I Mechanical power generated by a force F

[fog37]

TL;DR Summary

Mechanical power generated by a force F

Hello,

It is well understood that a constant horizontal force $F_{applied}$ applied to an object of mass $m$ over a distance $d$ in a direction that is not orthogonal to the direction of motion produces mechanical work $W= F d$.
This work is a mechanism to inject (or subtract) energy from the object (tow other two possible mechanisms are heat and electromagnetic radiation).
The force $F_{applied}$ produces an acceleration $a_{applied}$ and using one the kinematic equation we can determine the very exact time $t$ it takes the object to cover the distance $d$. Mechanical power is defined as $P = \frac {W}{t} = \frac {Fd}{t}$, so it seems that when a single force acts on an object, the power is always automatically determined since the the time $t$ over which the force acts during the displacement $d$ is fixed by the force. The force, through its acceleration, fixes the time it takes to produce the work $W$, correct?

Another equivalent expression for power is $P = F v$. Assuming the force $F$ is constant, the velocity $v$ will change linearly with time. This means that the expression $P = F v$ represents the instantaneous power. So in the case of a single constant force $F_{applied}$ the generated power increases linearly with time since $v$ increases linearly with time.

However, if the object was moving at a constant speed $v$ and several constant forces were acting on the object, the net force would be zero and the net generated power would also be zero.
The expression $P = F v$ would represent the power generated by a particular force $F$. In this scenario, the time $t$ can be made to vary depending on the magnitude of the constant speed $v$ so the power produced by each force can vary while in the case of a single applied force the time $t$ is a very specific one fixing the value of the generated power.

Thanks!

 

[Drakkith]

The force Fapplied produces an acceleration aapplied and using one the kinematic equation we can determine the very exact time t it takes the object to cover the distance d. Mechanical power is defined as P=W/t=Fd/t, so it seems that when a single force acts on an object, the power is always automatically determined since the the time t over which the force acts during the displacement d is fixed by the force. The force, through its acceleration, fixes the time it takes to produce the work W, correct?

I think you're saying that given a constant force on an object F and a set distance D, the work done is the same regardless of when that object moves through that distance, but the power isn't. Which is true.
If the object moves through D soon after the force is applied, say from T₁ to T₂, it will take longer to move through D than if it moves through D later, say T₃ to T₄, as it will have accelerated more and its velocity will be higher. So the work done during either time interval is the same, but T₃-T₄ is a shorter interval than T₁-T₂. Since P=W/t, and t is smaller, P must be larger.

However, if the object was moving at a constant speed v and several constant forces were acting on the object, the net force would be zero and the net generated power would also be zero.
The expression P=Fv would represent the power generated by a particular force F. In this scenario, the time t can be made to vary depending on the magnitude of the constant speed v so the power produced by each force can vary

The time doesn't matter in this case. It doesn't change the power at all. Say an object is moving at a constant 10 m/s with an applied force of 10N and an opposing force of 10N. Take any point in time and the power done by each force is 100 W and -100 W respectively. Or 100 W produced and 100 W dissipated. The only difference is that the total energy or work is different as you change your time interval.

while in the case of a single applied force the time t is a very specific one fixing the value of the generated power.

Assuming you hold F constant then both power and work increase as T increases, yes. But you could just as well hold the power constant by reducing the force over time.

 

[Dale]

@fog37 what is your question?

 

[A.T.]

However, if the object was moving at a constant speed $v$ and several constant forces were acting on the object, the net force would be zero and the net generated power would also be zero.
The expression $P = F v$ would represent the power generated by a particular force $F$. In this scenario, the time $t$ can be made to vary depending on the magnitude of the constant speed $v$ so the power produced by each force can vary while in the case of a single applied force the time $t$ is a very specific one fixing the value of the generated power.

If $F$ and $v$ are both constant, then so is $P$. So I have no idea what you mean by "power produced by each force can vary".

 

[russ_watters]

Maybe consider a car driver riding the brake or braking while coasting downhill?

 

[vanhees71]

It's very simple. If $\vec{F}$ is the force acting on a particle with velocity $\vec{v}$ the power is $P=\vec{v} \cdot \vec{F}$. So what's unclear about this?

