Question for the resident boffins.....

[MSTRS]

When your bike is stationary, it's weight (plus rider) is all on the two tyres' contact patches. However, once you start moving, centrifugal force comes into play. This means that (theoretically), the force throwing outwards at the tyre tread is altering the contact patch. What I am interested in knowing is, does this make your total unit 'lighter' in relation to it's stationary weight?? I realise that the arc of tyre etc below the axle is pressing outwards in a generally down direction, but not being at rightangles to the contact patch, only part of that force is applied to the contact patch. Of course, that means that the upper half of the arc is applying force away from the contact patch.

[scumdog]

When your bike is stationary, it's weight (plus rider) is all on the two tyres' contact patches. However, once you start moving, centrifugal force comes into play. This means that (theoretically), the force throwing outwards at the tyre tread is altering the contact patch. What I am interested in knowing is, does this make your total unit 'lighter' in relation to it's stationary weight?? I realise that the arc of tyre etc below the axle is pressing outwards in a generally down direction, but not being at rightangles to the contact patch, only part of that force is applied to the contact patch. Of course, that means that the upper half of the arc is applying force away from the contact patch.


I doubt that centrifugal force re the tyre has much bearing, the forces cancel each other out.
Lateral forces are more what affects your tyres grip. (I think)

[MSTRS]

Ahhh, but do they?? I am talking of when you are travelling in a straight line only.

[sAsLEX]

centrifugal force comes into play

Cant do it doesn't exist

[MSTRS]

Try swinging a bucket full of water then

[sAsLEX]

Try swinging a bucket full of water then

and? I will provide a force associated with centripetal acceleration, but no cenrifugal force.

and
The fictitious centrifugal force appears when a rotating reference frame is used for analyzing the system. The centrifugal force is exerted on all objects, and directed away from the axis of rotation.

[scumdog]

Try swinging a bucket full of water then

As said before, they cancel out unless you have a really out-of-balance rim.

Try swinging 80 buckets of water in an arc - all equally spaced out.

After all, if you spin a push-bike wheel while holding it by its axle does it leap violently up and down??

[sAsLEX]

Consider a ball that swings around a stationary pivot to which it is tethered by a light, strong rope. There is tension in the rope, pulling inwards on the ball (the centripetal force) and simultaneously pulling outwards on the pivot (the reactive centrifugal force). The tension is real, so these two forces still exist if we move to a corotating frame. However, in the rotating frame there is also a fictitious centrifugal force that pulls outwards on the ball. It is distinct from the reactive centrifugal force that pulls outward on the pivot.

This is what I meant, no cetrifugal force on the tread.

[MSTRS]

After all, if you spin a push-bike wheel while holding it by its axle does it leap violently up and down??
No, of course it doesn't. We are talking about a state of balance, but not sure that upper and lower quadrants cancel each other out

[sAsLEX]

below the axle is pressing outwards in a generally down direction, but not being at rightangles to the contact patch, only part of that force is applied to the contact patch. Of course, that means that the upper half of the arc is applying force away from the contact patch.


Your argument falls apart here, as the "force" you mention is equal in all directions on the tyre it cancels out completely. It does introduce gyroscopic implications though, but you are talking about going in a straight line so they dont come in to it.

[Skyryder]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder

[MSTRS]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder
Absolutely- that is the point I was trying to get clarified

[scumdog]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder

The scales would require some sort of downwards motion to cause a reading, if there was insufficient time for that motion to occur fully then I suppose you would get a lighter reading.
If the scales had a weigh-bridge of say 50 metres long you would then get a true reading.

[scumdog]

No, of course it doesn't. We are talking about a state of balance, but not sure that upper and lower quadrants cancel each other out

Lets re-look at the push-bike wheel thing, if it was on it's side and spinning it would weigh X weight, stand it vertical and I can not see it suddenly being lighter, it stays the same weight, X, ergo there is no force caused by the spinning that will 'lift' it and make it lighter.

Quadrants means in four sections, I cannot see how anything would affect the wheel in any particlar quadrant at random other than gravity.

[sAsLEX]

The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder


not too sure how you could get this. If an object is not accellerating *sp then all forces are balanced, wether at 0 or 900 kmhr. No matter how fast one goes the vertical forces are balanced or else at a certain speed you would fly without wings ( I am ignoring lift associated with the rider/bike here).

What force associated with an increasing speed would counter the downwards force of gravity and reduce the bikes weight!?

[Big Dave]

regardless of centrifugal forces - I reckon the bike gets lighter when in motion because of the curvature of the earth. mass wants to keep going straight ahead away from the axis against the pull of gravity - thus making it lighter.

you'd need some pretty trick scales but.

[MSTRS]

This is getting really heavy (no pun intended). If said bike could be put on a treadmill that was capable of giving a weight reading, then my theory could be tested.

[Highlander]

Duh. Course the bike gets lighter when it's moving. Nasa studies the phenomenum closely it is refered to as "FUEL BURN"

[MSTRS]

Duh. Course the bike gets lighter when it's moving. Nasa studies the phenomenum closely it is refered to as "FUEL BURN"
Yes.....well.....OK

[Ixion]

but, but, but you are ignoring the Einsteinein effects. Theory of Relavity, mass increases as speed increases. So the faster it goes the heavier it gets.

Wadda ya mean , insignificant . I thought that all the litre sprotsbikers regularly got their knee down at near light speed?

[sAsLEX]

Theory of Relavity,

General or Special?

[scumdog]

Maybe that is why planes fly - they go so fast that they get really-really light and then they fly?????

[Ixion]

General or Special?

Well, both really. Einstein has said that all of the consequences of Special Relativity can be found from examination of the Lorentz transformations, and since the velocity/mass realtionship is part of the Lorentz transformations, it must be contained within the Special Theory. (ie the relationship is not affected by gravity). But the General theory is an extension of the special theory (or, the special theory is a subset of the General theory). So, both.

EDIT: mr = m0 /sqrt(1 - v2/c2) defines it.

