02022013, 08:05 PM  #16 
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Location: Syracuse, NY USA
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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
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http://www.fuelly.com/driver/sendler/cbr250r sendler screwed with this post 02032013 at 07:34 AM Reason: Wrote P = M*V should have been P = F*V 
02032013, 07:19 AM  #17 
Studly Adventurer
Joined: Apr 2012
Location: Syracuse, NY USA
Oddometer: 737

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."
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02042013, 04:37 PM  #18 
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Location: Syracuse, NY USA
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C*V*T = Obvious
. http://en.wikipedia.org/wiki/Continu...e_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.
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02052013, 09:02 AM  #19 
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Joined: Apr 2012
Location: Syracuse, NY USA
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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.
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02062013, 09:54 AM  #20 
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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).
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