 

[sophiecentaur]

@fog37 what is your question?

The question is hidden inside a lot of muddled statements which seem to be attempting to find a hole in basic mechanics. 'We' all know that the way to deal with that sort of problem (and we all get there from time to time) is to find a reason that the textbooks are right and what we have ignored or mis-stated. The OP should think in terms of Net Work done, net KE gained and net losses. Once the object has reached a steady state, the KE will remain constant and input Work will equal to Power Loss.
Make a list of all relevant Energies / Powers.

 

[vanhees71]

Math is your friend in making things simple.

The above statement is simply derived by taking the time derivative of the kinetic energy:
$$\dot{T}= \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 \right)=m \vec{v} \cdot \dot{\vec{v}} = \vec{v} \cdot (m \vec{a})=\vec{v} \cdot \vec{F},$$
i.e., the power describes the momentary temporal change of kinetic energy due to the force $\vec{F}$ acting on the particle.

 

[fog37]

Thanks everyone. I am noodling with these concepts as I am trying to better understand how the gears in a bicycle, the force on the ground and the input and output power work.
For example, assuming negligible air resistance, rolling resistance, internal resistance, let's consider a bicycle in low and high gear. The cyclist introduces the same input power $P_{input} =F_{in} v_{pedal}$ in both cases.

CASE 1: low gear. The force of the foot on the pedal $F_{pedal} =1000N$ is transmitted by the chain to the back cassette to eventually produce a force on the ground $F_{ground}=440N$ pointing in the direction of motion.

CASE 2: high gear. The force of the foot on the pedal $F_{pedal} =1000N$ is transmitted by the chain to the back cassette to eventually produce a force on the ground $F_{ground}=220N$ smaller than $F_{ground}$ in low gear.

My question: does it feel very hard to pedal in high gear because the force on the ground is small? If $F_{ground}$ is small in low gear, the resulting bike's acceleration is small which means that the bike would speed up slowly. But why does it feel hard to pedal in high gear? From experience I know that the bike reaches a faster speed in low gear than in high gear for the same number of pedal rotations.

The output power is $P =F_{ground} v$. In high gear, it does not feel hard to pedal because the force on the ground is larger ($F_{ground}=440N$ in the above example).
The output power $P_{low gear} =F_{ground} v$ seems smaller than $P_{high gear}$ since the smaller $F_{ground}$ in high gear would produce a smaller increase in kinetic energy over time...

I am trying to piece together these concepts as clearly as possible...

Thanks!

 

[Drakkith]

My question: does it feel very hard to pedal in high gear because the force on the ground is small? If Fground is small in low gear, the resulting bike's acceleration is small which means that the bike would speed up slowly. But why does it feel hard to pedal in high gear? From experience I know that the bike reaches a faster speed in low gear than in high gear for the same number of pedal rotations.

Because for a given amount of acceleration, you have to apply less force on the pedals in low gear.

 

[Drakkith]

The output power Plowgear=Fgroundv seems smaller than Phighgear since the smaller Fground in high gear would produce a smaller increase in kinetic energy over time...

Are you sure? You haven't shown any numbers supporting this.

 

[A.T.]

Thanks everyone. I am noodling with these concepts as I am trying to better understand how the gears in a bicycle, the force on the ground and the input and output power work.
For example, assuming negligible air resistance, rolling resistance, internal resistance, let's consider a bicycle in low and high gear. The cyclist introduces the same input power $P_{input} =F_{in} v_{pedal}$ in both cases.

View attachment 293961

CASE 1: low gear. The force of the foot on the pedal $F_{pedal} =1000N$ is transmitted by the chain to the back cassette to eventually produce a force on the ground $F_{ground}=440N$ pointing in the direction of motion.

CASE 2: high gear. The force of the foot on the pedal $F_{pedal} =1000N$ is transmitted by the chain to the back cassette to eventually produce a force on the ground $F_{ground}=220N$ smaller than $F_{ground}$ in low gear.

Are you assuming the same speed over ground here?

My question: does it feel very hard to pedal in high gear because the force on the ground is small?

