Question for the resident boffins.....

**froggyfrenchman** · 26th February 2006, 18:44

Originally Posted by 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.

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

**Skyryder** · 26th February 2006, 20:11

Originally Posted by scumdog

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** · 26th February 2006, 20:23

Originally Posted by sAsLEX

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** · 26th February 2006, 21:07

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** · 26th February 2006, 21:14

Originally Posted by Jamezo

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

Whoa, what he said

**Skyryder** · 26th February 2006, 21:48

Originally Posted by Jamezo

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** · 26th February 2006, 21:59

Originally Posted by Virago

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** · 26th February 2006, 22:08

Originally Posted by Skyryder

..
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** · 26th February 2006, 22:21

Originally Posted by Jamezo

#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** · 26th February 2006, 22:50

wow, some bright sparks here!
very interesting read

**thealmightytaco** · 26th February 2006, 23:01

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** · 27th February 2006, 07:19

Originally Posted by Skyryder

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** · 27th February 2006, 07:41

Originally Posted by Ixion

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.

Originally Posted by TwoSeven

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

Originally Posted by TwoSeven

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** · 27th February 2006, 09:12

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).