First attempt with SPD's today in 30 mins

Will Snow

RealMan wrote:

Will Snow wrote:

First of all, ha! So who are these studies funded by, big companies such as shimano (who make mainly... err clipped in pedals) or boutique flat pedal companies, such as burgtech (who are so small they couldnt afford the huge expense of a scientific study). According to you, the latter.

The company RealMen LTD just did a study and found that RealMan is a 4000% better rider then anyone else in the world, and that toe clips give more power and control then any other type of pedal system in the world.

That was hard.

...how very mature.

Anyway, first of all realman, lets get the qualifications out of the way if I must. GCSE A physics, B alevel, sitting my degree level (BSc physics) exams atm.

Next point, you all call yourself phyicists! No one has drawn a diagram!

If I may simplify the point very much so:

Engine fixed distance from flywheel, with fixed power output. This represnts a rider, sat down on his bike, with the flywheel representing the cranks. They are connected by a chain. This represents the legs of the cyclist. The engine is started, a force is applied through the chain to the flywheel. However, does the chain that is returning from the flywheel, back to engine, which is in essence being pulled up (like the clipped in foot) provide extra power? If so, where does this come from? Assuming perfect system, i.e. perfect tension in chain etc.

If the chain only ran from the engine to the flywheel and not back again (an engineering nightmare I grant you), would the same amount of power be sent to the flywheel?



I think this is a fair model, though if anyone has a better idea, please pipe up.

Cferg

Will Snow wrote:

Next point, you all call yourself phyicists! No one has drawn a diagram!

Spoken like a true physicist Sheldon from the Big bang theory much? All jokes aside, I can't see anything wrong with that system seems like the best post in this thread, well makes most sense to me anyway.

bike-a-swan

My drawing was a scribble and would've made things clear. My critique of that model would be that a better representation is of two pistons working independently. I haven't really got anything to add to what I've already said yet- when I think of something I'll put it up.

cgarossi

I've done a diagram.

Please can someone explain to me if/why I am wrong? I am willing to be educated.

Will Snow

I think the argument is that you can pull up on the pedal, hence using different muscles etc., however the counter argument is that there is nothing to pull up against, by pulling up on the pedal all you are doing is pulling yourself off the seat.

The way I see the clips may be more efficient, as you can use different muscles, rather than with flats, therefore you dont get as tired as quickly. However I dont buy the claims of power coming from no where, in essence

NDawn

I smell a case for Mythbusters! In fact I have just submitted a thread on the mythbusters forum.
Let's hope they can settle this once and for all :P

bike-a-swan

Anyway, I would never call myself a physicist. I'm an engineer. I can't be expected to do bad drawings of things anywhere except the back of a napkin/menu/receipt in the pub.

cgarossi- based on that diagram, are you saying that the action of moving the crank away from you reduces the force you can put on it?

Will Snow

engineers are meant to be much better at drawing!! lol

cgarossi

Its a catch 22.

You must match the upward force in order to apply for downwards. Which then cancels out any upward force which then cancels downward force which then cancels upward force. etc etc etc.

The power is combined in the axle.

Will Snow

cgarossi wrote:

Its a catch 22.

You must match the upward force in order to apply for downwards. Which then cancels out any upward force which then cancels downward force which then cancels upward force. etc etc etc.

I think you would have to match the upwards speed, not force.

cgarossi

Yes exactly. If the power was the same, it would be super effecient. But thats hardly ever the case.

As soon as one force is greater than the other it dominates the turning of the cranks.

bike-a-swan

sorry, I really have lost you now. I think what you're saying is that the side moving at a greater angular velocity dominates and the other has no effect at all. That's not remotely relevant. They're moving at the same speed. The cranks connect the pedals, so they can't help it.

will snow- I can do technical drawings, or illegible scribbles on scrap paper. Little in-between!

cgarossi

Same speed yes. Of course.

But which crank is applying power?

Will Snow

i think the point that is trying to be made is similar to trying to pedal when your road speed doesnt match your gearing speed, you turn the cranks but no power is transfered to the wheel:- if you turn the crank pushing down faster than you pull up it will be the same effect. Not sure if I explained that very well.

I think in theory whichever crank is moving faster will apply to power, but no one can pull up with the speed (and power) they can push down with (with gravity and mass helping them)

bike a swan - haha spoken like a true engineer

Oxygen Thief

Loved reading all this again. Especially the link to those physics geeks, assuming it's as straight forward as pushing down and pulling up the same, that's it pure laws of physics with nothing else factored in whatsoever. Obviously leading very sad lives, never leaving the house, probably haven't seen a bike just sat in front of their computer masturbating over Isaac Newton.

supersonic

The more I think about it the more confused I get lol. Too many thought experiments going on.

