First attempt with SPD's today in 30 mins

supersonic

It has been, in large, a good discussion, even if there was a bit of mudslinging at the start. We ain't STW lol.

I certainly admit I didn't fully look at the picture, but was happy to go off and have a good think about it.

bike-a-swan

It's taken my mind off some extremely tedious work. I've enjoyed it! Nice to think about this sort of thing properly.

Anonymous

Right, here's my points of note.
considering a scenario when crank arms are horizontal

1) Maximum downwards force is achieved when entire bodyweight is supported on frontmost pedal

2) Lifting force is applied by using leg muscles to apply such a force that the distance between saddle and rearwards pedal is reduced

3) The lifting force in 2) means that at least some bodyweight is resting on the saddle, taking bodyweight from the frontmost pedal.

4) The forces applied whilst pulling (lifting) and pushing (bodyweight) do not directly cancel each other out. Rather it is the method employed to produce lifting force that makes the downwards force less effective

5) it is impossible to apply a lifting force to the pedal whilst standing. This is because in order to lift, the rider's body must be braced against something. In the case of bracing against the front pedal, what is actually achieved is a weight transfer to the frontmost pedal.

6) point 5) can be demonstrated by unclipping one foot and leaving it dangling in the air, whilst trying to lift the pedal with the clipped in foot. This must be done from a static starting point in order to negate any momentum playing a part.

Oxygen Thief

Been interesting following this I must say. Not that I'm much the wiser either way. The bit I take most from is that people do their best using their preferred method. That'll do for me.

alfablue

yeehaamcgee wrote:

Right, here's my points of note:

5) it is impossible to apply a lifting force to the pedal whilst standing. This is because in order to lift, the rider's body must be braced against something. In the case of bracing against the front pedal, what is actually achieved is a weight transfer to the frontmost pedal.

6) point 5) can be demonstrated by unclipping one foot and leaving it dangling in the air, whilst trying to lift the pedal with the clipped in foot. This must be done from a static starting point in order to negate any momentum playing a part.

Yeehaa, you are probably correct, just 2 questions, can lifting force not be achieved by bracing against the resistance of the downstroke; and whilst greater power may be achieved whilst standing, if choosing to sit then full rider weight cannot be put on downstroke so is the pedal lift therefore not a significant bonus?

Aidy

bike-a-swan wrote:

yeehaamcgee wrote:

then you could not put any more power than your weight through them.

This is the only thing on here I have issue with, I'm afraid it's wrong.

Ok, I'm going to try another shot at explaining why. I tested this earlier. Get your bike, sit on it, leaning against a wall. Set the cranks vertical. Put all your weight through the lower pedal. Then try and put more force through it. You can do this by pulling up in the opposing pedal.

Agreed?

That's not a particularly good experiment.

Here's a better one:

Start with the cranks of your bike horizontal, apply the rear brake such that you can put your entire weight on the leading pedal.

Now pull up with the rear-most pedal.

bike-a-swan

Yeah, that'll do too. I suggested the first because I thought it'd probably be easier to do. Either is good!

Anonymous

alfablue wrote:

Yeehaa, you are probably correct, just 2 questions, can lifting force not be achieved by bracing against the resistance of the downstroke; and whilst greater power may be achieved whilst standing, if choosing to sit then full rider weight cannot be put on downstroke so is the pedal lift therefore not a significant bonus?

Lifting force can't be produced when "bracing" against the frontmost pedal. I'm still trying to think of a simple way of explaining why - but it essentially boils down to the fact that what you're bracing against actually moves away, combined with the fact that you're actually just transferring your weight by removing load from the front pedal (effectively pulling yourself INTO the rear pedal)

I'm trying to think of something like a situation where you're pushing a car by pushing with your feet against another car which has it's handbrake off - but there's too many differences in that setup - when I come up with a workable analogy that's easier to visualise I'll explain further.

