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I have a kind of Oblique Pyramid. I use this structure up side down. Tip of the pyramid is looking towards down. All four legs of the pyramid are wire lines and connected to winding motors on the corners of the base. Base is square 10m x 10m. I am able to move the tip of the pyramid within the perimeter using motors. Some are winding some are unwinding while I am doing this.

So, I know exact lengths of the 4 lines ($a$, $b$, $c$, $d$) and the perimeter sizes (10m x 10m x 10m x 10m). Can I calculate the height of the tip of the pyramid with those variables? If yes, what would be the equation(s)? enter image description here

Sener
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    Welcome to Math SE! Please try to provide a picture of what you're talking about. You may provide a link or an attached picture by clicking on that "edit" hyperlink. It doesn't have to be a perfect picture. We don't expect you to be an artist. We just need something very basic. –  Jul 06 '16 at 14:21
  • If you can make a photo (this is a real structure?) it would be good too – Yuriy S Jul 06 '16 at 14:36
  • @Yuri, I have attached a link to the image I made, I hope it helps to imagine. Yes, it is a real structure hanging down from a sea vessel. – Sener Jul 06 '16 at 14:42
  • Thank you @KingDuken, I believe I will enjoy a lot here with lots of mathematics as I am a software engineer. – Sener Jul 06 '16 at 14:43
  • As a short answer to your question - yes, you can find $h$ from the data you have. Unfortunately, I don't have the time to write a solution right now, I'm sure someone will do it soon – Yuriy S Jul 06 '16 at 15:09

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Let $H$ be the projection of the vertex onto the base, and let $x$, $y$, $10-x$, $10-y$ be its distances from the base sides. Then by Pythagoras' theorem you get: $$ h^2=a^2-x^2-y^2=b^2-x^2-(10-y)^2=c^2-y^2-(10-x)^2. $$ Solve this set of equations to get $h$.

Intelligenti pauca
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  • thank you for your the equation. The problem is, I don't have values for variables $x$ and $y$. If I knew them, the issue has already been solved. Or am I missing something here? – Sener Jul 06 '16 at 19:52
  • These are THREE equations for THREE unknowns $x$, $y$ and $h$. – Intelligenti pauca Jul 06 '16 at 20:03
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    Since the OP is looking for a practical solution, and there seems to be no study value in having them try and find it, I'll just note that solving $x$ and $y$ and $h$ from the three equations (but simplifying the expression for $h \ge 0$ -- since only nonnegative $h$ makes physical sense -- using $x$ and $y$), we get$$\begin{cases}x = \frac{a^2 - c^2}{20} + 5\y = \frac{a^2 - b^2}{20} + 5\h = \sqrt{a^2 - x^2 - y^2}\end{cases}$$. – Nominal Animal Jul 06 '16 at 21:23
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    In fact, there are six triangles in the physical device. As you can see from this answer by Aretino, only three are needed for the mathematical solutions -- $d$ is not used here at all. A practical device would -- depending on whether it is to control a changing $h$, or just to measure $h$ -- use multiple combinations, and average them; for closed-loop control, it could minimize "slop" (backlash) by minimizing the differences between the possible solutions by changing the length of one side that minimizes the differences. But that is an interesting cross-discipline problem in and of itself. – Nominal Animal Jul 06 '16 at 21:30
  • Thanks a lot @Nominal for clarifying Aretino's equation. It makes now perfect sens to me and even further it just works OK in my system. – Sener Jul 08 '16 at 10:09