Feelings are subjective. You are assuming the same pedal force but there is less pedal movement in hight gear. That migth "feel" like it is resisting more to being pushed. Also, the acceleration of the bike is less for the same pedal force.

If $F_{ground}$ is small in low gear,

$F_{ground}$ is higher in low gear for the same pedal force. Look at you diagram/numbers.

 

[sophiecentaur]

I am trying to piece together these concepts as clearly as possible...

If that is the case then avoid using complicated examples like gears and bicycles. Just consider levers, accelerating a mass with a lever etc. etc.
'Personal Experiences' can produce misleading evidence which can also make it harder to reconcile theory with intuition.
Everything relevant to bicycles is there with levers. With a bike, you have a pedal - lever, a chain wheel - another lever - rear sprocket - another lever and the tyre on the rim of the wheel - yet another lever. Together, they constitute a single lever. Why make things more complicated than necessary?

 

[fog37]

Are you sure? You haven't shown any numbers supporting this.

I guess I was incorrect. Given a certain input power $P_{in}$ determine by the force on the pedal and by the cadence, the output power is actually the same in low or high gear.

I understand sophiecentaur's point but it is fun to understand those general concepts (levers, torques) in a real world application like a bike.

As far as the feeling of working harder in high gear than in low gear: it may just be the perception that with a single pedal rotation the acceleration is small (since the the ground force is smaller) but the speed increase is larger in high gear. Is that really true? That confuses me: the force on the ground is smaller in high gear than in low gear. The achieved speed in high gear is higher since we moves a larger distance in the same amount of time compared to low gear.

How does the smaller force in high gear manage to produce the larger speed in the same amount of time when compared to low gear? I tend to think that the large force generated in low gear would bring the bike to a certain speed in a shorter amount of time and that a larger speed would be achieved in low gear than in high gear over the same distance traveled by the bike...

 

[sophiecentaur]

How does the smaller force in high gear manage to produce the larger speed in the same amount of time when compared to low gear?

Your use of the "manage" sounds as if you actually doubt the Physics that's at work here if the situation doesn't make sense to you then you have to assume you are not looking at it in the right way. Here, you are changing more than one thing at a time and you are getting an apparent conflict. If the bike is doing work in taking you up a hill at a certain speed, moving up a gear will increase the force needed. It is pretty much impossible to know haw much work your body is actually doing when cycling. It's only after a long period of exercise that you will be aware of just how 'tired' you have become; no actual figures are available so you are stuck with a subjective effect. (Lab measurements are used in Sport Science and they mostly follow 'the theory'. See this link and many others)
You have accepted my point about studying a simple lever first and it is quite possible to express what happens on a bike to what a lever does. Clearly, as it involves wheels, a bike will run continuously and you can more easily relate Force times Speed for your feet directly to force times speed for the bike frame moving over the ground. But there is such a direct equivalence with a simple lever that there can be no surprises. Any confusion you may have will be down to how your body is experiencing the exercise. Sports coaches earn their money by reconciling what athletes feel to their actual performance.

 

[Drakkith]

As far as the feeling of working harder in high gear than in low gear: it may just be the perception that with a single pedal rotation the acceleration is small (since the the ground force is smaller) but the speed increase is larger in high gear. Is that really true?

What does "speed increase is larger" mean? Are you referring to acceleration, or to a starting speed and end speed? If you mean the latter, then see the rest of my post below.

That confuses me: the force on the ground is smaller in high gear than in low gear. The achieved speed in high gear is higher since we moves a larger distance in the same amount of time compared to low gear.

One problem you're having is that you're examining one or two static setups and then trying to apply your intuition to them. In reality a person riding a bike is a very dynamic system that goes through power changes, speed changes, and gear changes as the situation demands, and you aren't taking any of this into account.

Here's the thing. A multi-speed bike is like a car. It has an engine (you) that provides the force and power and it has a transmission that takes an input rotation and converts it into an output rotation to vary the output force, torque, and rotational speed. But why would we need to do this to begin with? Well, just like a car engine, our bodies cannot provide the same force on the pedals over an infinite RPM range. We start in low gear and quickly 'redline', where we just cannot pedal any faster. So we shift to a higher gear to bring our pedal speed back down to let us keep accelerating to a higher speed. The tradeoff is that we give up a larger force on the ground for a slower input RPM.

