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# Reverse trajectory calculation

Using the following rules:

x = vx * t
y = vy * t + c * t
sqr( vx * vx + vy * vy ) = 1

When the numbers c, x and y are known,
how do I calculate t, vx and vy for the lowest positive value of t?
how do I calculate vx and vy for a given value of t?
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JasonMewes
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1 Solution

Commented:
Since c, x, and y are known, for a given t:

from (1), we can have: vx = x/t
from (3), we can have vy = sqrt(1- vx*vx) = sqrt(1- (x/ t) * (x/t))

considering (2), after a series of transformation:
t=sqrt(x^2 + (y-ct)^2) or t = -sqrt(x^2 + (y-ct)^2)

It is can be inferred that the lowest positive value of t is location, (x,y), dependent. With x = 0, y = ct, t can be 0.
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Commented:
Alternatively (and a bit simpler) you have
vy = y/t - c
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Commented:
the bottom line for this problem is:

although it is routine to calculate the trajectory based on known initial condition, the answer to the reverse problem may not always be unique or finite.
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Commented:
I beleive there is some difficulty with the equationb
t=sqrt(x^2 + (y-ct)^2) or t = -sqrt(x^2 + (y-ct)^2)
The units do not check out.
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Commented:
x = vx * t
y = vy * t + c * t
sqr( vx * vx + vy * vy ) = 1
y = (vy+c) * t
vx = x/t
vy = y/(t+c)

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Commented:
oxo
are you sure your vy is correct? It looks to me that it has units trouble.
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Commented:
No more units trouble than the original rules
What are the units of y, vy, c and t?
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Commented:
If you know x, y, and c there is only one value of t which will fit.
(actually possibly two because there is a square root involved)
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Commented:
L      L/T      L/T      and    T

your last equation returns to y = vy *t + vy*c
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Commented:
sorry, you're right
vy = y/t - c
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Commented:
t = -2c*y +- (sqr (4 c^2*t^2 + 4 (V^2-c^2)*(x^2+y^2))/(2(V^2-c^2)))

Where V is the 1 in your velocity equation.

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Author Commented:
leiz, your solution works quite well for the problem I describes - so I will award you your points.
however, i made I mistake in the formulation, so I'm going to add another question with the correct formulas.
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