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Portland, Oregon
Joined April 2008
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Now using Twitter via an IRC Proxy... flic.kr/p/asA9ug #geek code.google.com/p/tircd/
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Replying to @bradfordcross
@bradfordcross You can't fault him completely! He did successfully get you to tweet a link to his site.
Replying to @christianp
@christianp Wow, awesome... I think that might be a better way of approaching it. Would appreciate more details!
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yo dawg, i heard you like servers in your server's servers, so i put a server in your server's server so you can have horrible i/o
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Replying to @kirillzubovsky
@kirillzubovsky ...or you will get bought very quickly!
Well that was a rather large rabbit hole...
Replying to @christianp
@christianp @mathpunk Yea it's actually on a sphere, so if there's a better spherical solution I'm all ears!
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Replying to @christianp
@christianp @mathpunk. Yes. Whether the line (segment) intersects the circle.
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Replying to @mathpunk
@mathpunk Alright here is my updated solution flic.kr/p/asft78
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Replying to @mathpunk
@mathpunk Yep my thoughts exactly. The circle is part of my original problem definition but is not needed to get e.
Replying to @mathpunk
@mathpunk ok here's my actual problem: flic.kr/p/ashXKJ
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Replying to @mathpunk
@mathpunk Actually I just realized I'm not looking for the distance AD, I actually want CD. More sketches coming...
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Replying to @mathpunk
@mathpunk Interesting... what do you think of this solution? flic.kr/p/arNB5V
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Replying to @andormaybe
@amarkor It gets better! No sin/cos/tan needed! d = (b^2 + c^2 - a^2) / 2c Where d is the missing edge segment and a,b,c are known edges (!)
Replying to @twaddington
@twaddington Hey cool! We're heading over there shortly! Hopefully will see you there!
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Replying to @marshallk
@marshallk The idea of software locked in a box cracks me up. What about updates? Code is a living thing!
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Waiting for your check at a restaurant is like being held hostage by your credit card.
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