I want to swap springs in some cherry blacks. Stock they start at 35, actuate at 60 and bottom at 80 on the cherry website. I want to go with TX longs at 70. Does anyone know where they start at and actuate at before bottom at 70. New to this, hope I explained it correctly.
Unfortunately aftermarket springs are usually just rated at bottom out force. So I don’t think someone could tell you that info without a force curve graph made with TX 70g long springs in a MX black. Overall they will feel lighter than the stock MX black springs, but since you went with the long springs they may feel a little heavier at the beginning of a keystroke. The ideal behind long (or slow curved) springs is that the extra length causes them to be pre compressed a bit in the switch, which in turn causes the starting & actuation weight to be closer to the bottom out weight than it would be on normal length springs. Hope that info helps out some!
Looking at most 70 g switches, it looks like they actuate at 52 so it would be safe to assume that is what it would be for the TX spring too. Now for the starting force I am not sure, its not something that is actively looked at from the searches I have done.
The fact that it is a long or slow curve spring will skew the starting force & actuation force higher than than a regular sized spring.
This is true, maybe try a few in the once you get it and if you dont like it you can always sell it, I am sure someone will want it and you will only lose a $ or 2.
If you really looking for a more detailed information about springs, look at sprit spring. They put more information about there springs.
Sprit still can’t tell you at what weight a spring will actually start or actuate in a switch unless they have directly tried it, because these are both dependent on the housing and stem. This is (part of) why aftermarket springs usually only mention bottom-out weight.
Could you calculate this systematiclly then? Given you have original spring and it’s force curve, by comparing two spring, should be possible for working our all of those numbers if I am not mistaken
Quite possibly, but I don’t think anybody has quite mathed it out and they’d need to have some fairly special measuring tools to do it right.