 model and I would get 63 mpg (averaged over many fill-ups) riding 2-up with luggage. My 2005 R1200GS gets only 42 mpg average under similar conditions.

cool stuff. At the end of the day, every engine has a "perfect" spot where it is doing the work required using the least amount of fuel. This "spot" is different based on how much work is being done. Driving 80 mph on flat ground will result in one rpm for max efficiency, and it will be different than 40mph climbing a 10% grade....etc. The idea behind the "continually variable transmission" is that the computer can keep the engine close to the perfect zone at all times, based on work being done (load). This is why cars with CVTs always will have the best mpg ratings (all other things equal). The most efficient working zone, from what I've read, is not peak HP or TQ (unless that is exactly what's required for the conditions). The VAST majority of losses are to heat, and thus end up exiting via the radiator and/or the exhaust. Most engines are 30% efficient, AT BEST....with the majority more like 20%. We all drive furnaces with just a small by-product of mechanical energy.

A reply from another site . . "Here's the problem... you can't disregard air resistance at higher speeds. That's like saying granite would float in water if you disregard gravity. In top gear, you use a significant amount of your power due to wind resistance and friction. Only the remainder is used to accelerate the vehicle. Since you have less available power to accelerate the vehicle, you will accelerate slower. It is also more "violent" in low gears because you have more torque at the wheels which is more likely to get the tire to lose traction or for the reaction to lift the nose of your vehicle upwards." But consider this: What about the turbo diesel trucks in the link which all have a basically FLAT power curve. Could you then choose from 3rd, 4th, or 5th which would give the same exact power and road speed? Just a different rpm and torque multiplication. And expect the exact same level of acceleration? Crazy topic huh. I think 3rd will accelerate much harder. . . . .

I don't think it can be known from the curves. The variable needed is fuel required to maintain speed, given a particular resistance. And yes, at higher speeds, the ONLY variable to consider is wind resistance (unless you're climbing, then you must deal with gravity, which accelerates against you at 30 feet per second per second). Around town, the only real consideration is resistance from inertia. SO.....a lighter vehicle gets better mpg. On the highway, all you care about is speed (wind resistance) and rate of vertical change (which requires horesepower)....so the more aerodynamic vehicle gets better MPG....or, you can just go slower (which is why scooters are rated at 100 mpg...yea, 100mpg at a constant 40 mph)

The HP vs. Torque vs. Fuel Efficiency will always be confusing because of comments already made: HP is torque per unit time; HP=torque/sec with torque as ft-lb or N-m and HP as Ft-lb/sec. The metric equivalent to horsepower is watts as J/s or N-m/s. The two are related and only separated by time. Torque is work or energy. Work per unit time (and energy per unit time) is Power. Power is in horsepower or watts. Force applied for a distance is work; i.e. torque; i.e. energy: Torque = force * distance: ft-lbs (feet*lbs) and N-m (Joules) What matters is the force needed over a distance, or energy, which affects gas mileage. Gas has energy and the conversion of this energy into motion takes force. It is this force that determines gas mileage fuel efficiency. As an extreme example: a bumble bee and a Mack truck; which uses the least energy to start and stop? Since motion is related to acceleration, a, and force is mass * acceleration (i.e. F=ma) then the more massive object will always use more energy to start and stop. The bumble bee wins. Recognizing energy use determines miles/gallon or E=force*distance and thus for the energy in a gallon of gas, the distance = Energy/force. So, to increase the miles per gallon the force is reduced. Reducing the force from acceleration, bearing drag, internal friction, aerodynamic drag, rolling resistance drag, etc. will increase MPG. Viewing MPG relative to torque and horsepower - which are both energy with and without a time component - will not lead to improving gas mileage. If the rider does not accelerate quickly and uses a fairing, the fuel efficiency will go up and a lighter motorcycle will get better fuel efficiency than a heavier one. Physics.