The reason a high gear feels 'harder' than a lower gear is that our input force is converted to a lower output force, which makes us accelerate slower. At too high of a gear you simply cannot apply enough force on the pedals to overcome friction and air resistance to keep up your speed, or to overcome gravity if you're trying to go up a hill in high gear, as your output force is only a fraction of what you're putting in. In these situations you shift to a lower gear to make things 'easier'.

Another thing is that shifting to a gear that's too high is similar to staying in low gear too long. Both result in a reduced force applied to the ground. In high gear this is because of the greatly reduced mechanical advantage, while in low gear it is caused by a greatly reduced input force on the pedals thanks to the very high pedaling speed. Note that in the latter case you still expend a lot of energy and get tired quickly even though your force on the pedals is not very high. This is because most of the force applied and energy spent goes into accelerating your feet and legs up and down, not into rotating the pedals. Hence why we have the bicycling exercise where you lie on your back, raise your feet and legs into the air, and pedal like you're riding a bike. Do that exercise as fast as possible and you'll find that you don't need pedals and a bike to tire yourself out.

 

[A.T.]

That confuses me...

You are throwing out one unjustified assertion after the other. Never specifying what assumptions you make like what is held constant in your comparisons. It's not even clear if you ask about basic mechanics or biomechanics and human leg efficiency.

 

[sophiecentaur]

You need to be careful about confusing cause and effect. You can jiggle about with equations and get all sorts of relationships in Physics but you always have to make sure that all the variables are taken into account. The 'demand' cannot affect what's 'available'.

Given a certain input power Pin determine by the force on the pedal and by the cadence (??), the output power is actually the same in low or high gear.

This is far too simplistic because it implies the input powers are the same (and the losses, of course). The reason that bikes have so many available gear ratios is that a cyclist cannot actually produce the same leg power over a wide range of speeds. If that were possible, we would all have just a 'three speed' with a very high, a medium and a very low gear. Reality is that the best and most comfortable arrangement ensures that the cyclist's legs are going at as near as possible the same rate, however fast the bike is travelling. If you want the best performance at a given speed then you will only get that over a very narrow range of gear settings.
The bottom line is available Power. That governs the speed that can be achieved (Speed = Power/Force or Power = torque times rotation rate etc).

 

[A.T.]

This is far too simplistic because it implies the input powers are the same (and the losses, of course). The reason that bikes have so many available gear ratios is that a cyclist cannot actually produce the same leg power over a wide range of speeds.

See also:
https://en.wikipedia.org/wiki/Muscle_contraction

 

[sophiecentaur]

See also:
https://en.wikipedia.org/wiki/Muscle_contraction

Humans are just not good subjects for simple physics experiments - certainly not when trying to understand Physics initially.

 

[fog37]

Thank you everyone. I certainly don't mean to be confusing and mix mechanics with biomechanics with my questions. I ride my bike and realized that my mechanical understanding of what is going on is limited.

In summary, assuming:

$F_{pedal}=$ input pedal force
$F_{ground}=$ output force
$R_W=$ wheel radius
$R_C=$ crank radius
$R_F=$ front sprocket radius
$R_B=$ back sprocket radius
$m=$ bike mass

The output force on the ground propelling the bike forward is $$F_{ground}= \frac {R_C R_B} {R_W R_F} F_{pedal}$$

The bike's acceleration is $$a= \frac {F_{ground}} {m}$$

 

[berkeman]

Thank you everyone. I certainly don't mean to be confusing and mix mechanics with biomechanics with my questions. I ride my bike and realized that my mechanical understanding of what is going on is limited.