[MSTRS]

Well, both really. Einstein has said that all of the consequences of Special Relativity can be found from examination of the Lorentz transformations, and since the velocity/mass realtionship is part of the Lorentz transformations, it must be contained within the Special Theory. (ie the relationship is not affected by gravity). But the General theory is an extension of the special theory (or, the special theory is a subset of the General theory). So, both.
F... me - would you like fries with that?

[sAsLEX]

Well, both really. Einstein has said that all of the consequences of Special Relativity can be found from examination of the Lorentz transformations, and since the velocity/mass realtionship is part of the Lorentz transformations, it must be contained within the Special Theory. (ie the relationship is not affected by gravity). But the General theory is an extension of the special theory (or, the special theory is a subset of the General theory). So, both.

True, just doing some light reading of Hawkings a Universe in a Nutshell at the mo, his sense of humour is quite a laugh

[Highlander]

Me thinks you'd hear some interesting concepts if you raised this question around the camp fire at a rally when fuelled by a few ales....:drinknsin

[Sniper]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.


Until you went too fast and flew over the scales.

[merv]

If your bike weighs say 200kg wet and you weigh 100kg also wet, then the total weight is 300kg and it remains that weight no matter how fast you go within the realms of the speed the bike can go that is. There is no way the bike can become lighter with speed at the contact points unless you go over a jump.

Or unless like Big Dave says because of the earth's curvature, but no bike can go fast enough to make much of a difference. If I could be bothered finding the time I could work that out for you.

Also taking the theory of relativitiy, as I've never been that fast I'm not sure what happens as you approach the speed of light. I've heard it slows down time, but I'm not sure about weight.

In basic engineering design the opposite is actually allowed for in that the live loading (i.e. at speed) is always increased over static loading. What I'm talking about here is loadings on bridges for example of traffic going over them where extra is added for impact of live loads i.e. the assumption that increased forces are applied at joints, bumps etc. The motorcycle designers would do the same in designing the strength of their wheels.

Any feeling of lightness when on the move is all about the gyroscopic effect of the wheels making the bike want to stay going as it is.

[Pixie]

Lets put it this way. If you weighed your bike when it is stationary and the scales showed say 200lb you would get a lighter weight if you rode over the scales. The faster you went over the scales the lighter the weight would be.

There's a formula for this just not too sure what it is.

Skyryder
No,you've got it all wrong.
The faster you rode over the scales,the shorter you and your bike will become,foreshortened in the direction of motion,and the more massive you and the bike will become.

[merv]

Going on about scales, the only reason they might read less is they are too bloody slow in their mechanism, but any decent in motion weighing system would read the correct weight in a tiny split second and it won't be any different than stationary.

[Ixion]

..
Also taking the theory of relativitiy, as I've never been that fast I'm not sure what happens as you approach the speed of light. I've heard it slows down time, but I'm not sure about weight.

,,.

I'm just waiting for someone to tell y' that if y'had a Suzuki you would have been that fast.

[yungatart]

Lets not confuse speed with velocity or weight with mass. If I remember my 4th form science lessons (and, lets face it, it was a loong time ago..),they are not the same thing. It could have a bearing on what you, the esteemed boffins, are trying to explain to the poor uneducated plebs like me..

[Waylander]

It was all very annoying when they tought us it was centafugal force one year then the next suddenly centrafugal force no longer exhists and centripetel is the word of the day. Can't these 'scientists' make up thier mind?

Damn braniacs getting revenge for all that nerd shit in highschool.

[Harry33]

:argh: This is all neat and stuff but does anyone know what the 13 herbs and spices are?

[MSTRS]

:argh: This is all neat and stuff but does anyone know what the 13 herbs and spices are?
Who cares.....it's still cat underneath.:lol:

[Skyryder]

The scales would require some sort of downwards motion to cause a reading, if there was insufficient time for that motion to occur fully then I suppose you would get a lighter reading.
If the scales had a weigh-bridge of say 50 metres long you would then get a true reading.

No you would not. Velocity or forward movement reduces weight at point of contact. This is one of the few things that still sticks from my science class, years ago. It's one of the reasons cars slide off the road when going round corners at speed. They become too light and lose traction and keep on travelling in the direction of momentum.

It's the reason racing cars have foils so that wind pressure helps to 'push' the car down on the tarmac.

Skyryder

[Skyryder]

This http://regentsprep.org/Regents/physics/phys06/bcentrif/centrif.htm
explains the difference between Centrapetal and Centrfugal Force. The two are not the same.

Skyryder

[Jamezo]

It was all very annoying when they tought us it was centafugal force one year then the next suddenly centrafugal force no longer exhists and centripetel is the word of the day. Can't these 'scientists' make up thier mind?

Damn braniacs getting revenge for all that nerd shit in highschool.
As already mentioned, "centrifugal force" is a disgusting laymans term for an observation of straight-line force about a rotating central frame of reference.

It's all a borgeious conspiracy.

[marty]

a vehicle (and everything for that matter) is being held on the earth at 1g. if you exceed 1g in a sidewise or upwards direction the vehicle will head there.

what is being said here means that a 4wd vehicle, on a 4wd dyno/treadmill, when it has it's wheels rotating, will weigh less? don't think so - it still will exert 1g downwards.

[scumdog]

No you would not. Velocity or forward movement reduces weight at point of contact. This is one of the few things that still sticks from my science class, years ago. It's one of the reasons cars slide off the road when going round corners at speed. They become too light and lose traction and keep on travelling in the direction of momentum.

It's the reason racing cars have foils so that wind pressure helps to 'push' the car down on the tarmac.

Skyryder

It's not because they get 'lighter' - it's because the lateral force exceeds the downwards force (gravity)

To look at it another way when a car is slowing down it should get heavier according to your theory - so if you slow down as you go around a corner the car should stick better and grip the road - not sledge straight ahead off the road?

And yes, airfoils do 'push' a car downwards - to overcome the sideways pull of lateral forces when cornering AND at high speeds to stop 'lift' due to airflow over the body.

[marty]

geez SD - that's a late night on the riviera!

