I would like any and all feedback or input on the thoughts expressed here. The text is somewhat complex, and maybe flawed(?) but it will be worth it!
My plan has been to position the springs at the hub, perpendicular to the seat stay, since it’s an easy way. Yet, it gives a
linear suspension (ie every part of the suspension travel has the same resistance) and is a bit bulky.
View attachment 238359
And I would like to build a
progressive suspension. Some might wonder
what the appeal with a progressive suspension is.
Well, in a progressive suspension construction, the spring travel relative to rear wheel travel changes as the suspension is compressed. Hence, it is 'plusher' on small bumps (increased small bump sensitivity) and stiffens up as the travel is compressed (meaning it won't bottom out as abruptly).
-You know that scissor jack I cut to pieces earlier?
It suddenly popped up in my mind again, and it got me thinking about
Pythagoras' theorem. You know, that
a2 + b2 = c2 stuff.
Yeah, I know, I’m a bit strange.
Anyway, after using the theorem, I found out that the lift of one turn of the handle changes depending on where in its travel the jack is.
-It is progressive!
In the end, I won’t be using the leftover jack parts, but I’ll use a similar construction to achieve a progressive suspension. And in the process I’ll be able to tuck the springs parallel to the seat stays and the rear fork.
View attachment 238360
My calculations indicate that the very first and the very last parts of the scissors’ travel aren’t optimal to use, as it is too weak initially and too strong at its end. So I plan to exclude the top and bottom 20mm of the theoretical maximum stroke, and place the maximum travel of my springs of 44mm, in the middle of this range. 44+20+20=84mm total possible horizontal travel (ie the length of the scissors’ legs).
So theoretically rear wheel travel is 2 x 86mm [2 x
a [fully extended legs]] x (44 / 84) [wanted use of travel as percent of total travel] = 88mm.
Opinions?