As for the situation where riding seated, you're quite right that pulling up can produce more power than a casual pedal. However, you will still not produce as much power as when having all your weight on the downstroke.
When seating, if you have just enough force going through the downwards pedal to remove your weight off the saddle, you're already putting down as much power as you can.
Having enough power down to remove your weight from the saddle results int he same scenario as standing.
This is a similar concept to maximum traction on tyres....
When braking, you are decelerating most rapidly at that point just just just just just before you loose traction - same goes for putting force through the pedal when "seated"

bike-a-swan

yeehaamcgee wrote:

the fact that you're actually just transferring your weight by removing load from the front pedal (effectively pulling yourself INTO the rear pedal)

you pull yourself down into the rear pedal. the reaction pulls the rear pedal up.

To cancel this out, you must put more force down through the front pedal.

Get a bike, try either my experiment or aidys, either proves the point.

Anonymous

The reaction is minimal to none, I'm using the analogy of pulling yourself into the pedal to try and explain why it's a weight shift, and not producing torque.

Aidy

yeehaamcgee wrote:

Right, here's my points of note.
considering a scenario when crank arms are horizontal

1) Maximum downwards force is achieved when entire bodyweight is supported on frontmost pedal

If you ignore coupled forces, yes. You can apply more if you have something to push against.

2) Lifting force is applied by using leg muscles to apply such a force that the distance between saddle and rearwards pedal is reduced

Okay...

3) The lifting force in 2) means that at least some bodyweight is resting on the saddle, taking bodyweight from the frontmost pedal.

Only if you ignore coupled forces again.

4) The forces applied whilst pulling (lifting) and pushing (bodyweight) do not directly cancel each other out. Rather it is the method employed to produce lifting force that makes the downwards force less effective

Using your argument, yes.
Using physics, no.

5) it is impossible to apply a lifting force to the pedal whilst standing. This is because in order to lift, the rider's body must be braced against something. In the case of bracing against the front pedal, what is actually achieved is a weight transfer to the frontmost pedal.

Fair enough.

6) point 5) can be demonstrated by unclipping one foot and leaving it dangling in the air, whilst trying to lift the pedal with the clipped in foot. This must be done from a static starting point in order to negate any momentum playing a part.

True enough, but misleading.

Will Snow

Aidy wrote:

Using your argument, yes.
Using physics, no.

Well that settles that then.

Anonymous

Aidy wrote:

yeehaamcgee wrote:

Right, here's my points of note.
considering a scenario when crank arms are horizontal

1) Maximum downwards force is achieved when entire bodyweight is supported on frontmost pedal

If you ignore coupled forces, yes. You can apply more if you have something to push against.

Precisely the problem, you have nothing to push against.

bike-a-swan

Regarding point 5)

Say you start with your weight equally distributed between the two pedals. weight on front = weight on rear. weight on front + weight on rear = total weight

You reduce the weight on the rear one, as you said. The weight on the front increases correspondingly. weight on front /=(can't find a 'does not equal' sign) weight on rear. weight on front + weight on rear = total weight

Eventually you get to the point where you have all the weight on the front.

weight on rear = 0 weight on front = total weight

Then, if you pull up on the rear pedal...
weight on rear = -1N (say) (assuming up = positive)
conservation of force in the vertical axis dictates that: weight on front + weight on rear = total weight. as before. therefore:

weight on front = total weight - weight on rear
weight on front = total weight - -1
weight on front = total weight + 1

weight (force) on front can be greater than total weight.

Aidy

Will Snow wrote:

Aidy wrote:

Using your argument, yes.
Using physics, no.

Well that settles that then.

Brief, but the coupled forces thing mentioned several times before kinda gives the game away

bike-a-swan

aidy, you seem to have this one, I'm going to bed. Don't let me down!

Anonymous

bike-a-swan wrote:

Regarding point 5)

Say you start with your weight equally distributed between the two pedals. weight on front = weight on rear. weight on front + weight on rear = total weight

You reduce the weight on the rear one, as you said. The weight on the front increases correspondingly. weight on front /=(can't find a 'does not equal' sign) weight on rear. weight on front + weight on rear = total weight

Eventually you get to the point where you have all the weight on the front.

weight on rear = 0 weight on front = total weight

Then, if you pull up on the rear pedal...
weight on rear = -1N (say) (assuming up = positive)
conservation of force in the vertical axis dictates that: weight on front + weight on rear = total weight. as before. therefore:

weight on front = total weight - weight on rear
weight on front = total weight - -1
weight on front = total weight + 1

weight (force) on front can be greater than total weight.