[marty]

the resultant component of weight does shift as the car moves around, at a standstill it is right below the c of g, but apply an acelleration in any direction (or multiple directions) and the resultant vector will move. the total mass doesn't change, so the weight of the vehicle doesn't change, unless there is a substantial (that is, measurable) change in altitude, to one which is beginning to realise a change in the earth's force due to gravity

[thealmightytaco]

Any centripetal force from the wheels act equally up as they do down so there wont be any weight change, unless you're nearing the speed of light and E=mc^2 is coming into play bu that's another story.

And the original q about whether the centripetal acceleration on the wheel will deform the tyre more = bigger contact patch; This is why you buy speed rated tyres, for cars at least, I'm not sure about bikes, I havent had to buy any yet, but the tyres can expand from the force and all but want to leave the rim if they're not up to task. But in real life (with a properly rated tyre) the extra deformation from centripetal force compared to you sitting on the bike would be minimal against a tyre inflated to 30psi. Really.

[scumdog]

geez SD - that's a late night on the riviera!

Yeah well home from quelling riots, rescuing virgins, busting drunks, etc etc and needed an 'unwind' before hitting the sack.:sleep: :zzzz:

[MSTRS]

And the original q about whether the centripetal acceleration on the wheel will deform the tyre more .....

Sorry, you misunderstand. That was not the original question. Skyrider seems to understand where I was coming from. Perhaps I worded it ambiguously.
The question (reworded) was - does the pressure exerted on the contact patch lighten as a bike gets faster??
Everything has a given mass, but it's weight is subject to gravity and other forces, so can *alter*. The braking scenario is one - whereby impetus denied translates to a greatly increased downward pressure on the front contact patch, thereby making the attached bike 'weigh more'. Also when accelerating, the same force applies to the rear contact patch, for as long as acceleration continues.

[scumdog]

Sorry, you misunderstand. That was not the original question. The braking scenario is one - whereby impetus denied translates to a greatly increased downward pressure on the front contact patch, thereby making the attached bike 'weigh more'. Also when accelerating, the same force applies to the rear contact patch, for as long as acceleration continues.

ONLY because the C of G of the bike alters, nothing to do with spinning wheels, there would be the same (more or less) downwards pressure on the front whel if it was locked-up under braking.

[Highlander]

there would be the same (more or less)

How does that work? if it is more or less how can it be the same? :confused:

[MSTRS]

ONLY because the C of G of the bike alters, nothing to do with spinning wheels, there would be the same (more or less) downwards pressure on the front whel if it was locked-up under braking.
Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???

[sAsLEX]

ONLY because the C of G of the bike alters, nothing to do with spinning wheels, there would be the same (more or less) downwards pressure on the front whel if it was locked-up under braking.

The coefficeint of friction would change once locked up, thereby reducing the friction force acting on the c.o.g therefor there would be less pressure.

[scumdog]

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???

It was more of a reference to the fact wherether the wheels were spinning or not the downwards force is (relatively) the same.

A wheelspin take-off should according to your theory provide more rearwards downpressure - yet in reality it would not. i.e. try it on a wet road with wheelspin vs a dry road with no wheelspin.

[sAsLEX]

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???


Its *baby* weight increased not in the vertical plane but in the horizontal actually. It didnt get heavier in that it would squish poor mummy holding it.

[MSTRS]

Now you are just getting silly, I wasn't talking about traction.:shifty: It is a good point however, since drag or friction has a bearing on how much pressure (weight) is exerted at the point of contact.

[MSTRS]

Its *baby* weight increased not in the vertical plane but in the horizontal actually. It didnt get heavier in that it would squish poor mummy holding it.
yep, but thought it would help illustrate what I was getting at

[sAsLEX]

yep, but thought it would help illustrate what I was getting at

nope just illustrates you are getting confused really, completely different case that one. Can you provide any sort of theory or formula for this weight increase as something increases in speed?

[scumdog]

How does that work? if it is more or less how can it be the same? :confused:

O.K. pedant, I was allowing for 'minor variations'!:nya:

[Highlander]

O.K. pedant, I was allowing for 'minor variations'!:nya:

I knew what you meant, just had to do something to bring this back to Earth. Can anyone else hear that whooshing sound?

[MSTRS]

nope just illustrates you are getting confused really, completely different case that one. Can you provide any sort of theory or formula for this weight increase as something increases in speed?
No I can't - and the theory was that a bike's 'weight' on the contact patches would decrease ever so slightly at speed (assuming acceleration has stopped) because of what I see as the lifting action of the tyre's weight being thrown outwards from the centre (axle).
Perhaps there is no lift because being circular the forces around the circumference keep everything equal as others have said.

[scumdog]

No I can't - and the theory was that a bike's 'weight' on the contact patches would decrease ever so slightly at speed (assuming acceleration has stopped) because of what I see as the lifting action of the tyre's weight being thrown outwards from the centre (axle).
Perhaps there is no lift because being circular the forces around the circumference keep everything equal as others have said.

That is what the rest of us are trying to tell you : the lifting as you call it is 360 degrees around the wheel, ergo it is pulling downwards, forwards and backwards just as much as it is upwards, therefore the forces cancel each other out.

As I said earlier, spin a bike wheel while hanging onto it by its axle - it certainly won't take off upwards into orbit.

[Virago]

Is more than a shift in C of G. Say 200kg bike at equilibrium exerts 100kg on each contact patch, then under braking the force on the front patch is waaaaay greater than the total weight of the bike (200kg). Remember the ad of the baby in the front seat passenger's arms and what 'weight' it became when the car had the impact???
You're still confusing the downwards "weight" with the forwards momentum.

The "baby in the arms" scenario explains that beautifully. If the baby weighs 5Kg, it will still have a downwards force of 5Kg after impact. But attempts by the mother to stop the forward momentum make the baby seem many times that weight, but in a horizontal direction.

A bike accelerating or deceleratiing is exactly the same. A bike weighting 200Kg with a 50/50 balance between the wheels, will exert 100Kg on each point of contact. Under braking, the centre of balance shifts forward, putting a weight of say 150Kg downwards force on the front wheel. But there is now only 50Kg downwards force on the back wheel.

The additonal force on the front wheel comes from absorbing the forward momentum (deceleration), but this force is horizontal, not vertical.