In summary, assuming:

$F_{pedal}=$ input pedal force
$F_{ground}=$ output force
$R_W=$ wheel radius
$R_C=$ crank radius
$R_F=$ front sprocket radius
$R_B=$ back sprocket radius
$m=$ bike mass

The output force on the ground propelling the bike forward is $$F_{ground}= \frac {R_C R_B} {R_W R_F} F_{pedal}$$

The bike's acceleration is $$a= \frac {F_{ground}} {m}$$

I've only skimmed the rest of the thread (so apologies if this has been mentioned), but you are leaving out the effects of the MOI of the wheels. On a light road racing bike, it might be close to negligible, but on my MTB with wide dirt tires (and thorn protection bands inside the tires and heavy-duty inner tubes), the MOI of the wheels is definitely not negligible. The resistance to acceleration definitely includes the effort to spin up the tires.

Also, I don't see you taking into account the difference between standing and sitting when accelerating or climbing hills. You can generate more power standing, but you also tend to want to be in a slightly higher gear to optimize your pedal strokes...

 

[fog37]

I am assuming that all resistive horizontal forces are not present and the only horizontal force acting on the bike is $F_{ground}$.
As mentioned, the output force exerted on the bike by the ground is $$F_{ground}= \frac {R_C R_B} {R_W R_F} F_{pedal}$$ leading to a bike's acceleration is $$a= \frac {F_{ground}} {m}$$

In regards to power being the product $P_{ground}= F_{ground} v$: for the same power $P$, if the force $F_{ground}$ is small the velocity $v$ is large. I am still a little confused about this result. A small force $P_{ground}$ produces a small constant acceleration $a$ which, over a certain time interval $\Delta t$ will produce a small speed $v=a \Delta t$.

How does a small force $P_{ground}$ applied to the bike correlates with the the bike's velocity $v$, which increases as the bike moves, being larger than when the force is smaller?
When the force $F_{ground}$ is small, the velocity is low for the same power $P$. I envision the bike as an object with a force $F$ applied to it. How does the velocity $v$ stay small?

In regards to berkeman's observation, I am certainly not trying to include other biomechanical details at this point. I will after I am more clear.

Thanks for the patience.

 

[sophiecentaur]

Also, I don't see you taking into account the difference between standing and sitting when accelerating or climbing hills. You can generate more power standing, but you also tend to want to be in a slightly higher gear to optimize your pedal strokes...

There's no end to this, though. So many people perform better and with no increased 'effort' when they ride in a group. However can you quantify that?

 

[berkeman]

There's no end to this, though.

Well, from my experience riding a MTB and constantly trying to get better and faster, the MOI of your MTB wheels is significant. Even to the extent of me considering taking the thorn protection strips out and using standard innertubes instead of heavy duty. I would definitely gain in acceleration, but the extra hassle of a few more flats in the middle of nowhere would offset that... (yes I do carry a flat repair kit and pump).

So many people perform better and with no increased 'effort' when they ride in a group. However can you quantify that?

That's called the Peleton...

 

[hutchphd]

From experience I know that the bike reaches a faster speed in low gear than in high gear for the same number of pedal rotations.

How does a small force Pground applied to the bike correlates with the the bike's velocity v, which increases as the bike moves, being larger than when the force is smaller?
When the force Fground is small, the velocity is low for the same power P. I envision the bike as an object with a force F applied to it. How does the velocity v stay small?

What I think you are failing to understand is that much more Energy (Power)x(time) is required to go from 20mph to 25mph than to go from 0mph to 5mph. In fact it requires 9 times as much even ignoring any drag (you work it out !). If your cadence and foot force are the same it will take 9 times as long

 

[sophiecentaur]

thorn protection strips

Serious stuff. In my case it was always nails and screws.
I guess the MI can just be added to a general 'Inertia' value.
BUT:

How does a small force Pground applied to the bike correlates with the the bike's velocity v, which increases as the bike moves, being larger than when the force is smaller?

The OP still seems to be having a problem relating the 'gear equation' to what you get from a certain amount off pedal force. It all goes back to simple levers and I already advised starting with that most simple model. There are no surprises here - just a need for proper interpretation of the theory. I find his verbal version of it more or less impossible to make sense of.

 

[fog37]

Hello everyone, things are making more sense.

I am left with this dilemma though: as we change from a low gear to a higher, even without lifting the pedal from the accelerator, the engine RPMs go automatically down...Why is that?

