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-   -   Exhaust Gains ( https://www.xt660.com/showthread.php?t=12069)

Gas_Up_Lets_Go 28-10-09 14:35

Quote:

Originally Posted by ag1s (Post 110183)
This is not entirely true,

Quote:

Originally Posted by Gas_Up_Lets_Go (Post 110168)
it's very simplistic to say the least.

Still doesn't give me the information I asked for though ?


What are the percieved and actual advantages of spending �400 - �1000 on a new set of pipes ?

ag1s 28-10-09 14:48

Quote:

Originally Posted by Gas_Up_Lets_Go (Post 110188)
Still doesn't give me the information I asked for though ?


What are the percieved and actual advantages of spending �400 - �1000 on a new set of pipes ?

The expenditure is relative, the advantage is absolute. You can grab any good quality second hand exhaust of a large volume for peanutes, make a mod to fit it and the result will be same or better than spending 400 or 1000. Preferably one made for a large capacity twin. Together with a pc3 of course and fliter otherwise its a bad idea.

The advantages are weight loss high up as others say and a snappier throttle.

Whether you need it, only you can tell, but usually some bikers (including myself) are infected by the 'mod' virus and there is nothing we can do about it!

uncle ricky 28-10-09 14:54

Quote:

Originally Posted by Gas_Up_Lets_Go (Post 110188)
Still doesn't give me the information I asked for though ?


What are the percieved and actual advantages of spending �400 - �1000 on a new set of pipes ?


It will make your bank balance alot lighter :042:

****** I said I was not getting involved anymore

ag1s 29-10-09 22:51

Quote:

Originally Posted by ag1s (Post 110183)
This is not entirely true, because the weight of lets say 8kg puts down a reverse torque that keeps the tail down. I just worked it out and it seems that, assuming an angle of 45 degrees of the horizontal to the exhaust can (apex being the front wheel-ground interface) you need a braking force of 10G for the inertia of the exhaust to negate its downward force! If this angle is 30 degrees the force is about 17.5 G

According to my calculations, the braking g-force is given by the formula

g=9.81/tan(a)

where 'a' is the afforementioned angle, and g is independent of the actual weight of the can, or its distance from the front wheel.

Not that anybody cares, but for clarity sake there is a decimal place wrong in the above G-force numbers, sorry! They are 1.0 G and 1.7 G respectively. Still unatainable probably by the tenere brakes.


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