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The three t,y pairs satisfy the quadratic equations:
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y0 = α t02
+ β t0 + γ
y1 = α t12
+ β t1 + γ
y2 = α t22
+ β t2 + γ
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To simplify the derivation, make a change of variables: x = t - t1
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x0 = t0 - t1
x1 = t1 - t1 = 0
x2 = t2 - t1
- Then the quadratic equations, in terms of x, are:
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y0 = a x02 + b x0 + c
y1 = c
y2 = a x22 + b x2 + c
- So now we know c = y1.
After solving for a and b, the peak can be found as the point where
the slope of the quadratic y = a x2 + b x + c
is zero, i.e. dy/dx = 2 a x + b = 0 for x = -b/(2a).
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Multiply the first equation by x2,
multiply the third equation by x0,
and subtract:
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x2 (y0 = a x02 + b x0 + y1)
- x0 (y2 = a x22 + b x2 + y1)
x2 y0
- x0 y2 =
a x02 x2 + y1 x2
- a x22 x0 - y1 x0
- Solve for a:
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a =
( x2 y0
- x0 y2
- y1 x2
+ y1 x0 ) /
( x02 x2
- x22 x0 )
=
( x2 ( y0 - y1 )
+ x0 ( y1 - y2 ) ) /
( x0 x2
( x0 - x2 ) )
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Now use a and c (y1) to solve for b:
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y0 = a x02 + b x0 + y1
b = (y0 - y1 - a x02) /
x0
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In terms of x, the peak is at:
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x = -b/(2a)
y = a x2 + b x + y1
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In terms of t, the peak is at:
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t = t1 - b/(2a)