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 All Forums
 Motorcycle Safety
 General Discussion
 High side assumption clarification
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rayg50
Male Moderator
2083 Posts
[Mentor]


NYC, NY
USA

Honda

Shadow Spirit 750DC

Posted - 12/15/2010 :  7:53 PM                       Like
My understanding of one cause of a high side is as follows.

A rider applies the rear brake to the point that it locks (ceases to rotate). The rear (wheel) of the bike moves out of line with the front (wheel) of the bike as the tire skids (loses traction). The rider releases the rear brake, the rear wheel regains traction as it begins to rotate again. The force of the regained wheel on the out of alignment bike causes the bike to buck tossing it and the rider into the air (I picture it as a martial arts sweep).

Please correct the above poorly worded explanation as needed.

Here is where I need clarification. I have assumed that if the clutch is fully in there can be no high side even if the rear is released. I have assumed that since there is no power going to the rear there can be no violent re-establishment of traction. Has my thinking been faulty? BTW, I am not siding that you can't or won't low side.

James R. Davis
Male Administrator
17377 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 12/15/2010 :  8:19 PM Follow poster on Twitter  Join poster on Facebook as Friend  
Your thinking is faulty, but clever.

The clutch MUST be pulled to the grip if you lock the rear wheel, or your engine stalls immediately. So, regardless of whether or not it remains pulled to the grip, the rear wheel does not have power driving it when a high-side occurs as you've described it.

The high-side torque is developed from the contact patch regaining traction in the direction the tire is sliding. If the clutch is fully pulled at that time, the bike CAN gain a little extra torque as it starts to drive the bike it the direction it is pointed (not the direction it is sliding), which is completely opposite your concept.
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