Not bad. However, what you achieve by attempting to lift the rear pedal is to actually remove the force applied to the frontmost pedal.

(incidentally, not equals sign can be represented as != just so we're on the same page )

Will Snow

Ok so it seems the problem breaks down like this:

On one side, people say that you can pull up against the pushing down force

The other side say that there is nothing to pull up against, as you are not braced against anything and by pulling up you simply reduce the pushing down force .

I think thats right? Until someone can prove the other one wrong, this isnt going anywhere. Bonus points for diagrams!!

Anonymous

What's your hypothesis, Will?

alfablue

Will Snow wrote:

The other side say that there is nothing to pull up against, as you are not braced against anything and by pulling up you simply reduce the pushing down force .

I think the issue that is being overlooked is that there is resistance to the pedal downforce and therefore a lag, in this lag an upward force can be applied, I often find myself doing this when trying to accelerate hard (sorry, no knowledge of physics or engineering). If the pedals could spin freely then I could understand how any upforce would be impossible, but they are not spinning freely (in the real cycling world, rather than imagining a bike on a workstand).

supersonic

Yes, I believe it is the resistance many have overlooked. You cannot have torque without resistance ie something to oppose the effort you are putting in.

Will Snow

im not sure I get where you are coming from alfablue, if there is a lag in the downstroke then why wouldnt there be the same lag in the upstroke?

Me Yeehaamcgee? I cant see that pulling up would get you more power, but I would say it may be more efficient, as you use more muscles (pully up muscles as well as pushy down muscles [phyicist not a biologist haha]) therefore your pushy down muscles wont get tired as quickly, yah de dahdy yah. Would explain why roadies and XCers use them.

Anonymous

Fair enough. I see the advantages of SPDs being that you can pedal through parts of the stroke that you just can't with flats, and being able to control the rear of the bike in any situation.
One of the things I do miss form my days riding SPDs was being able to pick up the back end of the bike in a two wheel drift and re-plant it before completely losing traction. When using flats, I have to have my feet in the right position to be able to pick it up.
Also, as others have noted, SPD shoes have a more direct shoe-to-pedal interface, minimising any power lost through squidgy soles in shoes. For extreme distance road riding that would surely be a great benefit, as any energy lost due to squidgy soles is going to hurt you over hundreds of Kilometres.

antfly

I don`t know if this has been linked to before but it`s interesting.
http://pt.wkhealth.com/pt/re/ppv/abstra ... 29!8091!-1

And this study http://www.ncbi.nlm.nih.gov/pubmed/18418807

alfablue

Will Snow wrote:

im not sure I get where you are coming from alfablue, if there is a lag in the downstroke then why wouldnt there be the same lag in the upstroke?

well there is a lag on the upstroke, therefore you can pull against it (this resistance). The other explanations seem to suggest that any up-pull results in the pedals running away from you negating any effect, but this would assume a freely spinning crank, which it is not, and therefore the resistance means there is something to pull against.

Will Snow

Agreed on all points really there, yeehaa. Circular systems are funny to work with anyway, and there are alot of variables here. Many other benefits of clipless, a few of which youve touched upon there. I dont use them because Im a wuss hahaha

Alfablue, do you mean pulling against the resistance of the road? If so I think I see where you are coming from

alfablue

Will Snow wrote:

Alfablue, do you mean pulling against the resistance of the road? If so I think I see where you are coming from

Yes, the resistance through the drivetrain, wheels, tyres and the road, the whole thing fighting against your turning force being converted into forward motion.

Will Snow

so far i think its fair to say that we have been dealing with an ideal system, i.e. no friction etc, and we've not had a whole heap of joy finishing that yet! I think the trick is to work out which way this resistive force would be working, and I think as it is resisting the movement of the bike, this would translate to a circular force rotating the opposite direction to the cranks, in essence just reducing the force input by the rider, so Im not sure how you could push against it...but I could be wrong, I might not be thinking about it right.

antfly

The studies suggest you are wrong.

Will Snow

who me? In what way, I thought the 'studies' were on my side? Or did I read them wrong?