My last example with the trucks was a bad one. And my statement that 3rd would pull harder than 5th given the same speed and power with just a different rpm along the super flat power curve was wrong. The acceleration would be the same. The starting speed was the same in each example and so was the power being applied. And so was the rear wheel torque in either gear. More on this coming right up. . Even though I was wrong in that statement, it helped me get back to the premise I am trying to prove. . First gear accelerates at a much higher value than 6th. And not because of air resistance at speed in 6th. . Because we are comparing changes in energy. . Lets use horsepower hours for energy to keep the terms we have already been using. Or, horse power seconds to put into the appropriate scale. If I have to come up with exact equations for mass and speeds to support my point, I will. But I think we can get the idea from the concepts, without resorting to using exact numbers which will waste a lot of time to throw together. . We have a CBR250R and the area under the power band we are using as we shift from the high side of the power peak to the low side with each gear is the same from 9,500 rpm to 7,500 and averages 20 hp. Let's say we start with 20 horse power seconds to begin with and 226kg at 6.7 meters/ second in 1st. So we pin the throttle while cruising at 7,500 rpm in 1st, 1 second later we have added 20hp seconds of energy to what ever we had before. Now we are up to 9.47 m/s and are carrying 40 hps. At the end of 2 seconds we will be carrying 60 hps and 11.6 m/s. We have doubled our energy vs the starting point in the first second and added another 33% to that in the second second. These are big changes of energy vs time. And big changes of speed vs time. So, a high acceleration. .276 G over the first second and .217 G over the second second. Now we continue to accelerate and shift up through the gears each time we reach 9.500 rpm until we hit 5th at the ten second mark. We now have 220 hp seconds of energy and 22.22 m/s. One more second later we have 240 hps and 23.21 m/s. We only gained 1 meter per second which is .1 G of acceleration. Less than half the acceleration we had between the first and second seconds of the run when the speed was low. . This has nothing to do with wind yet which will indeed have a big effect on such a small amount of horsepower and acceleration. It is because the energy we are putting in is just increasing linearly with time but the speed is squared on it's side of the equation. . . And this brings me to the real revelation I had from the truck's dynos and 3rd accelerating the same as 5th. Power is the metric of how fast you can gain speed as I have been saying. And in the end, the torque multiplication of the different gears doesn't really do anything to make something pull harder other than to match the engine power band to the needed wheel speed.

Still at it and working some of the bugs out of some misstatements earlier. The highest rear wheel torque will give the highest acceleration. . . . . The faster you go, the less torque you have as you shift up through the gears if power is constant. This is intuitive but it is also predicted by . Power = F*V . This holds true according to theory and measurement even for a variable pitch airplane prop which is like a cvt transmission. Check out the chart that shows the thrust of a certain variable pitch prop at constant horsepower and vs increasing speed which is calculated from . Thrust = ( HP * eff * 375 ) / air speed (mph) . and looks like this:. . . . . The lines rise at first because a real world prop has poor efficiency at low speeds. Efficiency actually rises throughout the increasing speeds from 65% at 100 mph to 85% at 240. Otherwise the slope of the thrust drop off would be even steeper. . The physics we need to understand are really very simple. Reciprocating engines make Torque and when looked at by multiplying that by the Revolutions (How many "Torques") in a certain amount of Time, you get the Power of the engine. . hp=Torque * RPM/ 5252 . . Rear wheel Torque is Force on the road. But it is interesting to note that the highest rear wheel Torque will occur, ie with a constantly variable transmission, which is set to the Power peak of the engine. Not the Torque peak. Due to the fact that it will be using more torque multiplication (lower gear) at the higher rpm of the power peak. . Force (from the rear wheel Torque) times the Distance the bike moved is Work. . Power is the measure of Work vs Time. If you do the same amount of Work in a shorter Time, you have used more Power. . P=W/T Work=Force*Distance so P=F*D/T and D/T is Velocity so P=F*V Given the same Power from the engine, The Force at the rear wheel must be less as the Speed between the road and the bike increases. . F=M*A Acceleration = Force/ Mass F, and therefore A, must be less and less as the speed difference between the driven wheel and the road, or the airplane propeller and the air, is increasing. . So this is the "Captain Obvious" over complicated way of saying the bike will pull harder in 1st than in 6th. . Mass ejection engines such as a rocket may be different. I'm having a tough time with this and I must admit that I intuitively think that for vehicles propelled by ejected mass, rather than a driven member that is coupled to a stationary medium like the ground or air, or water, that the speed that is valid for the calculation of Force to the frame of reference of the passenger, is the speed difference between the gases and the nozzle. Not between the gasses and the Inertial Frame of Reference of the Universe. Which might simply indicate that the term constant Power doesn't apply to such engines which will have fairly constant force at the nozzle despite the Velocity of the rocket. I'm having a hard time finding anything to contradict this. I was hoping to find some model rocketry pages with measured thrust data to compare to the well documented bench testing of the various engines which could show if speed of the rocket comes into play as it does with the bike. . ThrustCurve Hobby Rocket Motor Data . Some interesting info on rocket engines I dug up: The efficiency of a rocket engine changes, starting at near 0% on lift off because the only thing moving is the ejected matter. And increases with increasing speed up to the best efficiency when the velocity of the exhaust gas as referenced to the nozzle is equal and opposite to the Velocity of the rocket referenced to the Inertial Frame of Reference. The rocket is flying away from the ejected mass at mach 10 or whatever and leaving the mass which is exiting the nozzle at mach 10, essentially motionless in space behind it. And a rocket engine can continue to supply force to a vehicle that is traveling faster than the exhaust speed at the nozzle. But with decreasing efficiency as now the exhaust will again have speed and momentum of it's own left over and in the same direction as the vehicle. . Rocket engine - Wikipedia, the free encyclopedia . Rocket - Wikipedia, the free encyclopedia

