Image credil Voyager, NASA d (dist. from s/c to maon) I 8 x (distance to closest approach) Spacecraft The slew angle, 0, is related to r and x by tan(0) = r/x. (Important note, r is a constant!) Taking the time derivative (and remembering the chain rule), we get 8 (sec 0 )² = - 17/7/2 x ₁ or 8 = - * (cos 0)² Remembering our trigonometry, we can replace cos(0) and set x = -V (because x is getting x Vr Therefore, required 8 = x²+r² with smaller). "
Image credil Voyager, NASA d (dist. from s/c to maon) I 8 x (distance to closest approach) Spacecraft The slew angle, 0, is related to r and x by tan(0) = r/x. (Important note, r is a constant!) Taking the time derivative (and remembering the chain rule), we get 8 (sec 0 )² = - 17/7/2 x ₁ or 8 = - * (cos 0)² Remembering our trigonometry, we can replace cos(0) and set x = -V (because x is getting x Vr Therefore, required 8 = x²+r² with smaller). "
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If theta dot is the slew rate, what would be the maximum slew rate? How would you differentiate theta dot to get theta double dot?
Transcribed Image Text:Image credil. Voyager, NASA
r
d
(dist. from
s/c to maon)
0
V
x (distance to
closest approach)
Spacecraft
The slew angle, 0, is related to r and x by
tan(0) = r/x. (Important note, r is a constant!)
Taking the time derivative (and remembering the chain
rule), we get (sec 0 )² = -1/2 x ₁
x,
ė
x2
or 8 =
Remembering our trigonometry, we can replace cos(0)
with √22, and set x = -V (because x is getting
X
Vr
smaller).
Therefore, required 8 =
x²+r²
r
2 x (cos 0) ²
x2
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