After playing with the similar triangle idea, it's a messy way to go. I assume you've got stuff like this floating around, but just for obviousness, I'll post something here to draw fire^Wcomments.

Assuming that the slope is uniform for each vertical cross-section: 

| Zp (peak)
|\
| \
|  \
| a \
|    \
|-----\ (Xa,Ya)
|      \
|  b    \
|        \
|---------\ (Xb,Yb)
|          \    
|           \   
|    c       \  
|             \ 
|--------------\ (Xc,Yc)
|               \    
|                \   
|      d          \  
|                  \ 
|-------------------\ (Xd,Yd)
(Xp,Yp)

[download]
The height of each triangle is Hi = Zp-Zi. The base of each triangle is Bi = sqrt((Xp-Xi)**2 + (Yp-Yi)**2). The slope M of the hypotenuse is Mi = Hi/Bi, and since they're all the same, M = Hi/Bi, and: 
Ha   Hb   Hc   Hd
-- = -- = -- = --
Ba   Bb   Bc   Bd

[download]
Here's where I go wrong -- setting up the system of equations, and solving for Xp, Yp, and Zp. But since you need 3 equations for 3 unknowns, you may only need 3 points, not 4: 
Ha   Hb
-- = --
Ba   Bb

Ha   Hc
-- = --
Ba   Bc

Hb   Hc
-- = --
Bb   Bc

[download]
(But I'm fuzzy on the independence of those.)

But multiplying all of that out, and solving appropriately, is not something I'm good at. Now if I had a symbolic solver handy, it should pop out straight away (if I've done it right, which is debatable).

-QM
--
Quantum Mechanics: The dreams stuff is made of

In reply to Re: OT:Math problem: Grids and conical sections. by QM
in thread OT:Math problem: Grids and conical sections. by BrowserUk

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