"Robert Larsen" <robert@[EMAIL PROTECTED]
> schreef in bericht
news:481b307f$0$27377$d40e179e@[EMAIL PROTECTED]
> Hi group
>
> I seem to have lost all math skills :-(
>
> I am trying to determine the coefficients of a quadratic polynomial
> given the following data:
>
> x1,y1 - One point on the curve
> x2,y2 - Another point on the curve
> y3 - The Y coordinate of the top coordinate
>
> y3 > y1 and y3 > y2
> x2 > x1
>
> I believe this should yield exactly one solution but have no idea how to
> find it. Can you help me ?
> It should be used for calculating the path for an animation in a
> computer game.
>
> Sincerely,
> Robert
coordinates: (x1,y1) , (x2,y2) and top(??,q)
y = a(x-p)^2 + q
y1 = a(x1-p)^2 + q
--> a = (y1-q)/((x1-p)^2) (I)
[x1 = p is special case]
y = (y1-q) * (x-p)^2 / (x1-p)^2 + q
y2 = (y1-q) * (x2-p)^2 / (x1-p)^2 + q
(y2-q) = (y1-q) * (x2-p)^2 / (x1-p)^2
(y2-q) * (x1-p)^2 = (y1-q) * (x2-p)^2
(y2-q) * (x1^2 - 2px1 +p^2) = (y1-q) * (x2^2 - 2px2 + p^2)
p^2 * ((y2-q) - (y1-q)) + p * (-2x1*(y2-q) + 2x2*(y1-q)) + (x1^2 * (y2-q)
-
x2^2 * (y1-q)) = 0
p^2 * (y2-y1) + p * (-2x1 * (y2-q) + 2x2 * (y1-q)) + (x1^2 * (y2-q) - x2^2
*
(y1-q)) = 0
D = ( -2x1 * (y2-q) + 2x2 * (y1-q) )^2 - 4 * (2*(y2-y1)) * (x1^2 * (y2-q)
-
x2^2 * (y1-q))
D = 4 * (x1-x2)^2 * (y1-q) * (y2-q)
[D>0 (y1,y2 < q or y1,y2 > q) --> 2 solutions]
p = (2x1*(y2-q) - 2x2*(y1-q)) / (2*(y2-y1)) +/- sqr[D] / (2*(y2-y1))
--> p = ( x1*(y2-q) - x2*(y1-q) +/- (x1-x2) * sqr[(y1-q)*(y2-q)] ) /
(y2-y1) (II)
[y1 = y2 is special case]
example 1.
y = - x^2 + 4x + 1 take (0,1) , (3,4) , ([2],5)
p= (0*-1 - 3*-4 +/- 3*sqr[(-1)(-4)] ) / 3 (from II)
p = (12 +/- 6) / 3
p+ = 18 / 3 = 6
p- = 6 / 3 = 2
a(p=6) = (1 - 5)/(0 - 6)^2 = -1/9 (from I)
--> y = -1/9(x-6)^2 + 5
a(p=2) = (1 - 5)/(0 - 2)^2 = -1
--> y = -(x-2)^2 + 5 [ y = - x^2 + 4x + 1 ]
example 2.
y = -2x^2 + 5x - 3 take (2,-1) , (-3,-36) , ([1.25],0.125)
p = (2*-36.125 + 3*-1.125 +/- 5*sqr[-1.125*-36.125] ) / -35
p = (-75.625 +/- 31.875) / -35
p+ = -43.75 / -35 = 1.25
p- = -107.5 / -35 = 43/14
a(p=1.25) = (-1.125)/(2-1.25)^2 = -2
--> y = -2(x-1.25)^2 + 0.125 [ y = -2x^2 + 5x - 3 ]
a(p=43/14) = -1.125/(2 - 43/14)^2 = -0.98
--> y = -0.98*(x-43/14)^2 + 0.125
There will be 2 solutions.
One with p (= x_top) between x1 and x2.
special cases:
x1 = p (y1=q) --> a = (y2-q) / (x2-p)^2
y1 = y2 --> p = (x1 + x2)/2 --> a = 4(y1-q) / (x1-x2)^2
AvdB
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