An object simply cannot become "lighter" by travelling at speed. It's mass is fixed and unalterable. It's weight (mass x gravity) therefore cannot change either, unless you can magically change gravity. Only other external forces acting on the object can change the perceived weight.

[MSTRS]

That is what the rest of us are trying to tell you : the lifting as you call it is 360 degrees around the wheel, ergo it is pulling downwards, forwards and backwards just as much as it is upwards, therefore the forces cancel each other out.

As I said earlier, spin a bike wheel while hanging onto it by its axle - it certainly won't take off upwards into orbit.
:not: :not: I know that now. As said was just a theory I was asking about. But as usual *some* members went so far off on a tangent (think about that one) that I couldn't resist seeing if digging in my heels would increase the pressure exerted on them:Pokey:

[RantyDave]

the force throwing outwards at the tyre tread is altering the contact patch. What I am interested in knowing is, does this make your total unit 'lighter' in relation to it's stationary weight??
No, but it will make the contact patch slightly smaller since the centrifugal forces will stretch the tyre out.

And, for the pedants, there is indeed no such thing and centrifugal force. But it's so much easier to explain than acceleration towards the centre of a circle so shall we just leave it at that?

Dave

[MSTRS]

You're still confusing the downwards "weight" with the forwards momentum.

The "baby in the arms" scenario explains that beautifully. If the baby weighs 5Kg, it will still have a downwards force of 5Kg after impact. But attempts by the mother to stop the forward momentum make the baby seem many times that weight, but in a horizontal direction.

A bike accelerating or deceleratiing is exactly the same. A bike weighting 200Kg with a 50/50 balance between the wheels, will exert 100Kg on each point of contact. Under braking, the centre of balance shifts forward, putting a weight of say 150Kg downwards force on the front wheel. But there is now only 50Kg downwards force on the back wheel.

The additonal force on the front wheel comes from absorbing the forward momentum (deceleration), but this force is horizontal, not vertical.

An object simply cannot become "lighter" by travelling at speed. It's mass is fixed and unalterable. It's weight (mass x gravity) therefore cannot change either, unless you can magically change gravity. Only other external forces acting on the object can change the perceived weight.
My Primer One teacher told me I should not bother continuing wth Physics:weird: All I needed to know was that if I jumped out of a tree onto concrete, I would probably hurt myself. I believed her.

[Virago]

My Primer One teacher told me I should not bother continuing wth Physics:weird: All I needed to know was that if I jumped out of a tree onto concrete, I would probably hurt myself. I believed her.
I recall reading that Eintein's teachers told him the same thing..........

[scumdog]

My Primer One teacher told me I should not bother continuing wth Physics:weird: All I needed to know was that if I jumped out of a tree onto concrete, I would probably hurt myself. I believed her.


NOW you've said something scientific that is true!!

(if you spun as you fell would you hit the ground harder or softer.........?)

[MSTRS]

(if you spun as you fell would you hit the ground harder or softer.........?)
Just more accurately.....

[Virago]

......(if you spun as you fell would you hit the ground harder or softer.........?)
.....and would the blood spurt in an arc or a parabola....?

[MSTRS]

A bike accelerating or deceleratiing is exactly the same. A bike weighting 200Kg with a 50/50 balance between the wheels, will exert 100Kg on each point of contact. Under braking, the centre of balance shifts forward, putting a weight of say 150Kg downwards force on the front wheel. But there is now only 50Kg downwards force on the back wheel.


Must resist....mus....resi....res...oh bugger it.....
In the case of a 'stoppie' does the back wheel lift off the ground because the whole bike is rotating around the front axle....or....does the front wheel get so heavy that it 'steals' so much weight off the back wheel, thereby giving it a negative weight and enabling it to float up off the ground???:rofl: :wari: :eek5: :dodge:

[Virago]

regardless of centrifugal forces - I reckon the bike gets lighter when in motion because of the curvature of the earth. mass wants to keep going straight ahead away from the axis against the pull of gravity - thus making it lighter.

you'd need some pretty trick scales but.
Okay, just got out of the shower (it's where I do my best thinking).....

What sort of horizontal "escape velocity" would be required to effectively launch an object into "orbit", ie render it "weightless"? I'm guessing around 30,000km/hr?

Sooooooo........,

If a bike is travelling at 300km/hr (1% of 30,000), would it be 1% "lighter", due to the curvature of the earth?

And......

Here's where it gets tricky. Due to the Earth spinning on its' axis, we effectively have some centripedal force trying to "spin us off", ie making us effectively lighter. Would the direction travelled by the bike make any difference to the situation?:blink:

This is getting too complicated - must be time for a ride....

[Virago]

Must resist....mus....resi....res...oh bugger it.....
In the case of a 'stoppie' does the back wheel lift off the ground because the whole bike is rotating around the front axle....or....does the front wheel get so heavy that it 'steals' so much weight off the back wheel, thereby giving it a negative weight and enabling it to float up off the ground???:rofl: :wari: :eek5: :dodge:
Negative weight - you've got it!

It's like how doughnuts aren't fattening - the hole in the middle has effectively "negative" calories, which balance out the rest of the doughnut......

[MSTRS]

Negative weight - you've got it!

It's like how doughnuts aren't fattening - the hole in the middle has effectively "negative" calories, which balance out the rest of the doughnut......
:yeah: :drinknsin :woohoo: :rofl: I so want to bling you for 'this one'.....but site has other ideas. Consider your self blung

[Highlander]

wwhhhoooosshhh

[MSTRS]

wwhhhoooosshhh
.....sound of toilet flushing at high speed, making contents too heavy to remain in the bowl.....

[scumdog]

What a great thread! A lot of crap typed out but it mainly (more or less?) stuck to the same topic - which for KB is pretty good considering most threads veer off (spinning too fast?) at a *tangent in about 4 or less postings.:laugh: :woohoo:


*a man who has been on the sun-bed a lot???<_<

[MSTRS]

Who would have thought it possible??? I was impressed too. And no-one mentioned beer.....whoops

[TwoSeven]

The question (reworded) was - does the pressure exerted on the contact patch lighten as a bike gets faster??


