Comments on: Let's do some math
http://ask.metafilter.com/212472/Lets-do-some-math/
Comments on Ask MetaFilter post Let's do some mathMon, 09 Apr 2012 07:39:38 -0800Mon, 09 Apr 2012 07:44:32 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Let's do some math
http://ask.metafilter.com/212472/Lets-do-some-math
A is the top of a vertical line 30 feet in length. B is at the base of that line. C is 63 feet from B in a horizontal right angle. What is the distance from A to C. <br /><br /> Asking for my stepfather who wants put an antenna (for his ham radio) amongst some trees and needs to know how far the wire will reach out from the top of the mast and then back to roof level. Or at least this is the information he has supplied to me in text form.<br>
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I have no idea why he is asking me for help with this because I'm woefully terrible at math. Will you all help me? Pretty please?post:ask.metafilter.com,2012:site.212472Mon, 09 Apr 2012 07:39:38 -0800ephemeristamathresolvedBy: Etrigan
http://ask.metafilter.com/212472/Lets-do-some-math#3065533
AB^2 + BC^2 = AC^2<br>
(30)^2 + (63)^2 = AC^2<br>
900 + 3969 = AC^2<br>
4869 = AC^2<br>
sqr(4869) = AC<br>
69.8 = ACcomment:ask.metafilter.com,2012:site.212472-3065533Mon, 09 Apr 2012 07:44:32 -0800EtriganBy: shiny blue object
http://ask.metafilter.com/212472/Lets-do-some-math#3065535
Isn't this just a right triangle? I'm assuming B and C are at the same "height", or both are at ground level. So AB is the vertical leg, BC is the horizontal leg, and AC is the hypotenuse. If that's the case, AC is 69.8 feet.<br>
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(Or, what Etrigan said in words instead of equations.)comment:ask.metafilter.com,2012:site.212472-3065535Mon, 09 Apr 2012 07:45:44 -0800shiny blue objectBy: ephemerista
http://ask.metafilter.com/212472/Lets-do-some-math#3065536
That is what I asked him. Isn't it just a right triangle? He wasn't forthcoming and I'm terrible at picturing these sorts of things, so after a few half-hearted attempts to draw what he was describing I gave up and turned to you all. I knew what would take me a while would take someone here all of 1 minute!comment:ask.metafilter.com,2012:site.212472-3065536Mon, 09 Apr 2012 07:47:41 -0800ephemeristaBy: ephemerista
http://ask.metafilter.com/212472/Lets-do-some-math#3065540
He says thank you very much. I'm marking shiny blue object's the best, because once I txted him the solution with words he figured out what was messing him up. Asked and answered in less than 15 minutes! Thanks again!comment:ask.metafilter.com,2012:site.212472-3065540Mon, 09 Apr 2012 07:55:05 -0800ephemeristaBy: Protocols of the Elders of Sockpuppetry
http://ask.metafilter.com/212472/Lets-do-some-math#3065624
Don't forget catenary sag, which will add less than a foot to the length of wire and lead to about 4 inches of sag at the middle.comment:ask.metafilter.com,2012:site.212472-3065624Mon, 09 Apr 2012 09:00:28 -0800Protocols of the Elders of SockpuppetryBy: ephemerista
http://ask.metafilter.com/212472/Lets-do-some-math#3065664
He said he added in sag, but didn't know the term catenary, so thanks!comment:ask.metafilter.com,2012:site.212472-3065664Mon, 09 Apr 2012 09:38:57 -0800ephemeristaBy: w0mbat
http://ask.metafilter.com/212472/Lets-do-some-math#3066294
Yeah this just requires your standard <a href="http://en.wikipedia.org/wiki/Pythagorean_theorem">Pythagorean theorem</a> and the <a href="http://www.google.com/search?rls=en&q=sqrt((30*30)%2B+(63*63))&ie=UTF-8&oe=UTF-8#hl=en&safe=off&client=safari&rls=en&sclient=psy-ab&q=sqrt((30%5E2)%2B+(63%5E2))&oq=sqrt((30%5E2)%2B+(63%5E2))&aq=f&aqi=&aql=&gs_l=serp.3...4956l5715l1l6866l2l2l0l0l0l0l221l308l1j0j1l2l0.frgbld.&pbx=1&fp=1&biw=1310&bih=940&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.,cf.osb&cad=b">Google calculator is handy for solving it</a>. Extra marks to Protocols for accounting for sag though.comment:ask.metafilter.com,2012:site.212472-3066294Mon, 09 Apr 2012 17:04:49 -0800w0mbat