Mechanical power generated by a force F

[russ_watters]

Because that's what the gears do! They fix the ratio between pedal (or engine) rotation and wheel rotation. That is a rigid link (except there is a freewheel on most but not all bikes) The rpm in a car is mechanically forced to match when the clutch is let out...if you don't decelerate the engine you will burn the heck out of the clutches (an automatic transmission does this for you)

Minor caveat: the other potential breaking point is the tires' contact with the road. If you pop the clutch it may spin the tires slightly until the rpm drops.

 

[russ_watters]

I see. I think I have seen dirt bike riders change gears without using the clutch and without letting go of the throttle. Probably not a good practice for a lasting engine but I guess it may be effective when racing...

That seems unlikely to me, but it is not uncommon to do it if you let off the throttle/gas.

 

[Drakkith]

To summarize, we start the car from rest and need a lot of force to cause it to move.

No, we don't need a lot of force to cause the car to move. We only need a little force, just enough to overcome friction assuming the car is on level ground, to get the car to accelerate. We only want a lot of force if we want a large acceleration. First gear is simply the gear ratio in which the tires spin slowest for any give engine RPM, which is necessary when moving at a slow speed if you don't want to burn up a clutch. It also happens to give the largest acceleration for a given input force, which is advantageous if you want to quickly accelerate. This is why I usually shift out of 1st gear around 15 mph unless I'm rapidly accelerating, in which case I'll wait until 25 mph or so to shift to 2nd gear.

It is time to shift to 2nd gear and the engine's RPM automatically decrease...The same happens with the bicycle: our legs don't go round as fast when you change up a gear on a bicycle even if we don't slow down our pedaling...

Of course you slow down your pedaling. You have to. If you didn't, the wheel would jump from spinning at one speed to a higher speed as soon as you shifted gears, implying your wheel lost traction with the ground or you have a magic bicycle that can change speeds instantly.

 

[sophiecentaur]

I see. I think I have seen dirt bike riders change gears without using the clutch and without letting go of the throttle. Probably not a good practice for a lasting engine but I guess it may be effective when racing...

Not particularly bad practice; the design allows for it, I think. Also would a spectator actually know what the rider was doing with the throttle?

Motorcycle gearboxes are all(?) of the Constant Mesh design. All the pairs of gears are in mesh all the time. There is a shaft on which all the gears are fixed and another on which only one gear at a time is fixed and the others are free to rotate. Forks move the gears on an off 'dog clutches' which are very chunky splines along each shaft and engage just one gear onto its shaft. The graunchy sound you get with clutchless changes is due to a dog clutch trying to engage. It doesn't involve the gears themselves. Sounds worse than it is.

Changing of gears on a motorcycle always involves just one step up or down so the speed change is small. This is especially true when there are many available ratios and the high performance engine is being run in a narrow rev range. A clutch is not essential for any gearbox change, except when starting from rest. You don't have to cut the throttle at all. Just moving the throttle a small amount will slow or speed up the driven shaft, once the previous dog is disengaged and the fork will slip the next pair of dogs into contact. There's much closer 'contact' with what's going on on a motorcycle.

Motorcycle boxes need to be compact and the moving mass of an m/c engine is lower than for a car so they don't use 'synchro' rings to help bring the shafts to the right speed for engagement. I imagine the low mass of a cycle makes for a potentially small shock to the transmission.
I used to drive an old van in the 60's that had a 'crash box' with no synchro. Matching gear speeds when changing down used to involve 'double de-clutching' to blip the input shaft and then let it slow down so that it would (slip in to) engage. But that wasn't really necessary if you got the throttle set right - as with an m/c.

 

[fog37]

No, it feels hard because you are trying to reach the same acceleration as in the low gear and thus increase the pedal force. Increasing the pedal force actually brings you to the same power input as in low gear (assuming we compare at the same speed).

Thank you jack action! I am close to getting it.
Let me see if I can correctly paraphrase your explanations:

We start from rest in low gear (1st gear). We start pedaling faster and faster. Initially, we are able to apply a sizeable force $F_{pedal}$ on the pedals. given the 1st gear large gear ratio $G_1$ , our pedal force $F_{pedal}$ causes a large force by the ground $F_{ground}$ that largely accelerates the bike forward. The force $F_{ground}$ is not constant in time: it decreases because $F_{pedal}$ progressively decreases while the bike's gains speed. The fact that $F_{ground}$ decreases while the bike speed increases means that the input power $P= F_{ground} v_{bike}$ also decreases with time. I guess the product $P$ does not remain constant with $v_{bike}$ increasing while $F_{ground}$ decreases...

Eventually, the bike reaches its maximum speed $v1_{max}$ in 1st gear with the pedal(s) spinning too fast for us to create a large pedal force. The pedal force $F_{pedal}$ is now small because we cannot physiologically generate a large force with the pedals spinning that fast. The bike now coasts at that speed $v1_{max}$.

At that point, to increase the bike's speed from $v1_{max}$, we shift to 2nd gear which has a lower gear ratio $G_2 < G_1$. The lower gear ratio and the fact that the bike is traveling at $v_1 {max}$ causes the pedals to automatically/instantly spin slower (than right before shifting to 2nd gear) to match the speed $v_1{max}$ at the new gear ratio $G_2$. However, we could still be applying, at that lower pedaling speed, the same pedal force $F_{pedal}$...BUT even with that same pedal force, the actual force $F_{ground}$ would be smaller. And we want a large $F_{ground}$, as large as it was in 1st gear. So we end up pushing harder on the pedals to re-establish that original $F_{ground}$. Is that correct?

