Can someone help me out explaining acceleration to a friend?

[Vic]

Originally posted by: jdoggg12

Originally posted by: Vic
The OP's friend is right. I think everyone arguing here forgot about the potential energy stored in the rotating mass of the drivetrain. There is a reduction in acceleration once the clutch is depressed, and that stored energy is relatively small and will be depleted within about a second, but you can accelerate smoothly through a shift if you do it properly. You can increase this effect (by increasing the potential energy) by engaging the clutch a split-second before you come off the gas, which results in that desired high note upon shifting.

You're saying that the energy 'stored' in the drivetrain is going going to be exerted as acceleration? This is wrong. Even if there IS that energy, then it will be exerted on the disengaged clutch plate clutch plate (now spinning freely b/c its not being 'bogged down' by the motor) since it will offer less resistance than it would take to keep accelerating the bike.

Goddamned this new forum upgrade. I typed a long post only to have the whole thing disappear because the thread was moved.

Anyway, I was referring to the rotating mass of drivetrain components post-clutch (I think you were confused and thought that by "drivetrain" I was referring solely to the engine). And while the clutch might free the engine from the transmission, it does not free the transmission from the drive wheel. Those components (transmission, chain/shaft, rear wheel, etc) all have mass, and when rotating store kinetic energy similar to how a flywheel does. This energy WILL be depleted as acceleration prior to deceleration forces acting on the vehicle body as a whole.

 

[jdoggg12]

I know, but think of it like this... (This is specific to your example vic)

You have an RC car motor connected to a rubberband. The rubber band is stretched out to a propeller. You turn on the motor, it spins and builds up energy, then once that energy is enough, it will turn the propeller. Now the connection of the motor to the rubber band is basically like the clutch. If you cut the band right there as the motor is spinning faster and faster, all that stored up energy will unwind at THAT point. It wont cause the propeller to turn faster b/c the force going through the transmission (rubberband) has been cut, not increased.

See what i'm saying?

 

[Vic]

Originally posted by: jdoggg12
I know, but think of it like this... (This is specific to your example vic)

You have an RC car motor connected to a rubberband. The rubber band is stretched out to a propeller. You turn on the motor, it spins and builds up energy, then once that energy is enough, it will turn the propeller. Now the connection of the motor to the rubber band is basically like the clutch. If you cut the band right there as the motor is spinning faster and faster, all that stored up energy will unwind at THAT point. It wont cause the propeller to turn faster b/c the force going through the transmission (rubberband) has been cut, not increased.

See what i'm saying?

I do, but the difference is that the energy stored in the rubberband will be dissipated by its flying apart at the point where it was cut. The drivetrain components do not have that option, they are mechanically linked to the wheel. Nor can you have energy go out one side (supposedly with lesser resistance) and not the other. I repeat, the whole thing is mechanically connected. Only the engine is freed by the clutch (and the stored energy of its rotating mass is dissipated through friction and compression, but even it continues to accelerate briefly, resulting in that "high note" I referred to earlier).
Energy is indestructible. It will go somewhere.

 

[jdoggg12]

Yes, but as soon as you disengage the clutch, it doesn't matter how much energy is in the transmission. There is no force being applied.

Another example: you're crouching on the ground holding a sphere up on your shoulders, like Atlas. The ground is the motor, holding you up. With all of your strength, you're at equilibrium holding that ball at a crouch. Engaging the clutch, in this example, would be the equivalent of opening a trap door under your feet. All of that energy you are exerting (you are the transmission) is expelled in the path of least resistance (the now open clutch). At no point in time are you going to push the ball any higher (acceleration) b/c you're already at your max by the time the trap door opens

 

[Vic]

I think the confusion here is that you're assuming that the transistion from positive acceleration to zero or negative acceleration would be instantaneous. It would not. As RGUN already pointed out, that in and of itself would take additional force (braking).
There is, yes, an abrupt reduction in positive acceleration upon clutch disengagement, but some does remain as stored kinetic energy in the drivetrain is dissipated AS FORCE (specifically torque) through the drive wheel. Your rubberband and your atlas/trapdoor analogies don't work, and there is no side of "least resistance." I repeat, the components share a fixed mechanical connection, so you can't have one side spin faster than the other. They all spin together.

 

[Howard]

Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: Howard
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN

I suspect you havent taken physics, Im a B. Eng with a specialization in solid mechanics.... This is my area of knowledge</end quote></div>
Then why don't you know what jerk is?</end quote></div>

What has made you believe that I dont?

Because you're equating force with jerk.

 

[Pacfanweb]

Not reading the whole thread, but I do believe if you were accelerating at WOT and clicked the trans in neutral or disengaged the clutch, the inertia of all the rotating masses in the car would briefly continue your acceleration.

Plus, many times I've accelerated hard and done just this...clicked the trans into neutral and watched my speedo keep climbing before it stabilizes and starts slowing down....and the only way the speedo can go higher is if the wheels/axles/output shaft goes faster.

So yeah, RGun is right.

 

[jdoggg12]

True, but have you ever locked up your tires? I have and it was NOT an instant drop from showing 80mph - 0. So it might've been the gradual change in current average speed.

 

[Vic]

Originally posted by: jdoggg12
True, but have you ever locked up your tires? I have and it was NOT an instant drop from showing 80mph - 0.

Yes, but that's because the wheels don't automatically go to 0 when you lock them up. They just decelerate faster than the tires' coefficient of friction with the road surface allows for.

 

[jdoggg12]

haha, no.... i was on the beach, going 80mph and locked all 4 discs up :laugh:

 

[SVT Cobra]

I tried this in my car now, and it does accelerate 1 or 2 mph when pushing the clutch in real quick after accelerating hard.

 

[Zenmervolt]

Originally posted by: Pacfanweb
Plus, many times I've accelerated hard and done just this...clicked the trans into neutral and watched my speedo keep climbing before it stabilizes and starts slowing down....and the only way the speedo can go higher is if the wheels/axles/output shaft goes faster.

A speedometer has a lag time. It's impossible to record speed instantaneously, and a speedometer will always have a slight reaction period. I think that all you're seeing is the speedo's slight reaction delay and nothing more.

ZV

 

[Zenmervolt]

Originally posted by: Vic
I think the confusion here is that you're assuming that the transistion from positive acceleration to zero or negative acceleration would be instantaneous. It would not. As RGUN already pointed out, that in and of itself would take additional force (braking).
There is, yes, an abrupt reduction in positive acceleration upon clutch disengagement, but some does remain as stored kinetic energy in the drivetrain is dissipated AS FORCE (specifically torque) through the drive wheel. Your rubberband and your atlas/trapdoor analogies don't work, and there is no side of "least resistance." I repeat, the components share a fixed mechanical connection, so you can't have one side spin faster than the other. They all spin together.

There's no way the inertial force of the drivetrain is greater than the frictional forces acting upon the vehicle. At least at speed. Perhaps at parking-lot velocities, but not at normal road speeds.

ZV