First attempt with SPD's today in 30 mins

cgarossi

Ok. Im not a physics expert but I beleive I do have a grasp of it:

I beleive it should look something like this:

DF = (DownwardResistance - DownwardPedalForce)
UF = (UpwardResistance - UpwardPedalForce)

T = ((DR - UF) - DF) + ((UR - DF) - UF)

Where DF = Downward Force, UF = Upward Force, DR = DownwardResistance, UR = UpwardResistance, T = total power output.

I stand to be corrected, but no way does T = Down + Up

RealMan

Why are you still talking about resistance? What has that got to do with anything?

cgarossi

Because pulling up with one pedal removes resistance from the downward stroke. And vice versa. Which has an effect on total power output.

RealMan

No it doesn't. The resistance is still there. But I still don't know why you want to talk about resistance.

bike-a-swan

I need to sit down and do a proper bit of work on this, but I'm currently running flow simulations which reduces my usable time to conveniently post length chunks between checking things. Until then, here's my tuppence worth on first impressions:

supersonic has a a good point regarding the fact that all the force you need to apply has to be reacted somewhere, and in this case that must be through your weight. The only thing that makes me think I'm missing something here- what happens if you cycle along sitting down (i know it's a different case), and only pedalling on the upstroke? The force is reacted by the saddle, not your weight. If this makes any difference in the full case, I don't know, as I said, not sure.

cgarossi, you're being unfair to realman. I think I agree with some of your posts, but I can't be sure because they're poorly explained. Writing something unclearly then insulting people when they don't understand them is not a good way to go about this sort of thing.

realman, don't pull more people into an already scarily large argument just for numbers sake. It really doesn't help the discussion.

Anyway, I'm actually curious now, anyone got any new, well reasoned, and un-insulting new explanations?

cgarossi

Well i've tried to explain it.

I've yet to see you try and explain why T = U + D?

cgarossi

I applogise if I insulted anyone, but that white flag he posted was an insult in itself

Oxygen Thief

Why don't the designers of these things put some coin in to some research of the matter. It's obviously a massive thing.

RealMan

bike-a-swan wrote:

Anyway, I'm actually curious now, anyone got any new, well reasoned, and un-insulting new explanations?

Its not really that new, but the laws of physics?

rubins4

jonbonjovial wrote:

Why don't the designers of these things put some coin in to some research of the matter. It's obviously a massive thing.

Just in case conclusive proof is discovered that spds dont have any effects at all! :shock:

cgarossi

Well I beleive Realman is just plain wrong. Sorry.

supersonic

So far we have talked about just up and down motion. It has been hinted on before that the perfect pedaling technique is one where the force you apply is at a tangent to the cranks, and than pedaling is not of course just up and down.

If the cranks were at say 6 and 12 and we pushed and pulled [and pulling in this case is distinctly different from pulling 'up' with spds] the force would be cumulative.

Lifer

It's rotational force not reciprocal, therefore the upstroke and downstroke do not cancel each other out.

Aidy

cgarossi wrote:

Ok. Im not a physics expert but I beleive I do have a grasp of it:

I beleive it should look something like this:

DF = (DownwardResistance - DownwardPedalForce)
UF = (UpwardResistance - UpwardPedalForce)

T = ((DR - UF) - DF) + ((UR - DF) - UF)

Where DF = Downward Force, UF = Upward Force, DR = DownwardResistance, UR = UpwardResistance, T = total power output.

I stand to be corrected, but no way does T = Down + Up

How have you managed to get Power output from summing a collection of forces?

cgarossi

Hmm.

I dont think so because moving the crank moves the opposite crank.

cgarossi

Aidy wrote:

cgarossi wrote:

Ok. Im not a physics expert but I beleive I do have a grasp of it:

I beleive it should look something like this:

DF = (DownwardResistance - DownwardPedalForce)
UF = (UpwardResistance - UpwardPedalForce)

T = ((DR - UF) - DF) + ((UR - DF) - UF)

Where DF = Downward Force, UF = Upward Force, DR = DownwardResistance, UR = UpwardResistance, T = total power output.

I stand to be corrected, but no way does T = Down + Up

How have you managed to get Power output from summing a collection of forces?

