Calculate sides of pyramid

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This may sound like a stupid question, but how do you calculate the sides of a pyramid-type structure.Lets say the base is 27"x27" and the peak needs to be 12".  How do I calculate each side to be 28" up to the peak?I used to have an app for android that would help me with a great GUI but cannot seem to find one for apple.  Besides all that, I'd like to know how to do it without the assistance of apps.Math was never my strong suite...Sent using Tapatalk
Reply:Try here see if it helps you, BTW viewed on Apple mini.http://www.cleavebooks.co.uk/scol/calpyr.htm
Reply:Sides of a right triangle triangle... A squared + B squared = C squared, A and B being the two sides touching the 90 deg angle, C being the hypotenuse.Because the pyramid sides are tipped, you have to 1st work out the triangle that will give you a right triangle  with the length you want included. One side will be 12" the height of the pyramid. The 2nd one needs to be done with math.To find the line that would connect the center at the base to a corner you need to draw another triangle using info you already have. Connect the center at the base at a right angle to the base of a side. Total side is 27", so the middle would be 13 1/2" from there to an end. Distance from a side to the center of the base is also 13 1/2". So the line from center to the corner is 13.5 squared x 13.5 squared + C squared.  c = 19.092 or really close to 19 3/32"Now to find the side you want, it's 19.092 squared + 12 squared ( height of the pyramid) = C squared. C = 22.55 roughly or really close to 22 35/64"There are other ways to come up with the same answer as well.Hope this makes sense. If not I'll try to draw out the angles..No government ever voluntarily reduces itself in size. Government programs, once launched, never disappear. Actually, a government bureau is the nearest thing to eternal life we'll ever see on this earth! Ronald Reagan
Reply:Originally Posted by DSW... So the line from center to the corner is 13.5 squared x 13.5 squared + C squared.  c = 19.092 or really close to 19 3/32"...
Reply:The link pepi provided is the clearest example.Enter in the first two boxes and hit "calculate".By hand, pythagorean theorem would be 12^2 + 13.5^2 = c^2  gives you 18.06" for your flat triangle height.Cut four triangles with the base 27" wide and 18.1" tall, lean together and done Dave J.Note: your base length and peak height drive all the other measurements.Last edited by MinnesotaDave; 03-21-2013 at 11:56 AM.Dave J.Beware of false knowledge; it is more dangerous than ignorance. ~George Bernard Shaw~ Syncro 350Invertec v250-sThermal Arc 161 and 300MM210DialarcTried being normal once, didn't take....I think it was a Tuesday.
Reply:My inclination was to calculate the "slant height".  Makes it easy to lay out with a square.Actually, you don't even need math.  Use a square to draw an "L", 13.5" on one side and 12" on the other.  Then just use a ruler or tape to measure the diagonal from one end of the "L" to the other.Last edited by Oldendum; 03-21-2013 at 12:13 PM."USMCPOP" First-born son: KIA  Iraq 1/26/05Syncrowave 250 w/ Coolmate 3Dialarc 250, Idealarc 250SP-175 +Firepower TIG 160S (gave the TA 161 STL to the son)Lincwelder AC180C (1952)Victor & Smith O/A torchesMiller spot welder
Reply:Trig is real handy once you get the hang of it. Slice the pyramid in half so that you have a right triangle. Then you'll have two quantities that you know:a = (1/2) x 27" = 13.5"b = 12" (vertical height of pyramid)Plug these numbers intohttp://www.carbidedepot.com/formulas-trigright.aspto calculate the "longest leg" (hypotenuse) of the right triangle, which is always labeled as leg "C" and which would be~ 18"Last edited by Kelvin; 03-21-2013 at 12:52 PM.
Reply:Originally Posted by OldendumMy inclination was to calculate the "slant height".  Makes it easy to lay out with a square.Actually, you don't even need math.  Use a square to draw an "L", 13.5" on one side and 12" on the other.  Then just use a ruler or tape to measure the diagonal from one end of the "L" to the other.
Reply:Originally Posted by MinnesotaDaveAn excellent point Dave J.
