First attempt with SPD's today in 30 mins

Anonymous

Nope. You are assuming the rider is fixed in space. Remember that your weight is also supported by your legs.

Oxygen Thief

Yes and F = m x a. The a in this case is g.

supersonic

What about at constant motion, or when the bike is accelerating?

The study I linked shows a net reduction in torque in any scenario compared to pushing down ie all your weight on the crank.

bike-a-swan

I knew I shouldn't have checked back. You pull up on the other pedal, thats the U. That's why if you manage a perfectly spinny pedalling technique there will be no effect to compress your suspension, the forces in the vertical direction cancel out. In rotation they reinforce each other.

Anonymous

Sonic, the net reduction comes from one foot removing the ability of the other foot to apply pressure.
It's not forces cancelling out, it's far simpler than that. It's common sense, and presumable, physiology.

antfly

bike-a-swan wrote:

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!

Does this mean SPDs work?

Anonymous

bike-a-swan wrote:

I knew I shouldn't have checked back. You pull up on the other pedal, thats the U. That's why if you manage a perfectly spinny pedalling technique there will be no effect to compress your suspension, the forces in the vertical direction cancel out. In rotation they reinforce each other.

Right, now you've just lost a load of credibility . The up and down pedalling technique is FAR from the only thing that causes suspension movement. Think about the the chain trying to pull the rear hub closer to itself and compressing the suspension (when the pivot single pivot, or VPP is above the BB) or extending the suspension (when the pivot is a low one)
Another commonly misunderstood aspect of mountain bike physics and mechanics.

supersonic

I am saying that the actual outcome is the opposite of what many have expected ie an increase in force [and therefore torque]. Even worse than the output being the same lol.

bike-a-swan

Oh, I accept how it all works practically, and don't really care. I use spds because I like the fact the bike sticks to my feet. My interest has descended into a theoretical exercise. I was undecided for a while (see earlier posts), but when I really sat down and worked it out the maths just works out on one side.

antfly

jonbonjovial wrote:

Yes and F = m x a. The a in this case is g.

G causes a but g isn`t a itself. There is no force without a.
Where`s Stephen Hawking when you need him?

bike-a-swan

yeehaamcgee wrote:

bike-a-swan wrote:

I knew I shouldn't have checked back. You pull up on the other pedal, thats the U. That's why if you manage a perfectly spinny pedalling technique there will be no effect to compress your suspension, the forces in the vertical direction cancel out. In rotation they reinforce each other.

Right, now you've just lost a load of credibility . The up and down pedalling technique is FAR from the only thing that causes suspension movement. Think about the the chain trying to pull the rear hub closer to itself and compressing the suspension (when the pivot single pivot, or VPP is above the BB) or extending the suspension (when the pivot is a low one)
Another commonly misunderstood aspect of mountain bike physics and mechanics.

Please don't descend to accusing me of no credibility, the thread has managed to remain a reasonably civilised discussion since you and realman stopped mudslinging. I used an illustration, which necessarily means referring to some sort of real world case. I am well aware the chain tension and who knows how many other things affects the suspension as well, I am discussing the effects of these forces only.

Will Snow

antfly wrote:

jonbonjovial wrote:

Yes and F = m x a. The a in this case is g.

G causes a but g isn`t a itself. There is no force without a.
Where`s Stephen Hawking when you need him?

g is acceleration due to gravity, 9.81ms^-2

bike-a-swan

antfly wrote:

bike-a-swan wrote:

ooh, mech eng.

Point though- m........
...maths stuff...
...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!

Does this mean SPDs work?

Theoretically. For practical tests please refer to that thing super put up. Or your own bike.

bike-a-swan

http://xkcd.com/386/

supersonic

If the chain tension is constant, wont the suspension reach an equilibrium?

