The four equations are the immediate offspring of plugging in values for x and y and z into that first equation I posted: (x-x')^2+(y-y')^2=m(z-z')^2. So for the equation for the top left corner (x'=-1, y'=1, z'=whatever height value was given), we plug those values into the above equation. Alas, I saw symmetry where there was none, and so I set m=1 which allowed much cancelling out. But even without m==1, when you multiply out the squares and then subtract any two equations, all the squares cancel out. | [reply] |
Point 1: (1, 0) at height h1
Eqn 1:
(x-1)2 + y2 = m(z - h1)2
x2 - 2x + 1 + y2 = m(z2 - 2zh1 + h12)
**** step 4
Point 2: (0, 1) at height h2
Eqn 2:
x2 + (y-1)2 = m(z - h2)2
x2 + y2 - 2y + 1 = m(z2 - 2zh2 + h22)
Point 3: (-1,0)) at height h3
Eqn 3:
(x+1)2 + y2 = m(z - h3)2
x2 + 2x + 1 + y2 = m(z2 - 2zh3 + h32)
Point 4: (0, -1) at height h4
Eqn 4:
x2 + (y+1)2 = m(z - h4)2
x2 + y2 + 2y + 1 = m(z2 - 2zh4 + h42)
-------------------------------------------------------
(Eqn 2 - Eqn 1)/m:
2(x - y)/m = -2zh2 + h22 + 2zh1 - h12
(Eqn 3 - Eqn 4)/m:
2(x - y)/m = -2zh3 + h32 + 2zh4 - h42
Combine to get:
-2zh2 + h22 + 2zh1 - h12 = -2zh3 + h32 + 2zh4 - h42
2z(h1 - h2 + h3 - h4) = h12 - h22 + h32 - h42
z = 0.5(h12 - h22 + h32 - h42)/(h1 - h2 + h3 - h4)
**** step 1
-------------------------------------------------------
(Eqn 3 - Eqn 1)/4:
x = 0.25m(-2zh3 + h32 + 2zh1 - h12)
**** step 2
(Eqn 4 - Eqn 2)/4:
y = 0.25m(-2zh4 + h42 + 2zh2 - h22)
**** step 3
And now we can use step 1 to find z, use steps 2 and 3 to substitute into step 4 to come up with a quadratic equation in m. Solve for m (remember that that's the slope squared). Substitute m back into steps 2 and 3 to find x and y.
Whew!
Update: Typos fixed. Had a - where I needed a +, and 4 when I needed 2. | [reply] |
Great job, Tilly! I just got it over here and then checked "Newest Nodes" before posting.
You beat me to the punch!
I'll try to transform my coordinates into yours for this post....
<UPDATE>After all the excitement of thinking I had a different way of acheiving the solution than Tilly, it turns out it's basically the same. I'm leaving the post alive because it may be nice to have the solution to that final quadratic equation, which I include at the end.</UPDATE>
Once getting z as you did, I plugged back into the original equation (x-x')^2+(y-y')^2=(z-z')^2/m^2 using each point for (x',y') and h1..h4 for z'. This gave me four equations (xx means x^2):
xx-2x+1+yy =aa/mm
xx+2x+1+yy =cc/mm
xx +yy-2y+1=bb/mm
xx +yy+2y+1=dd/mm
I subtracted the first two to get 4*x*m*m=h3^2-h1^2.
I subtracted the last two to get 4*y*m*m=h4^2-h2^2.
Dividing these two values: 4*y*m*m/(4*x*m*m)=(h4^2-h2^2)/(h3^2-h1^2)
Simplifying: y/x=(h4^2-h2^2)/(h3^2-h1^2)
Then I plugged back into the original equation with just one point (1,0):
Let k=y/x=(h4^2-h2^2)/(h3^2-h1^2) (from above)
Let n=x*m*m=(h3^2-h1^2)/4 (from above)
Original equation:
(x-x')^2+(y-y')^2=(z-z')^2/m^2
(x-1 )^2+(y-0 )^2=(h1 )^2/m^2
(x-1 )^2+(xk-0)^2=(h1 )^2*x/n
(x-1 )^2+(xk-0)^2=(h1 )^2*x/n
multiplying it out:
x^2-2x+1+x^2*k^2 =x*h1^2/n
combining terms:
x^2*(1+k^2)-x(2+h1^2/n)+1=0
let a=1+k^2
let b=2+h1^2/n
x1=(b+sqrt(b^2-4a))/2a
x2=(b-sqrt(b^2-4a))/2a
y1=k*x1
y2=k*x2
Sadly, only one of (x1,y1) or (x2,y2) is a valid solution. | [reply]
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There appears to be an error in your calculations or transcription of them.
Specifically, the sign of the
y2 component of equation 1, changes between the original form
(x-1)2 + y2 = m(z - h1)2
___________^
and it's expansion:
x2 - 2x + 1 - y2 = m(z2 - 2zh1 + h12)
________________^
I don't follow the steps you used in the next stage of the reduction, (Eq.2 - Eq.1)/m, so I am unable to filter the correction through the rest of the workings.
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