To summarize, we start the car from rest and need a lot of force to cause it to move. First gear is the gear that provides the most force at the tire to accelerate the car. As we press on the accelerator, the engine's rotation increases and the torque also increase up to a point and the decreases. In first gear, the car eventually reaches a maximum speed $v_{max}$ and the engine's RPM are very high. The torque at that $v_{max}$ in first gear is matched by the net resistive torque. 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...

Thanks

 

[berkeman]

I am left with this dilemma though: as we change from a low gear to a higher, even without lifting the pedal from the accelerator, the engine RPMs go automatically down...Why is that?

Car? What car? I thought this thread was about bicycles...?

 

[sophiecentaur]

Car? What car? I thought this thread was about bicycles...?

It wouldn't work on a regular bike because there is a freewheel. It's easier to hop into a car. (But you can brake with a fixed wheel on a bike.)
The answer has to be that the engine turns into a compressor, pushing air out of the exhaust at a higher pressure than the intake. Work is done to achieve that and the energy source is the KE of the car as it slows.

 

[hutchphd]

I am left with this dilemma though: as we change from a low gear to a higher, even without lifting the pedal from the accelerator, the engine RPMs go automatically down...Why is that?

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)

 

[fog37]

Car? What car? I thought this thread was about bicycles...?

yes :) I started pondering about bike propulsion and ended up thinking about car propulsion. They are very different (cars use combustion engines) but I believe they share a lot of commonalities in terms of how the transmission is used...

 

[fog37]

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)

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...

 

[jack action]

does it feel very hard to pedal in high gear because the force on the ground is small?

But why does it feel hard to pedal in high gear?

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).

As far as the feeling of working harder in high gear than in low gear: it may just be the perception that with a single pedal rotation the acceleration is small (since the the ground force is smaller) but the speed increase is larger in high gear. Is that really true? That confuses me: the force on the ground is smaller in high gear than in low gear. The achieved speed in high gear is higher since we moves a larger distance in the same amount of time compared to low gear.

The speed increase is not larger if the acceleration is the same (i.e the force on the ground is the same). If the bicycle speed is the same, the power is also the same in both cases, so is the distance traveled in a given period of time. Again, the same acceleration means a higher pedal force in high gear.

If the wheel force is lower in high gear, then the pedal force will be the same but the pedal rpm will be lower and thus the pedal power will also be lower. The lower force at the wheel means a smaller acceleration and thus less distance traveled in a given period of time.

So if you are riding and change from low gear to high gear, the pedal rotation gets smaller instantaneously and so does your input power. At that point, you can still apply the same pedal force, but the sudden loss in acceleration feels like you are braking. Your natural instinct is to push harder on the pedal to reach the same power input, giving the same acceleration, all of that at the expense of increasing the pedal force (i.e. it feels harder).

How does the smaller force in high gear manage to produce the larger speed in the same amount of time when compared to low gear?

It doesn't. You either have the smaller wheel force and a lower acceleration or you increase the pedal force to create the same wheel force and the same acceleration.

The high gear only gives you the potential to reach a higher wheel speed (given that you have a maximum speed at which you can pedal). But the acceleration at a higher speed will necessarily be lower than whatever you would have gotten at a lower speed because of the lower wheel force (assuming the same power input that is).

 

[sophiecentaur]

I am left with this dilemma though: as we change from a low gear to a higher, even without lifting the pedal from the accelerator, the engine RPMs go automatically down...Why is that?

Your problem is your experimental method. You are not doing what you should always do in an experiment and that is to change only one thing at a time. Also you are recounting what things 'felt like' and not the actual measurements.
In this case, keeping the pedal in the same place doesn't actually keep anything constant - except the position of your foot. In particular, it doesn't keep the Power constant. Cars are a bit like humans; we don't have easy methods of measure the power they develop but we have do do stuff in a lab with a rolling road plus fuel metering and torque measurement etc..
Using an electric motor instead of a car engine will tend to get results that agree better with the theory you are having a problem with. Sounds a bit boring but you really have to believe all these basics and try to fit you experience to the theory - not the other way round. This is unlike the guys on the frontiers of Science, like Cosmologists and Fundamental Particle Physicists who keep having to change their models slightly.

 

[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.