Cferg

But the forces on the cranks should also be constant which is going to be very difficult to maintain.

bike-a-swan

ok, trying the suspension illustration was perhaps a bad idea, it's going to lead to discussions about vpp and things that I freely confess I only have a qualitative understanding of. Think about it on a hardtail then. You pedal in a 'push, push, push' manner and the fork goes up and down. Spin smoothly and it doesn't, because the net force in any direction remains constant. Look, basically, you get a couple-
http://en.wikipedia.org/wiki/Couple_(mechanics)
(sorry to use wikipedia, but it's faster and easier than typing it myself). no net force. just rotation.

Will Snow

what the deuce has suspension got to do with spds?

bike-a-swan

poorly chosen example, I'm afraid, trying to illustrate how you can achieve rotational motion with no net force in a linear axis. Perhaps not well enough explained at first.

Now I am definitely going to bed.

Will Snow

i still wanna see clip in gloves. Never loose grip on the bars again!! Increase steering efficiency!! haha

supersonic

I shall have another think tonight.

I think that we all may be half correct. And half wrong. I got an idea...

Anonymous

bike-a-swan wrote:

poorly chosen example.

Precisely why I called you out on it.

bike-a-swan

Not an incorrect example, just perhaps not the best way to illustrate my point in written form.

RichMTB

poorly chosen example, I'm afraid, trying to illustrate how you can achieve rotational motion with no net force in a linear axis. Perhaps not well enough explained at first.

But an object rotating at constant speed is actually accelerating!

wesk

So with all this theory floating around, I thought I'd try an experiment last night.
I clipped in on one foot only, then I tried to pedal. Guess what, I could pull up without exerting any force down with the other leg. I then did it with the other leg (just to make sure that my dodgy knee wasn't creating an artificial result!)
So, you CAN pull up and I'm positive it can increase your power. However, I'm not convinced it will increase your efficiency......
Additionally, I keep seeing that the maximum force you can put through the peddles is the weight of the rider. Erm, OK if the rider is a dead weight but I'm pretty sure we all have muscles. Think you can't exceed weight in force standing up? Why do bar ends help you climb.....?
It's actually a simple experiment to find out what makes a differeance. I read about a tyre one being done (properly for once)

http://www.bicicletta.co.za/Downloadabl ... trated.pdf

sustitute flat / spd for tyre swap around there you go.

However, assuming that the resuts showed that there is no differance wither way, I'd still stik to spd's 'cos I like them.

Anonymous

Try Sonic's experiment. Stand with one foot on the pedals, clipped in, the other one dangling.
Now try pedalling.
You can pull up with your bum on the saddle to brace yourself, but not when standing.

Meanwhile, maximum downwards force is achieved when standing, or, if seated at the point where your weight is actually fully supported by your "pushing" leg.
So, if you cannot pull up without anything to brace your eight against, you aren't actually gaining any power by pulling up.

antfly

yeehaamcgee wrote:

So, if you cannot pull up without anything to brace your eight against, you aren't actually gaining any power by pulling up.

Why not?

Anonymous

If your weight is supported by your leading foot, you have nothing to brace yourself against to pull.
You cannot pull up without something to brace yourself against.

By pulling up, all you are actually doing is transferring your weight to the frontmost foot - which gains nothing. using your whole bodyweight on your frontmost foot already achieves peak power, so all you're doing is putting peak power through the frontmost pedal, which can be done anyway without pulling - by simply unweighting your trailing foot.

Aidy

yeehaamcgee wrote:

Try Sonic's experiment. Stand with one foot on the pedals, clipped in, the other one dangling.
Now try pedalling.
You can pull up with your bum on the saddle to brace yourself, but not when standing.

Meanwhile, maximum downwards force is achieved when standing, or, if seated at the point where your weight is actually fully supported by your "pushing" leg.
So, if you cannot pull up without anything to brace your eight against, you aren't actually gaining any power by pulling up.

Try this experiment.

Put the bike in a workstand, using one hand spin the pedals.
Get a friend to apply the rear brake to the precise point where you can no longer rotate the pedals.

Now use both hands, one on each pedal.

antfly

Let me guess {save us the bother}, you can turn the pedals with two hands easier than one?