I think the question would be both yes and no, depending on how you look at things.

If you are looking at the weight of the bike+rider on a non-moving machine, then I think no - assuming that the weight bias (and a few other factors) never changes.

If you are looking at the driving force applied to the rear tire, then the answer is yes. Force = power/linear velocity so if you always drive at peak power, the force must be getting smaller as you get faster.

[froggyfrenchman]

regardless of centrifugal forces - I reckon the bike gets lighter when in motion because of the curvature of the earth. mass wants to keep going straight ahead away from the axis against the pull of gravity - thus making it lighter.

you'd need some pretty trick scales but.

And THAT is my newest excuse for pulling wheelstands everywhere i go.:Police:

[Skyryder]

It's not because they get 'lighter' - it's because the lateral force exceeds the downwards force (gravity)

To look at it another way when a car is slowing down it should get heavier according to your theory - so if you slow down as you go around a corner the car should stick better and grip the road - not sledge straight ahead off the road?

And yes, airfoils do 'push' a car downwards - to overcome the sideways pull of lateral forces when cornering AND at high speeds to stop 'lift' due to airflow over the body.

Yep that's it. Lighter is the wrong word. However in the sense of using scales lighter seemed the approriate word at the time.

Skyryder

[Skyryder]

not too sure how you could get this. If an object is not accellerating *sp then all forces are balanced, wether at 0 or 900 kmhr. No matter how fast one goes the vertical forces are balanced or else at a certain speed you would fly without wings ( I am ignoring lift associated with the rider/bike here).

What force associated with an increasing speed would counter the downwards force of gravity and reduce the bikes weight!?

Laterl force. The bikes weight does not alter only the registed weight from the scales. See Scumdogs post #41 and mine 78I think

Skyryder

[Jamezo]

If you are saying what I think you are saying, that is quite wrong.

The tyres of a banked over and cornering bike *do* exert more force on the road than that of a bike that is stationary or moving in a straight line. (but keep reading.......)

The increased force that your tyres have with the road (and we know this is the case, you can see how your suspension compresses in a corner), can be split up into two component vectors:

#1, the horizontal component, the lateral part of the force applied by the tyres, which causes the bike to accelerate towards the centre of the arc of motion, and is directly proportional to the magnitude of this acceleration.

#2, the vertical component, which is always (for a bike which is not presently accelerating upwards or downwards, so lets just say is it in a stable mid-corner) the same, exactly equal to the normal weight force of the bike, the same force that would be on the tyres/road in straight-line motion.

so we can see, that a bike, even in mid-corner, will have the same weight force distributed to the ground. if a large set of scales were constructed to be able to test this, it would be readily confirmed.

but *everything* changes, *if* the bike is momentarily undergoing vertical (in relation to the ground/road, the frame of reference) acceration.

this happens both when a bike is entering and exiting a corner. while a bike is entering a corner, the centre of mass of the bike/rider system undergoes an acceleration towards the ground, and subsequent decceleration when the maximum lean angle is approached. the converse happens when exiting a corner, with an acceleration away from the ground, and subsequent decceleration as the upright is approached.

during periods of acceleration towards the ground (starting the entry to the corner), and decceleration to the upright (coming out of the exit of the corner), the downwards component of the force transmitted through the tyres is reduced.

this fully complies with, and can concurrently be explained by, newtons laws of motion, the acceleration of the mass towards the ground "uses up" (ie. scalarifically negates (yes I made up a word, sue me)) some of the normal downwards weight force of the bike from gravitational acceleration.

as per the previous statement, if a set of scales were constructed to allow a bike to corner on them, this would experimentally show to be the case.

now that is The Word on bike weight-cornering-whatever. you guys have no excuses now.

[Coyote]

If you are saying what I think you are saying, that is quite wrong...
Whoa, what he said

[Skyryder]

as per the previous statement, if a set of scales were constructed to allow a bike to corner on them, this would experimentally show to be the case.

now that is The Word on bike weight-cornering-whatever. you guys have no excuses now.

No ones talking about cornering. If you travel in a straight line, lateral momentum is going to alter the gravity forces on the object. Lets look at this from another angle.

Say you are the passanger in a car travelling at 50kph and you drop a lead weight of 50 kg's out of the window onto the road. Now lets say from the point frome where the weight left your hand to the point of where it landed on the road is 50 meters.

Now lets say the second time you did this you were travelling twice the speed 100kph and you dropped the 50kg weight out of the window. All things being the same as the first experiment the distance travelled would be double. Same weight same gravity forces double the speed and double the distance.

Conversely the opposite would happen. Travel at 50kph drop a weight of 100kg and that would be 50% shorter than the first experiment or thereabouts. In all cases both weight and graity forces are the same but the differential in distance is governed by speed.

The same applies to mass when it is at velocity. Gravity is unable to exert the same amount of force to a moving object as it can to a stationary object and as such if the weight can be measured it will 'appear' to be lighter..

Its why rockets fly and planes glide. One uses wings the other sheer speed to keep it in the air. When the speed slows the rocket falls to the ground.

Skyryder

[TwoSeven]

An object simply cannot become "lighter" by travelling at speed. It's mass is fixed and unalterable. It's weight (mass x gravity) therefore cannot change either, unless you can magically change gravity. Only other external forces acting on the object can change the perceived weight.

What about Issac Newtons law of universal gravitation.

Gravity decreases with the square of the distance, so the force of gravity applied to a motorcycle is dependant on its distance from the earths surface.

So a motorcycle with a mass of 160kg + a rider mass of 90kg would have a total weight of 2450N on the earths surface. If the motorcycle was lifted 1 earth radius upwards (about 6378km), then its weight would only be 612.5N (a bucket load less).

Also, I think that an object of a mass traveling at a velocity can exert more force than that exerted by gravity. In which case I understand that it will leave the earth. I think the formula is v = sqrt(2gR).

This kind of indicates that a said motorcycle traveling at 11.1km/s (about escape velocity) would weigh about 75% less after 9.5 minutes of travel.