I told you I was no expert Replace PedalForce with PedalPower?

bike-a-swan

super, what do you think of this?

Put someone on a turbotrainer in a gravity free environment, not restrained in any way, on flat pedals. If they pushed down on the forwards pedal, they'd also begin to float away and stop being able to pedal.

Same condition, now clipped in. Downwards force on leading pedal can be reacted be pulling up on trailing pedal with the same force. Net force on person goes to zero, there is still rotational input to the cranks.

I'm not completely sure where I'm going with this, but I think it suggests you could exert more force than your weight. Does it ring true?

supersonic

Not sure about the gravity free thing lol, will have to think about it.

My maximum force statement was just for that particularly scenario of the down stroke/upstroke around the 3 and 9 oclock area, when standing on the pedals.

llamafarmer

RealMan wrote:

bike-a-swan wrote:

Anyway, I'm actually curious now, anyone got any new, well reasoned, and un-insulting new explanations?

Its not really that new, but the laws of physics?

I really don't think it's as simple as you make out RealMan (or you cgarossi) - It isn't a simple system, there's a ton of variables. The force exerted on the pedals isn't uniform all the way round the stroke, nor is it always perfectly perpendicular to the crank and through the pedal axle. You're never applying your full body weight to the pedals because you have to balance yourself on the bike and that force is effected by the handlebars and whether you're seated or standing.

cgarossi

Yeah. I was thinking on an all things being equal basis.

But I really object to the idea that its Downward + Upward.

Cferg

bike-a-swan wrote:

super, what do you think of this?

Put someone on a turbotrainer in a gravity free environment, not restrained in any way, on flat pedals. If they pushed down on the forwards pedal, they'd also begin to float away and stop being able to pedal.

Same condition, now clipped in. Downwards force on leading pedal can be reacted be pulling up on trailing pedal with the same force. Net force on person goes to zero, there is still rotational input to the cranks.

I'm not completely sure where I'm going with this, but I think it suggests you could exert more force than your weight. Does it ring true?

Your weight is the effects of gravity, where your weight (w) is a force with the direction of gravity. Weight = mass x gravity. If gravity = 0 then you can have no weight. Therefore you'd be floating anyway and would be completely weightless in a system which has no gravity. I can see what you're getting at but it couldnt happen.

biff55

good grief !
you people have way too much time on your hands.

Lifer

cgarossi wrote:

But I really object to the idea that its Downward + Upward.

It's rotational force! The downward and upward forces aren't countering each other.

bike-a-swan

That's my point. If you had no weight, could you power a rotating system? If you pushed on something unrestrained you'd float away. If you pushed on something with one hand and pulled with the other, you wouldn't move because the net force would be zero. However, if you add a distance between the action points, there will be a torque around some central position, surely?

Eranu

335D is clearly the answer to this debate...end of.

cat_with_no_tail

biff55 wrote:

good grief !
you people have way too much time on your hands.

Oh good god

+

Lifer

bike-a-swan wrote:

However, if you add a distance between the action points, there will be a torque around some central position, surely?

Bingo.

RealMan

cgarossi wrote:

Yeah. I was thinking on an all things being equal basis.

But I really object to the idea that its Downward + Upward.

Downward and upward doesn't exist in circular motion. Only clockwise and anti clockwise. Both pedals are going in the same direction.

http://www.singletrackworld.com/forum/t ... physicists

supersonic

bike-a-swan wrote:

That's my point. If you had no weight, could you power a rotating system? If you pushed on something unrestrained you'd float away. If you pushed on something with one hand and pulled with the other, you wouldn't move because the net force would be zero. However, if you add a distance between the action points, there will be a torque around some central position, surely?

If you were stood on the pedals I don't think you could turn the cranks.

If you were strapped to the bike, then yes.

Though I am tired, so let me think about it a little more lol.

supersonic

There is no need to keep linking to STW.

But one post stands out:

'It's about frames of reference and what the people are pulling and pushing against, that's what's complicating the thought process I think.'

Downward and upward doesn't exist in circular motion. Only clockwise and anti clockwise. Both pedals are going in the same direction.

But we have been talking about the direction of forces, which are important.