First attempt with SPD's today in 30 mins

rubins4

They would be, yes. the opposite reaction (as alluded to ealier) is being applied through the 2 different objects (the people appling the inital force)

But when pedalling, the reaction to the upward force is working in opposition to the inital downward force, cancelling the reactive force out.

That makes sense in my head anyway...


I think me and ss are barking up the same tree here. so i will bow out at this point, as must do some revision

[Duncan Banatyne] ...so; I'm out. [/Duncan Banatyne]

Oxygen Thief

rubins4 wrote:

RealMan wrote:

The fact the the table or shelf is stationary is moot, it makes no physical difference.

The fact it is stationary is moot, but the fact it is a fixed object that you are exerting a force on isnt. On the bike, the object you are exerting your force on is not fixed, but is working not in opposition as such, but 'counter' to the direction of your initial force. So, I believe that by adding to the upstroke you are taking away from the downstroke, so no additional force can be applied.

Your right, the anology of the scales would only apply if you was using one pedal, one foot on top and one on the bottom! Your 'physics professor' must be whack realman

stuart_c-2

rubins4 wrote:

Stuart- I would say that is the same for flats or spds. You wouldnt be able to pull up on the bars to aid the upstroke with spds and achieve more force tho.

(I appreciate thats not quite what you are saying, and i think you are right, but this thread is getting a bit complicated to be having side discussions going on! )

Rubins, I don't think the pulling of the bars aids the upstroke, I think it aids the downstroke. I'm trying to remember how exactly I pull when really gong for it but I think basically when you push down on the pedal you're body wants to lift up but by pulling on the bar on the same side as the downstroke it holds your body down and allows more force to transfer? I think.?

Subplots are great, it's what all the best drama's are made of.

RealMan

supersonic wrote:

But place a bar on the weighing scales, and stand on it. Now pull up on it. Nowt happens. No matter how hard you pull, your feet will create an exact opposite force and balance it out. This is my anology for the bike, and how the mass is supported [again I will say for the sprinting out of the saddle case]

Yes that's right, but its not an analogy for the bike, because your weight is being supported by that bar. When standing, your weight is only on one crank.

jonbonjovial wrote:

Your 'physics professor' must be whack realman

I don't like him much to be honest, but he does know his physics. He would probably explain this quite similar to how I am explaining it, but with more shouting and name calling.

cgarossi

The analogies used in this thread don't work because there are two planes of motion in cranks. Upwards AND downwards.

Walls have only one, the direction of the most dominant force

The same for suitcases and trains.

A force applied to a crank has a reaction on both the upward and downward planes. Therefore any opposite force applied to any part of the crank will counter act any other force. An upward force will move the opposite pedal away with the same amount of force.

rubins4

lol, subplots.

Yeah, I agree with what you say Stuart. If you can oppose the reaction to your pedal strokes when 'honking' :P it by pulling up on the bars, then the additional force will be applied to your downward stroke, I also think.

Oxygen Thief

And the wall anology! WTF?! That would be the equivalent of one foot puching the pedal down and the other being underneath taht pedal pulling it! What level of physics do you actualyl do realman because that's an absolutely shocking and embarrassing anaology if your anything above GCSE, lower even!

RealMan

cgarossi wrote:

The analogies used in this thread don't work because there are two planes of motion in cranks. Upwards AND downwards.

Walls have only one, the direction of the most dominant force

The same for suitcases and trains.

A force applied to a crank has a reaction on both the upward and downward planes. Therefore any opposite force applied to any part of the crank will counter act any other force. An upward force will move the opposite pedal away with the same amount of force.

I'm sorry but you just don't understand circular motion. If there are two planes of motion in cranks, up and down, what about left and right? Or what about diagonally left down, and diagonally right up? Ad infinitum...

supersonic

Yes that's right, but its not an analogy for the bike, because your weight is being supported by that bar. When standing, your weight is only on one crank.

My analogy is to do with the forces and the net sum. Stand one foot on the bar, one foot on the scales, makes no difference.

cgarossi

Surely if you understood it Realman, you would not be using wall analogies. :roll:

bails87

cgarossi wrote:

The analogies used in this thread don't work because there are two planes of motion in cranks. Upwards AND downwards.

Walls have only one, the direction of the most dominant force

The same for suitcases and trains.

A force applied to a crank has a reaction on both the upward and downward planes. Therefore any opposite force applied to any part of the crank will counter act any other force. An upward force will move the opposite pedal away with the same amount of force.

What about the revolving door one? Looked at from above it'll be like looking at the cranks side on.

supersonic

bails87 wrote:

cgarossi wrote:

The analogies used in this thread don't work because there are two planes of motion in cranks. Upwards AND downwards.

Walls have only one, the direction of the most dominant force

The same for suitcases and trains.

A force applied to a crank has a reaction on both the upward and downward planes. Therefore any opposite force applied to any part of the crank will counter act any other force. An upward force will move the opposite pedal away with the same amount of force.

What about the revolving door one? Looked at from above it'll be like looking at the cranks side on.

It is how to do with how things are supported. And where the force is coming from.

cgarossi

True.

