Sarbian changed the topic of #kspmodding to: Welcome to #kspmodding - the channel for discussing, and learning about, modding Kerbal Space Program. Code of Conduct: https://git.io/vSQh6 | Always provide logs. | *** PSA: https://kerbalspaceprogram.com/api/index.htm | Someone is writting a Phd dissertation on KSP modding please fill out this survey https://goo.gl/forms/1vrkPYH4az8l32sb2 - I have met him and it s legit
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<Sarbian> and I am back now
<Thomas> o/
<Sigma88> 0/
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<Sigma88> how do I show fps on ksp?
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<Sarbian> GCMonitor has a DPS counter overlay
<Sarbian> err, FPS
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<Sigma88> dps? this is not wow
<Sigma88> :D
<Sigma88> (most of the time it isn't https://i.imgur.com/G5jS3Jb.png)
<Orum> a circumscribed equilateral triangle has a height of h = 3/4*r, right?
<Orum> god, I wish I had a more regular use for trig in my everyday life
<Sarbian> Sigma88: I killed so many kerbal that is may be WoW in some way :)
<Sigma88> !wa height of equilateral triangle inscribed in a cicle of radius = 1
<Qboid> Sigma88: Seems that Wolfram is unable to understand that.
<Sigma88> hehehe sarbian, I understand :D
<Orum> just thinking, a side of such a triangle should be sin(120)/sin(30), halve that to make a right triangle, and then use tangent to find the height, which should be sin(120)/(2*sin(30))*tan(60)
<Orum> oh, wait, I think I had an extra /2 in my calculations...
<Sigma88> !wa circumscribed equilateral triangle height
<Qboid> Sigma88: Seems that Wolfram is unable to understand that.
<Sigma88> oh cm on
<Orum> wolfram is dumb if you don't phrase things in a very particular way :)
<Sigma88> !wa equilateral triangle distance between vertex and barycenter
<Qboid> Sigma88: Seems that Wolfram is unable to understand that.
<Sigma88> !g height of a circumscribed equilateral triangle
<Qboid> Sigma88: https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-inscribed-triangle-area/v/area-of-inscribed-equilateral-triangle-some-basic-trig-used [Area of inscribed equilateral triangle (video) | Khan Academy] (22100 results found, took 0.31s)
<Sigma88> you probably can get to it quicker by mental math
<Sigma88> :)
<Orum> yeah, just getting a number that doesn't make sense
<Orum> OH wait, it should be inverse tangent
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<Orum> nope, nevermind, I'm just an idiot, I had things right in the first place
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<Orum> h=r*sin(120)/sin(30)/2*tan(60)=r*sqrt(3)/2*tan(60)=1.5r
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<Sigma88> !wa sin(120)/sin(30)/2*tan(60)
<Qboid> Sigma88: ((sin(120°))/(sin(30°)))/2 tan(60°) = 3/2
<Orum> which means a regular tetrahedron should have a height of....sqrt(1.5^2-0.5^2)
<Orum> *r, of course
<Orum> which is sqrt(2)*r
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