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I want an animation of a circle whose radius starts at x and ends up at
infinity. If x were zero then I would use `tan(clock*pi/2)`. But x is
not zero. What should I use instead? Please answer my dumb maths
question, thanks.
Mike
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On 21/03/29 00:20, Mike Horvath wrote:
> I want an animation of a circle whose radius starts at x and ends up at
> infinity. If x were zero then I would use `tan(clock*pi/2)`. But x is
> not zero. What should I use instead? Please answer my dumb maths
> question, thanks.
>
> Mike
(1-1/(1-clock))-(1-x)
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On 21/03/29 00:56, Zeger Knaepen wrote:
> On 21/03/29 00:20, Mike Horvath wrote:
>> I want an animation of a circle whose radius starts at x and ends up
>> at infinity. If x were zero then I would use `tan(clock*pi/2)`. But x
>> is not zero. What should I use instead? Please answer my dumb maths
>> question, thanks.
>>
>> Mike
>
>
> (1-1/(1-clock))-(1-x)
you could also use tan, starting at atan(x) going to pi/2
Or log(1-clock)+x
Any function that goes to infinity at a known point could be used, I
suppose, but those are the three I can think of.
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On 3/28/2021 7:00 PM, Zeger Knaepen wrote:
> you could also use tan, starting at atan(x) going to pi/2
You mean this?
tan(clock * (pi/2 - atan(x)) + atan(x))
Not sure what to do.
Mike
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On 21/03/29 01:18, Mike Horvath wrote:
> On 3/28/2021 7:00 PM, Zeger Knaepen wrote:
>> you could also use tan, starting at atan(x) going to pi/2
>
> You mean this?
>
> tan(clock * (pi/2 - atan(x)) + atan(x))
>
> Not sure what to do.
you want a function that starts at x at clock=0 and goes to infinity at
clock=1, and for some reason you want to use tan() :)
you have:
tan(u)=x <=> atan(x)=u
and tan(v)=infinity <=> v=pi/2
but you want tan(f(clock)) with f(clock) a function that gives you the
desired results.
So you need to convert clock (which goes from 0 to 1) to the range
atan(x) -> pi/2
Let's figure this out:
you can't just add atan(x) to clock, or it would go from atan(x) to
1+atan(x)
So you also have to scale the clock-value.
Let's do that first, the range should be between atan(x) and pi/2, and
we know that atan(x) is smaller than pi/2.
So the difference is (pi/2)-atan(x), that is the scale. The offset is
indeed atan(x)
So we get: fClock=((pi/2)-atan(x))*clock+atan(x)
and then tan(fClock) should give you the desired result.
I think.
:)
I haven't used POV-Ray in a while, so you might have to translate some
parts into POVese
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On 3/28/2021 7:52 PM, Zeger Knaepen wrote:
> So we get: fClock=((pi/2)-atan(x))*clock+atan(x)
>
> and then tan(fClock) should give you the desired result.
>
> I think.
I think that's the same as what I came up with.
Mike
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Le 2021-03-28 à 18:20, Mike Horvath a écrit :
> I want an animation of a circle whose radius starts at x and ends up at
> infinity. If x were zero then I would use `tan(clock*pi/2)`. But x is
> not zero. What should I use instead? Please answer my dumb maths
> question, thanks.
>
> Mike
This one goes from 1 at zero to infinity at pi/2 :
1/cos(clock*pi/2)
Just add 1 to the value returned by the tangent :
tan(clock*pi/2)+1
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