Reply:I think some of us are working out different dimensions.I think the dimension he wants is line AB/AD. Most of you are working out the dimensions for line BF/DF because you are using 1/2 the distance from BD which is BE/DE and then using 12" as your height ( EF) This however doesn't take into account that the side is tilted down on a pyramid.To get that line you need to 1st determine the length of a line running from BC. Then you can work out the diagonal on the sloped face of the pyramid..No government ever voluntarily reduces itself in size. Government programs, once launched, never disappear. Actually, a government bureau is the nearest thing to eternal life we'll ever see on this earth! Ronald Reagan
Reply:Myself, I'd just solve for AE. Height is 12 (AC on your diagram), base is 13.5 (CE on your diagram) and the hypotenuse directly gives you the height of the side (AE). . 27 at the base, draw the peak at that distance on centerline, and you should have the exact side triangles you need in 1 step.- TimLast edited by tadawson; 03-21-2013 at 06:59 PM.
Reply:I think what I'm looking for is the slope.For example:12"x12" square w/6" height to the peak.FE?Sent using Tapatalk
Reply:Hi wfldmike, there's too many numbers in your question. Solving C^2 = a^2 + b^2 only will give the solution for a 4 sided polyhedron where the base and 3 upright sides are all triangles.A pyramid like in Egypt is a polyhedron with five sides (the base is a square side - the rest are triangles). All polyhedron's have a base (a square one here I hope) and a vertex (point at the top here). So you need base, height & number of sides for various solutions (dimensions, area & volume). From your dimensions if I'm right on the Egypt thing...We find the center of the base 27/2 = 13.5we know the height 12 or 28 (which is it)??? - I'll assume 12 for heightwe find the polyhedron slant height with s(squared) = 12(squared) + 27/2(squared)s(squared) = 144 + 182.25s(squared) = 326.25s = 18.062That bisected one side of the pyramid giving us another right triangle to solve for the length from corner to top...18.062 becomes side "b" in a new triangle - side "a" remains 13.5 from the original so we solve for "c" again. So...c(squared) = 18.062(squared) + 13.5(squared)c(squared) = 508.5 - or stated differently; c = (the square root of) 508.5c = 22.55 (rounded up .0011)So your base is 27 and your 2 vertical sides are 22.55 for each of 4 sides, and 27 by 27 is the square base.MattLast edited by Matt_Maguire; 03-21-2013 at 09:25 PM.
Reply:That helps quite a bit to my original question.  My first welding project was a fire pit that I built using the app...I uploaded the pictures.  For some reason it seemed way easier last summer Sent using Tapatalk
Reply:By the way...thanks everyone for the replies.Sent using Tapatalk
Reply:Originally Posted by mfldmikeFE?
Reply:Hello mfldmike, sorry I'm a little bit late on this one but this sketch might bring a lot of this information to light for you and others. Best regards, Allan Attached Imagesaevald
Reply:Right, and I'm thanking all of you that have done that.Sent using Tapatalk
Reply:Originally Posted by aevaldHello mfldmike, sorry I'm a little bit late on this one but this sketch might bring a lot of this information to light for you and others. Best regards, Allan
Reply:Hello Dave that's definitely true. If you wished to swing arcs off of the bottom of the base and intersect them at the top, solving for the hypotenuse on the blue triangle would provide that information. That's where I was coming from. Best regards, Allanaevald
Reply:Originally Posted by aevaldHello Dave that's definitely true. If you wished to swing arcs off of the bottom of the base and intersect them at the top, solving for the hypotenuse on the blue triangle would provide that information. That's where I was coming from. Best regards, Allan
Reply:Be sure to paint an eye at the top.Calculate? Attached ImagesLast edited by Oldendum; 03-22-2013 at 07:47 AM."USMCPOP" First-born son: KIA  Iraq 1/26/05Syncrowave 250 w/ Coolmate 3Dialarc 250, Idealarc 250SP-175 +Firepower TIG 160S (gave the TA 161 STL to the son)Lincwelder AC180C (1952)Victor & Smith O/A torchesMiller spot welder
Reply:The framing square is cheat'n! Matt
Reply:This is the app I was talking about.  This is the one I used over last summer. It's super helpful, but it's nice to know how to do it without the use of calculators. Sent from my SCH-I605 using Tapatalk 2 Attached Images