Here is the last piece we needed tying Power to Energy as posted by DieselMaxPower: . "I'll add some info, even though you've basically got it. So a simplified equation for your last post is P = M*A*V It took several relationships, and if you actually did it through torque you would have several factors of 2*pi and the wheel radius to carry around and eventually cancel. Now I know most people don't know calculus, so you will have to trust me on this: E = 1/2 M V^2 d(E)/dt = M V (d(V)/dt) d(E)/dt is power, and d(V)/dt is acceleration. So check this out P = M V A You get the EXACT same equation using calculus. And acceleration is: A = P/(MV), so as you go faster, your acceleration decreases (assuming constant P). This doesn't really add too much to your current understanding, but if you want to more advanced stuff it will really help. For instance, we may want to take into account the force it takes to spin the wheels. This gets so ridiculously tedious looking at forces. Instead, add it to the energy equation E = 1/2 M*V^2 + 1/2 I omega^2 d(E)/dt = M*V*(d(V)/dt) + I omega (d(omega)/dt) I is the mass moment of inertia and omega is the angular velocity of the wheel, and now d(omega)/dt is the angular acceleration of the wheel. In the simplest case (tires aren't slipping on the road, or slip by a constant percentage), d(omega)/dt is proportional to A: P = M*V*A + I*omega*c*A P = A*(M*V+I*omega*c) A = P/(M*V+I*omega*c) A = P/(M*V+I*V*d) where c and d are constants. We can change omega to V by the same argument that we could change d(omega)/dt to A. So acceleration decreases with increased velocity (assuming constant power), but now we get a more realistic estimate. Some of the power produced goes into spinning the wheels. Rockets get into a whole new mess, which I would actually say makes more sense by looking at momentum instead of energy. If you're interested, the Reynolds transport theorem for momentum should give you some good information. Most simple problems don't need calculus, even though the theorem in its purest form needs it. As a bonus, looking at model rockets is one of the most common simple problems."

C*V*T = Obvious . http://en.wikipedia.org/wiki/Continuously_variable_transmission . A continuously variable transmission makes the concept of combining the equations for power and energy very obvious. . Power = Mass * Acceleration * Velocity . When you grab full throttle with a CVT, the transmission is controlled to change down to a lower ratio and wind the engine up to the power peak rpm. Where the rpm stays for the whole run until you let off the throttle. The CVT then continuously varies the gear ratio longer and longer to keep the same engine rpm while the vehicle Velocity is increasing from the Acceleration. . The the throttle is held fully open by the rider and the engine will be kept at the same rpm by the transmission so the power from the engine is the same throughout the run even though the Velocity is increasing. The transmission is trading torque multiplication for speed as it moves up through it's range of gearing by squeezing the belt up in the front to make the engine pulley bigger and the wheel pulley smaller. The Power at the wheel is the same the whole time but there is less and less rear wheel Torque as Velocity increases. . P = M*A*V . Power stayed the same. And Mass stayed the same. So as the Velocity increases, the Acceleration and the Velocity must trade. One goes down if the other one goes up. Because the Energy that is added by the engine Power over time is linear. 1, 2, 3, 4. Going back to the Energy equation . E = 1/2M*V^2 . Energy would have to increase exponentially to keep the same Acceleration as Velocity increases. 1, 4, 9, 16.

There is another misconception about gearing that is related to energy. Motorcycle guys often change the final gear ratio by changing sprockets since it so easy and cheap to do. I went 15% longer for better fuel economy. Some guys go shorter for better acceleration and to get what they believe will be closer gear ratios. The interesting thing is that the rpm change between gears stays exactly the same either way. If you run 1st to redline at 10,500 and the rpm drops to 6,600 when you grab 2nd, that difference will remain the same regardless of what changes you make to the final drive. You would have to change the gear sets in the transmission to make the rpm drop to be closer between gears. But why do the gears FEEL closer with a lower final drive and FEEL so far apart with my long gearing when you are getting the same rpm drop as before? Because each shift point is now closer together in Energy with the shorter gearing.

A racer brought up a good point about going shorter on the final drive. You can make the rpm drop closer together by going shorter in the final drive. Throw away 6th by making it more like what 5th was. Make 5th into 4th, 4th into 3rd, ect. Now you are more often using one gear up from what you were using before where most transmissions group the ratios closer together at the top. . I cruise long distance on the highway so I use 6th. And 7th (what was 6th before I went longer on my sprockets).