However I did find this link which is interesting:

http://docs.google.com/viewer?a=v&q=cac ... D59Qr0WwEA

Feck, that is big.

Anyway if you go to the conclusion it says:

- Maximum torque was produced when actively trying to push down on the pedals
- Minimum torque is produced when actively trying to pull up on the pedals.
- Pulling up on the pedals smoothed out the torque curve ie a more even distrubution throughout the cycle.
- Your preferred method of pedaling is usually the most efficient out of all the methods employed.
- Actively pulling is the least efficient despite the more even torque.

Point 4 is very interesting.

Apologies if I have summed that up wrong, I am knackered lol.

flet©h

Ok, I'm new but I just couldn't resist with all this bogus science floating about. I do have a degree in mechanical engineering though so hopefully I should have some idea.

Lets try an experiment.

Stand on some scales. What happens? The weight reading is equivalent to the force applied though the soles of your feet.
Stand on the scales, lift one foot. What happens? The weight reading stays the same.
Fix your feet to the scales and pull up with one foot, what happens? The weight reading stays the same.
Put a handle bar in front of the scales attached to the floor. Unfix you feet from the scales. Pull up on the bar. What happens? The weight reading goes up.
Fix your feet back on the scales, pull up on the bar while pulling up on one foot. What happens? The weight reading stays the same as your previous attempt.

This should demonstrate that no mater how much you push and pull you can't exert more force through the soles of your feet. So SPDs don't let you exert more force. It's that simple. The circular nature of the cranks is just a red herring

However that has absolutely nothing to do with efficiency. I suspect the efficiency saving with SPDs may come when the pedals are vertical but I don't have a handy example to demonstrate that one.

supersonic

Hi Fletch

The scales was the analogy I used earlier

Welcome to the forums, I hope you post more.

Will Snow

@supersonic

i think i will read that tommorow haha but good find

Oxygen Thief

IP address check please! haha. Only joking. Well said fletch.

antfly

But is force just measured by weight? Force is mass x acceleration, is it not?

supersonic

Your mass x g = weight.

antfly

I know, but there is no force to measure without acceleration. Scales measure weight.

Will Snow

force is defined as change of momentum over time. That is the universal definition. There can be force without acceleration

Anonymous

It's also worth noting that anything moving in a circle is constantly accelerating.

EDIT:
Oh crap, just realised what thread this is! It's STILL going on?

Oxygen Thief

Scales measure your weight but there's force in your weight, it's g. You have no weight without g. So there's the force your looking for on the scales.

Will Snow

i hope the OP had fun on his SPD's

antfly

Force is an influence that causes a mass to change it`s velocity or accelerate. You have mass without g.

bike-a-swan

ooh, mech eng.

Point though- maximum net force you can apply to the cranks in any direction is not the question (at least, not for me). That, yes, is limited by your weight. However...

co-ordinate system. Up is positive. Looking at crankset from driveside. Clockwise (ie drive) is positive.

Imagine the cranks at 12 and 6. Rest all your weight (assumed vertical) on the 6'o'clock. Ok, force through that pedal is now your weight, call it W. To exert any more force down through that pedal you need something to pull against. If you pull up on the 12 by a further amount of force, say U, you can apply the same amount of force down through the 6 pedal, D. Where D = -U The net force on the bike remains your weight, W + D + U = W +D-D = W, ok?

Because the acting lines of the forces are through the pivot centre of the cranks, there is no rotation.

Bear with me...

Now, with the pedals at 3 and 9. Weight on the forwards pedal (3) W. Then pull up on the rear, as before, U. That means there is an upwards force on the rear and a corresponding extra force can be applied down on the front, D, with the net force through the pivot centre remaining your weight (W + D + U = W), as before. However, because the acting points of the forces are not through the a pivot centre of the cranks, you generate a moment. The moment applied by 3 is (W + D) multiplied by the crank length L ie. (W+D)L. Moment from 6 is UL. Total moment applied to the cranks = moment from front + moment from rear. Remember in this case the sign convention changes- both forces are operating in teh clockwise direction around the pivot, so the sign is now the same for U, L and D.

Total moment = (W+D)L +UL
= (W+D+U)L

With flats the largest moment you can apply is your weight, so the largest moment is only WL. Net force acting to push the bike (or rider) up or down remains only W in both cases, but the moment on the pedals for one is larger than the other.

And on that note, I'm going to bed. Enjoy!

Will Snow

but what are you pulling up against? Unless you have unholy buttock gripping power haha