It seems to me that an object travelling at speed away from the earth would be getting lighter - or have I gotten this wrong ? :)

[Ixion]

..
Conversely the opposite would happen. Travel at 50kph drop a weight of 100kg and that would be 50% shorter than the first experiment or thereabouts. In all cases both weight and graity forces are the same but the differential in distance is governed by speed.

...
Skyryder

Mr Galileo was of a different mind.

[TwoSeven]

#1, the horizontal component, the lateral part of the force applied by the tyres, which causes the bike to accelerate towards the centre of the arc of motion, and is directly proportional to the magnitude of this acceleration.


Do you have a formula for this ?





#2, the vertical component, which is always (for a bike which is not presently accelerating upwards or downwards, so lets just say is it in a stable mid-corner) the same, exactly equal to the normal weight force of the bike, the same force that would be on the tyres/road in straight-line motion.


F = mg

So for a 160kg bike + 90kg rider, then F = 250 * 9.8 = 2.45kN

Is this correct ?

[kiwifruit]

wow, some bright sparks here!
very interesting read

[thealmightytaco]

I think the question (reworded) is referring to weight transfer is it?

The bike nor tyre's will never weigh more but the weight each wheel is carrying at any one time is always changing. Race car's have brake balance adjustment for that business, aint nothin' new. Up to 75-80% weight on the front, or on a bike 100% if you're doin' a stoppie say, and then 0% on the back, and vice versa for pullin' wheelies. But the bike never weighs more due to centripetal forces because they act equally up and down and forward and back and everywhere, unless you cut a hunk out of your tyre, but you ain't doin' that.

[Lou Girardin]

No you would not. Velocity or forward movement reduces weight at point of contact. This is one of the few things that still sticks from my science class, years ago. It's one of the reasons cars slide off the road when going round corners at speed. They become too light and lose traction and keep on travelling in the direction of momentum.

It's the reason racing cars have foils so that wind pressure helps to 'push' the car down on the tarmac.

Skyryder

That's aerodynamics,or lift,the reason aircraft fly.

[Jamezo]

Mr Galileo was of a different mind.

Ding! That is correct.

Skyryder, you need to rethink your ideas of distance, weight, force, velocity and acceleration.


Do you have a formula for this ?

For a simplified situation, in which the velocity and corner radius remain constant, the horizontal component can be calculated thusly:

F = m * v^2 / r

Where:
F is the lateral force component of the tyre/road interface force
m is the mass of the bike/rider system
v is the velocity of the centre of mass of said system
r is the radius of the corner, measured between the centre of the turning circle and the centre of mass of the bike/rider system



Quote:
#2, the vertical component, which is always (for a bike which is not presently accelerating upwards or downwards, so lets just say is it in a stable mid-corner) the same, exactly equal to the normal weight force of the bike, the same force that would be on the tyres/road in straight-line motion.



F = mg

So for a 160kg bike + 90kg rider, then F = 250 * 9.8 = 2.45kN

Is this correct ?

Yes, that is the downwards component of the force for a non-vertically accelerating bike/rider system of that mass.

That force will be distributed (variously) between the two tyres, but will still add up to the same amount in the described circumstances.

[TwoSeven]

oh, yip, that was the formula I had, just didnt understand what you wrote :)

So a 160kg bike + 90kg rider travelling at 35km/hr on a 10m radius turn would roughly generate: about 241kg of force.

I assume the answer is in kg - I've never been quite sure. To convert to kN is it ok to multiply by 9.8

I've not taken weight bias of the machine into account. Too lazy to look the figures up and not sure how to write a formula to calculate dynamic weight transfer (yet).

[sAsLEX]

about 241kg of force.



N is the SI unit for force. in that one N is the force required to accerellerate 1kg at 1ms^2 for 1 second

[Jamezo]

Almost, except that force is a momentary value.

Applying 1N of force to a 1kg object for 1 second results in a kinetic energy transfer of 0.5J (Joules) (Energy = Force * distance, Distance = average velocity * time, in this case 0.5 ms^-1 * 1s = 0.5m. 1N * 0.5m = 0.5J).

So it takes 0.5J to accelerate a 1kg object at 1ms^2 for 1 second.

It takes 1N to accelerate a 1Kg object at 1ms^2, full stop (ignoring relativistic effects lol).

[TwoSeven]

So the answer to my question was ???

[MSTRS]

If this was the question....


It seems to me that an object travelling at speed away from the earth would be getting lighter - or have I gotten this wrong ? :)
...then the answer is 'Yes':Pokey:

[sAsLEX]

It takes 1N to accelerate a 1Kg object at 1ms^2, full stop (ignoring relativistic effects lol).

true dont know how I let that second slip in there, had only just got up so a little sleepy

[TwoSeven]

I suspect that there is a third force in the equation as well. The horizontal force I think needs to be balanced by the lean angle of the machine.

[Hitcher]

Does the light in the fridge go out when you shut the door?

[Jamezo]

If you were referring to 'does a motorcycle (or indeed any wheeled contrivance) weigh less as it travels rapidly around the curvature of the earth', then, yes, the answer is yes.

Ok, just for kicks, I shall figure out exactly how substantial this effect might be (the answer, which is rather obvious, is 'not very')

Let us compare the accelerations of the normal gravity force, and the 'driving around the earth very fast' acceleration:

A(G) = 9.81 ms-2, fairly standard

A(D) = v^2 / r, bog standard rotational motion formula

We can take the radius as 6,378m, the equitorial radius of the Earth.

Let us take the speed, for example, as 50ms^-1, or 180km/h.

So the acceleration towards the centre of the earth caused by a motorcycle travelling over the curvature of the earth at sea level and 180km/h can be calculated as:

A(D) = v^2 / r

= 50^2 / 6,378,000
= 0.00039197 ms^-2

So the change is about 0.004% of the total acceleration force of the bike.

Interestingly enough, the orbital velocity at sea level (the point at which these accelerations balance perfectly, and the ground falls away at the same rate gravity attracts it :O), works out to be:

A(G) = A(D)

9.81 = v^2 / r

v = sqr-rt(9.81 * r)
= sqr-rt(9.81 * 6,378,000)

= 7910ms^-1, or roughly 28,476km/h.

