trying to figure the shift throw out beforehand with a diagram and some rusty geometry/algebra so I know just how much to lengthen the linkage under the shifter as related to degree of swing on the shifter. Someone check my math!
Assuming the stock shift throw degree is “X” and the desired shift throw is “Y”.
Assuming the stock Length above the fixed point (shift handle) is L1 and the length to alter is L2.
We know that the angle of the shift throw is inversely proportional to the length below the pivot. In other words, the shorter the shift knob, the higher the degree below the pivot.
So the equation would look like
L2/X =L1/Y,
For example, if we measure the stock configuration degrees of movement of the stick handle, and we know how much we want to reduce it by which gives us our desired degree of movement, lets say a shift that moves half the distance.
So for fun, if L1 =10”, X = 30 degrees, Y= 15 degrees, then,
L2/30=10/15,
L2= 20” (your total length below the fixed point)
Im not sure Im setting this up properly, and if I really remember how to do equations like this, but we should be able to exactly calculate just how far to modify the linkage under the shifter by the number of degrees we want the shifter to move. I feel like I did something wrong here though, because the relationship is reverse in real life, the shifter handle is longer than the linkage bar directly underneath
shiftthrow.jpg (Size: 54.17 KB / Downloads: 86)