Flex!

[koba]

I was browsing some old pic's and found this.

This is with old standard tyres that come on an EN125.

Not 100 sure who took the photo, may have come from a bunch that was on the kartsport site from Hamish Edgar.

[F5 Dave]

One of these wheels is not like the other

[Nobbsie]

Just in terms of flex i found this really quite interesting, might be old hat for some people but here it is none the less
http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=14516

[TZ350]

Just in terms of flex i found this really quite interesting, might be old hat for some people but here it is none the less
http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=14516

Ok this is my 2 cents worth.

When you follow that link and read what people are saying, keep in mind Newtons 3rd law of physics ""For every action there is an equal and opposite reaction"".

In real life that means that to compress a fork spring you need an equal (but opposite) force on each end of the fork leg.

And the smallest of changes in the vertical downwards force by the suspension is felt by the tire as a change in the upwards force by the road regardless of what ever else is going on in the suspensions spring and damping system.

Damping affects how the bikes kinetic energy (weight x velocity) and original static weight transfers to the front tire, the tires adhesion to the road and how quickly you can store energy in the fork spring or release it.

For handling and other reasons forks are attached to the frame at an angle and if you pushed downwards on the handle bars some of the force of that push is transmitted to the tire contact patch horizontally and some vertically.

When braking there are two forces working on the tire contact patch. One is pushing the tire along the road horizontally and the other is pushing the tire vertically down into the road.

Any force pushing the tire horizontally along the road does nothing for braking.

Only the force pushing the tire vertically down onto the road helps braking.

Actual braking force possible is a function of the coefficient of friction between the tire and the road, and the amount of force pushing the tire directly down onto the road.

Double the contact patch and you double the braking possible before the high static friction between tire and road becomes lower sliding friction, and by doubling the downward force you also double the braking possible. Both reasons why cars out corner and out brake bikes.

Increase both and you dramatically increase the braking and cornering forces possible.

For all out braking effect from a given size of contact patch, the more force you can put vertically downwards on the tire the better.

As static friction is much higher than sliding friction. A braked but still turning wheel has much more static friction between the tire and road and therefore possible braking effect than a locked and sliding tire, sliding friction is lower than static friction. A wheel still turning at road speed is best, anything slower and the tire is skidding a bit.

What some of the suspension is about, is controlling the transfer of the bikes weight and as much of the forward momentum as possible to downwards vertical force on the front tire while still allowing the front tire to track over irregularities in the road surface without losing contact, ie bouncing up as would happen with a rigid suspension, so do not let your forks bottom out under heavy braking.

The best suspension system would be one that converted all the bikes forward momentum and weight to a vertical downward force shared equally between both front and rear tires as this would double the tire contact patch area doing the braking and therefore braking force possible but sadly that type of suspension has not been invented yet.


So basicaly we are stuck with making the front tire do most of the braking work.


If all that has not been mind numbing enough for you, you can follow the link below and get into the maths.

http://hyperphysics.phy-astr.gsu.edu/hbase/crstp.html

Assuming proper operation of the brakes (http://hyperphysics.phy-astr.gsu.edu/hbase/pasc2.html#brake), the minimum stopping distance for an automobile is determined by the effective coefficient of friction (http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html#fri) between the tires and the road. The friction force of the road must do enough work (http://hyperphysics.phy-astr.gsu.edu/hbase/work2.html#wsl) on the car to reduce its kinetic energy (http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html#ke) to zero (work-energy principle (http://hyperphysics.phy-astr.gsu.edu/hbase/work.html#wepr)). If the wheels of the car continue to turn while braking, then static friction (http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#sta) is operating, while if the wheels are locked and sliding over the road surface, the braking force is a (lower) kinetic friction (http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin)force.

[Yow Ling]

So with the maximum weight applied to the tyre the braking would be at its optimum

293315

All this time I been trying to keep my weight under control.

[TZ350]

Yes ... that should do it, now he can use lots more back brake before the rear wheel skids and locks up.

[speedpro]

I'm pretty sure that would be one "overloaded" Yamaha Scorpio

[AllanB]

I'm pretty sure that would be one "overloaded" Yamaha Scorpio

Hek I did not notice the motorcycle wedged up her arse!

[Pickle]

Might even be able to do a power wheelie with all that weight over the back wheel