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Old 11-27-2022, 04:32 PM   #26
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Join Date: Jan 2017
Location: Los Angeles
Posts: 1,156

Originally Posted by Justin View Post
hmm yeah that's nice. and the math can be simplified a bit since e^(2*ln(x)) = x^2, reduces down to ((m-1)/m)^2... not that it's that complex to begin with. we'll do that, :exp=2000 or whatever for the midpoint.
True, ((m-1)/m)^2 can be the base of the exponential b^x. I left it in e^ax form to show the relationship between "m" and "a" since Tale was advocating for setting "a" directly.
I see some value in that since e^ax is widely used and "m" (sorta like a midpoint percentage) is just something I worked out. No idea if it's a standard or known method.

But as you see "m" has two advantages:
1) It's intuitive, you can deduce the exponential shape easier when reading m=0.05 compared to a=5.89.
2) There's a simple relationship between m and the actual slider value at the midpoint (mid-slider-travel).

Very important, m HAS to be >0 and <1 or it doesn't work properly. I'm sure this is obvious, but just in case.

Originally Posted by Justin View Post
we'll do that, :exp=2000 or whatever for the midpoint.
Exponentially-scaled sliders would be so great, thank you if this happens regardless of the input parameters.
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