Better start fitting that Hayabusa turbo now eh?

[sAsLEX]

Interestingly enough, the orbital velocity at sea level (the point at which these accelerations balance perfectly, and the ground falls away at the same rate gravity attracts it :O), works out to be:


or roughly 28,476km/h.



but escape velocity is
v = sqrt(2gR)
The value evaluates to be approximately:

11100 m/s
40200 km/h
25000 mi/h

[Virago]

.......Interestingly enough, the orbital velocity at sea level (the point at which these accelerations balance perfectly, and the ground falls away at the same rate gravity attracts it :O), works out to be:

A(G) = A(D)

9.81 = v^2 / r

v = sqr-rt(9.81 * r)
= sqr-rt(9.81 * 6,378,000)

= 7910ms^-1, or roughly 28,476km/h......
And to think I took a rough guess and said 30,000km/h - how wrong can a person get?:lol:

[scumdog]

Now as you know for a fixed number of revolutions a smaller diameter wheel travels slower than a big one - covers less ground for a given number of turns.
So does that mean you go slow-fast-slow as you go through S bends - if you keep the same revs?:confused:


After all the centre of the tyre is in effect larger wheel than the shoulder of same tyre....:shifty:

[MSTRS]

Thinking about this, I come up with.....
If you are a motorcyclist who does NOT accelerate through and out of a corner then one of two things will be the case.....you will run wide or you ride a Harley (or both):doobey:

[Ixion]

Motards ? .

[scumdog]

Thinking about this, I come up with.....
If you are a motorcyclist who does NOT accelerate through and out of a corner then one of two things will be the case.....you will run wide or you ride a Harley (or both):doobey:

Ah, but how do you KNOW that your accelerating and not just keeping the same speed and the revs go up to compensate for the 'reduction in tyre diameter??

And by your theory if I stop braking while going around corners I would run off the INSIDE of the bend???:confused:

[Jamezo]

but escape velocity is
v = sqrt(2gR)
The value evaluates to be approximately:

11100 m/s
40200 km/h
25000 mi/h

nah, escape velocity is a hypothetical initial vertical velocity which would enable you to escape completely from the earths gravitational acceleration, to a hypothetical point an infinite distance from the earth, but in reality, just so far away that the inverse square relation of distance to gravitational acceleration renders it irrelevant.

what I calculated was in essence the orbital velocity at sea level, quite a different kettle of fish.

[Hitcher]

what I calculated was in essence the orbital velocity at sea level, quite a different kettle of fish.
Now THAT would be a rush! One would produce an impressive rooster tail too.

[marty]

No ones talking about cornering. If you travel in a straight line, lateral momentum is going to alter the gravity forces on the object. Lets look at this from another angle.

Say you are the passanger in a car travelling at 50kph and you drop a lead weight of 50 kg's out of the window onto the road. Now lets say from the point frome where the weight left your hand to the point of where it landed on the road is 50 meters.

Now lets say the second time you did this you were travelling twice the speed 100kph and you dropped the 50kg weight out of the window. All things being the same as the first experiment the distance travelled would be double. Same weight same gravity forces double the speed and double the distance.

Conversely the opposite would happen. Travel at 50kph drop a weight of 100kg and that would be 50% shorter than the first experiment or thereabouts. In all cases both weight and graity forces are the same but the differential in distance is governed by speed.

The same applies to mass when it is at velocity. Gravity is unable to exert the same amount of force to a moving object as it can to a stationary object and as such if the weight can be measured it will 'appear' to be lighter..

Its why rockets fly and planes glide. One uses wings the other sheer speed to keep it in the air. When the speed slows the rocket falls to the ground.

Skyryder

mm, gravity is constant at sea level, and always acts vertically down. it doesn't matter if the item is moving or not. a rocket needs some upward thrust component, or small wings (faster it goes the smaller the wings it needs). show me a non-winged, non vectored thrust cylindrical rocket that flies horizonatally please. and anyway, when a plane slows, it will also fall to the ground. it doesn't get any lighter or heavier just cause it's flying. it still weighs the same - it's just that the vertical component of lift is greater than the opposing vertical component of weight. and on your 50kg vs 100kg lead block analagy, apparantly fat guys will freefall out of a plane faster than skinny guys.

you need to start backing up some of your comments with researched facts.

[MisterD]

mm, gravity is constant at sea level, and always acts vertically down.

Rather the direction of gravity defines "down"


it doesn't matter if the item is moving or not. a rocket needs some upward thrust component, or small wings (faster it goes the smaller the wings it needs). show me a non-winged, non vectored thrust cylindrical rocket that flies horizonatally please.

You're contradicting yourself Marty, the faster it goes the smaller wings it needs....to the point where if it goes fast enough it doesn't need them at all.

It would be difficult to do, but in theory you could travel fast enough that the acceleration due to gravity would just maintain altitude. In effect a very low orbit. I think this was calculated earlier in the thread.

[scumdog]

It would be difficult to do, but in theory you could travel fast enough that the acceleration due to gravity would just maintain altitude. In effect a very low orbit. I think this was calculated earlier in the thread.

It would require some form of upwards thrust to counter-act the pull of gravity.

Even a bullet drops as soon as it leaves the barrel (If the barrel is horizontal)

[Jamezo]

Rather the direction of gravity defines "down"



You're contradicting yourself Marty, the faster it goes the smaller wings it needs....to the point where if it goes fast enough it doesn't need them at all.

It would be difficult to do, but in theory you could travel fast enough that the acceleration due to gravity would just maintain altitude. In effect a very low orbit. I think this was calculated earlier in the thread.

The 'wing size/air speed' relationship is a matter of aerodynamics, most certainly different to a hypothetical atmospheric orbit. marty's comments hold perfectly true.


As for the fat/skinny guys skydiving, this phenomenon is explained by the nature of air resistance: The rate at which a falling object reaches its terminal velocity is related to the ratio of the weight force (gravity) to the drag force (air resistance). At low speeds, the fat person has a higher net downwards force, and a very similar drag force to the skinny person, giving them a better ratio, and a slight edge on the initial acceleration.