In that case, a person going through the door would have to push on a door panel while another pulled.

I think you'd find one would be tripping up all the time or being squashed

Who ever was pushing/pulling hardest would be encountering the entire resistance of the door. The dominant force wins out, every time.

Dirtydog11

rubins4 wrote:

They would be, yes. the opposite reaction (as alluded to ealier) is being applied through the 2 different objects (the people appling the inital force)

But when pedalling, the reaction to the upward force is working in opposition to the inital downward force, cancelling the reactive force out.

That makes sense in my head anyway...


I think me and ss are barking up the same tree here. so i will bow out at this point, as must do some revision

[Duncan Banatyne] ...so; I'm out. [/Duncan Banatyne]

Are you saying that you wouldn't produce any power on the upstroke?

Surely if your power output was X amount on the downstroke and X amount on the upstroke. Total power would be would be X + X.

RealMan

cgarossi wrote:

True.

In that case, a person going through the door would have to push on a door panel while another pulled.

I think you'd find one would be tripping up all the time or being squashed

Who ever was pushing/pulling hardest would be encountering the entire resistance of the door. The dominant force wins out, every time.

Have you any background in physics? It really does not work like that.

cgarossi

Surely if your power output was X amount on the downstroke and X amount on the upstroke. Total power would be would be X + X.

This is what this whole thread is about

It would actually be:

(R - Df) + (R - Uf)

Where R is resistance.

rubins4

Dirtydog11 wrote:

rubins4 wrote:

They would be, yes. the opposite reaction (as alluded to ealier) is being applied through the 2 different objects (the people appling the inital force)

But when pedalling, the reaction to the upward force is working in opposition to the inital downward force, cancelling the reactive force out.

That makes sense in my head anyway...


I think me and ss are barking up the same tree here. so i will bow out at this point, as must do some revision

[Duncan Banatyne] ...so; I'm out. [/Duncan Banatyne]

Are you saying that you wouldn't produce any power on the upstroke?

Surely if your power output was X amount on the downstroke and X amount on the upstroke. Total power would be would be X + X.

I see what your saying, but I believe that any force applied to the upstroke is taken away from the downstroke, because of the circular nature of cranks. As we have discovered already, the situation is not the same a a train pulling from the front and another pushing from the rear.

Alternatively, you have just settled the arguement once and for all.

Regardless, I'm out. (again)

bails87

cgarossi wrote:

True.

In that case, a person going through the door would have to push on a door panel while another pulled.

I think you'd find one would be tripping up all the time or being squashed

Who ever was pushing/pulling hardest would be encountering the entire resistance of the door. The dominant force wins out, every time.

I know it's not a practical design for revolving doors

My point was that rather than a wall/suitcase, it's a rotating object, like the cranks.

Super: I've been basing my points on a seated rider, I think you've been assuming the rider was standing, and I think I understand the argument from that point of view.

cgarossi

RealMan wrote:

cgarossi wrote:

True.

In that case, a person going through the door would have to push on a door panel while another pulled.

I think you'd find one would be tripping up all the time or being squashed

Who ever was pushing/pulling hardest would be encountering the entire resistance of the door. The dominant force wins out, every time.

Have you any background in physics? It really does not work like that.

Please provide a clear explanation as to why not.

Oxygen Thief

Realman, do you mind me asking your physics background?

RealMan

http://www.singletrackworld.com/forum/t ... st-1427761

rubins4

He's got 100 meter swimming badge. :P

Jokes

Getting another load of layman in on a subject isnt gonna conclude anything by the way, nor does it reveal any of your credentials on the topic.

RealMan

No I don't actually (I'm terrible at swimming).

No, I don't mind. I have an A in GCSE Physics. And I'm just coming up to my final exams at A level physics, most likely that I will get a B.

Mostly cause I'm terrible at particle based physics, wave-particle duality and quarks and stuff like that. Its practical real life situations like this where I excel at.

ride_whenever

what would happen if you were cycling on the ceiling clipped in?

cgarossi

Realman, you really think total output via cranks is Downward + Upward? That simple?

Oxygen Thief

I also have A @ GCSE and have B @ A2 level. But I've been out of school for a years now :-/ But still I now that you can not make anologys between push/pull forces being added up on a singular object eg a wall and on two objects working about a pivot.

bails87

ride_whenever wrote:

what would happen if you were cycling on the ceiling clipped in?

Depends if your flux capacitor is working properly.

Oxygen Thief

cgarossi wrote:

Realman, you really think total output via cranks is Downward + Upward? That simple?

He does yeah! That's embarrassing for an A level physics student!

RealMan

Would just like to once again refer you to this, the guys on STW are quite clever (well, most of them).

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

They're cretins. Why on earth even bother trying to explain it to them?

if they are too stupid to understand that if you push in one direction and pull in the same direction that you have increased the force of just one of these alone then thank God I dont frequent there?

In the words of Ripley

'Did IQ's drop while I was away?'

They're not all as nice and polite as me though.. :?

Lifer

Think about an engine with two cylinders attached to a crank. Combustion chamber at the top of one and the bottom of the other. If both spark plugs ignite at the same time do you think that there would be no power produced?