You mention that we are inherently trying to achieve in 2nd gear the same power $P$ that we were generating in 1st gear. Why are we trying to do that? So the objective is the same power or the same force $F_{ground}$?
The power in first gear is $P_1 = F1_{ground} v1_{bike}$ and the power in 2nd gear $P_2 = F2_{ground} v2_{bike}$. Power $P$ can be viewed as the change in kinetic energy $\frac {\Delta KE} {\Delta t}$ or equivalently as $\frac {F_{ground} \Delta x} {\Delta t}$ so the fact that $F_{ground}$ is smaller should be compensated by the bike's velocity being much larger than in 1st gear keeping the input power the same...?

Thanks!

 

[sophiecentaur]

Thank you jack action! I am close to getting it.
Let me see if I can correctly paraphrase your explanations:

We start from rest in low gear (1st gear). We start pedaling faster and faster. Initially, we are able to apply a sizeable force $F_{pedal}$ on the pedals. given the 1st gear large gear ratio $G_1$ , our pedal force $F_{pedal}$ causes a large force by the ground $F_{ground}$ that largely accelerates the bike forward. The force $F_{ground}$ is not constant in time: it decreases because $F_{pedal}$ progressively decreases while the bike's gains speed. The fact that $F_{ground}$ decreases while the bike speed increases means that the input power $P= F_{ground} v_{bike}$ also decreases with time. I guess the product $P$ does not remain constant with $v_{bike}$ increasing while $F_{ground}$ decreases...

Eventually, the bike reaches its maximum speed $v1_{max}$ in 1st gear with the pedal(s) spinning too fast for us to create a large pedal force. The pedal force $F_{pedal}$ is now small because we cannot physiologically generate a large force with the pedals spinning that fast. The bike now coasts at that speed $v1_{max}$.

At that point, to increase the bike's speed from $v1_{max}$, we shift to 2nd gear which has a lower gear ratio $G_2 < G_1$. The lower gear ratio and the fact that the bike is traveling at $v_1 {max}$ causes the pedals to automatically/instantly spin slower (than right before shifting to 2nd gear) to match the speed $v_1{max}$ at the new gear ratio $G_2$. However, we could still be applying, at that lower pedaling speed, the same pedal force $F_{pedal}$...BUT even with that same pedal force, the actual force $F_{ground}$ would be smaller. And we want a large $F_{ground}$, as large as it was in 1st gear. So we end up pushing harder on the pedals to re-establish that original $F_{ground}$. Is that correct?

You mention that we are inherently trying to achieve in 2nd gear the same power $P$ that we were generating in 1st gear. Why are we trying to do that? So the objective is the same power or the same force $F_{ground}$?
The power in first gear is $P_1 = F1_{ground} v1_{bike}$ and the power in 2nd gear $P_2 = F2_{ground} v2_{bike}$. Power $P$ can be viewed as the change in kinetic energy $\frac {\Delta KE} {\Delta t}$ or equivalently as $\frac {F_{ground} \Delta x} {\Delta t}$ so the fact that $F_{ground}$ is smaller should be compensated by the bike's velocity being much larger than in 1st gear keeping the input power the same...?

Thanks!

I can’t identify anything actually wrong with all this. It’s just that Physics attempts to condense a rambling set of occurrences that we find in real life and come up with the basic (one short line if possible) essence.

Expanding it all, as you’ve done, makes everything very confused. How does all you wrote get any nearer the point than just describing what the basic statements that people have already made, higher up the thread?

If the above means that you’ve got it then fine. But if you use the same approach to the next problem and the next, you’ll find Physics pretty impossible to get on with. Aim for simplicity and make life easier for yourself.

 

[jack action]

BUT even with that same pedal force, the actual force $F_{ground}$ would be smaller. And we want a large $F_{ground}$, as large as it was in 1st gear. So we end up pushing harder on the pedals to re-establish that original $F_{ground}$. Is that correct?

Yes.

You mention that we are inherently trying to achieve in 2nd gear the same power $P$ that we were generating in 1st gear. Why are we trying to do that?

We are fighting the braking effect we feel when the acceleration suddenly drops with a smaller $F_{ground}$ after changing gears.

So the objective is the same power or the same force $F_{ground}$?

I would tend to say the same force, but since the vehicle speed is the same, it also coincides with the same power as well.

From a racing point of view, one might say that you are trying to get the same power since maximum power gives maximum acceleration; so you were using maximum power in the first gear and still want to use maximum power in the second gear.

But all of this is just stating the same thing in 2 different ways.

The power in first gear is $P_1 = F1_{ground} v1_{bike}$ and the power in 2nd gear $P_2 = F2_{ground} v2_{bike}$. Power $P$ can be viewed as the change in kinetic energy $\frac {\Delta KE} {\Delta t}$ or equivalently as $\frac {F_{ground} \Delta x} {\Delta t}$ so the fact that $F_{ground}$ is smaller should be compensated by the bike's velocity being much larger than in 1st gear keeping the input power the same...?

$v1_{bike} = v2_{bike}$. The bike velocity does not change during shifting.

But $v1_{pedal} \neq v2_{pedal}$ and if $F1_{pedal} = F2_{pedal}$, then $F1_{ground} \neq F2_{ground}$ which leads to an instanteneous deceleration after shifting. You have no other choice but to increase $F2_{pedal}$ if you want to bring back $F1_{ground} = F2_{ground}$.

You might also be interested in When Vehicle Power Dictates Acceleration.