As terminal velocity is approached, the overriding factor will be the proportionality of weight force to the drag coefficient. I'm not an expert on human aerodynamics, so I have little idea how body shapes of given weights correlate to drag coefficients.

[MSTRS]

Ah, but how do you KNOW that your accelerating and not just keeping the same speed and the revs go up to compensate for the 'reduction in tyre diameter??
The wear on the trailing edge of the tread pattern on the rear tyre tells me I am

And by your theory if I stop braking while going around corners I would run off the INSIDE of the bend???:confused:
If you can manage a corner without braking (scary thought, I know) then you will have your answer:msn-wink:

[scumdog]

O.K. is it engine torque that makes it easier to go around left hand bends than right hand bends??

And MSTRS, do you know the feathering occured when you went around bends or just when you hit the straight again?? After all the feathering that was not in the centre of the tyre would have occured as the bike was going slower on the smaller diameter shoulder part of the tyre.

[TwoSeven]

If you were referring to 'does a motorcycle (or indeed any wheeled contrivance) weigh less as it travels rapidly around the curvature of the earth', then, yes, the answer is yes.

Ok, just for kicks, I shall figure out exactly how substantial this effect might be (the answer, which is rather obvious, is 'not very')

Let us compare the accelerations of the normal gravity force, and the 'driving around the earth very fast' acceleration:

A(G) = 9.81 ms-2, fairly standard

A(D) = v^2 / r, bog standard rotational motion formula

We can take the radius as 6,378m, the equitorial radius of the Earth.

Let us take the speed, for example, as 50ms^-1, or 180km/h.

So the acceleration towards the centre of the earth caused by a motorcycle travelling over the curvature of the earth at sea level and 180km/h can be calculated as:


I dont believe you have used the right formula for your statements that went with it. But then I am not familiar really with that formula.

My question is that you stated that the motorcycle is accelerating towards the centre of the earth, but in your formula you give the value of 'r' as being the same. For something to accelerate towards the centre of something 'r' must get smaller.

As I understood newtons law, the wieght can only change if the distance from the centre of the earth changes, or another gravitational body pulls on it. So maybe it might weigh less on a full moon or something :)

[MisterD]

The 'wing size/air speed' relationship is a matter of aerodynamics, most certainly different to a hypothetical atmospheric orbit. marty's comments hold perfectly true.


Thought experiment: Remove gravity, then get a rocket and fire it horizontally. It would travel at a tangent to the earth's surface and therefore appear to climb. Now reinstate gravity, the balance between it's pull towards the earth and the rockets velocity determines whether the curved path the rocket follows hits the earth, runs parallel to it (orbit) or spirals outwards.

[MSTRS]

O.K. is it engine torque that makes it easier to go around left hand bends than right hand bends??
Not torque - tis the camber of the road (serious answers from me dooooo happen)


And MSTRS, do you know the feathering occured when you went around bends or just when you hit the straight again?? After all the feathering that was not in the centre of the tyre would have occured as the bike was going slower on the smaller diameter shoulder part of the tyre.
Feathering is towards the edge so has to happen when leaning over, acceleration has to be happening because the wear is on the trailing edge. Heavy braking will cause the leading edge to wear (on the rear tyre and assuming you don't lockup/lowside)).

[marty]

i give up.

[TwoSeven]

Does the light in the fridge go out when you shut the door?

You'd have to write a formula to prove which door you are shutting and if it is connected to the fridge. For example, if I shut the front door to the house, the status of the fridge light does not change.

Also, inertia on the fridge door [assuming the requirement is narrowed to determine which door is being talked about] would be important. It could possibly bounce back open again, in which case the state of light offedness would be very short.

[thealmightytaco]

You're all nuts. Centripetal force acts equally up down everywhere for a fully enclosed rotating object like a wheel so nothin' gonna happen there, and if anyone here can notice the say...

160kg*0.000392ms^2= 63 gram!!

...63 gram change in the weight of their bike when doing 180k around the surface of the earth, I'd be very surprised.

[Virago]

Does the light in the fridge go out when you shut the door?
(Muffled shout) - YES!

[Ixion]

Define "out". Do you have proof that it was ever "not out". Define "fridge" Define "door". Define "shut" . Do you have proof that the door was ever "not shut". If the "light" "goes out", where does it go to.

EDIT: That damned cat isn't in there is it? Dead or alive.

[MSTRS]

EDIT: That damned cat isn't in there is it? Dead or alive.
Might be...what's it's name?

[Ixion]

Schrodinger's cat :mad: :argh:

[MSTRS]

He a member here??? I was thinking of MidnightMike's and whether it has a name yet:laugh:

[Jamezo]

I dont believe you have used the right formula for your statements that went with it. But then I am not familiar really with that formula.

My question is that you stated that the motorcycle is accelerating towards the centre of the earth, but in your formula you give the value of 'r' as being the same. For something to accelerate towards the centre of something 'r' must get smaller.

As I understood newtons law, the wieght can only change if the distance from the centre of the earth changes, or another gravitational body pulls on it. So maybe it might weigh less on a full moon or something :)

Nope, anything that is not moving at a constant velocity is accelerating. The direction of this acceleration for an object moving in a circular fashion is tangential to the velocity, and directed towards the centre of the arc of motion.

Thus the bike accelerates towards the earth, as its velocity changes while travelling around the earth.

This is the acceleration which partially negates some of the weight force by 'allowing' the bike to accelerate in the same direction as gravity is trying to make the bike.

[scumdog]

Thought experiment: Remove gravity, then get a rocket and fire it horizontally. It would travel at a tangent to the earth's surface and therefore appear to climb. Now reinstate gravity, the balance between it's pull towards the earth and the rockets velocity determines whether the curved path the rocket follows hits the earth, runs parallel to it (orbit) or spirals outwards.

True, true, BUT if you extended the rocket so it was 20km long you would find the nose was further away from the earth than the centre section, ergo it is actually pointing AWAY from the earth, in other words upwards, albeit at a very slight angle.
So there!

Oops about the 'find'

[MisterD]

BUT if you extended the rocket so it was 20km long you would fine the nose

You bloody cops! Any excuse to fine something.....