Nina del Ser - Mathematics
18
| octopus
and their angular speeds Β and Ω respectively. 5 can therefore be modified to :
r x Bcos Βt bcos Β Ω t
r y Bcos Βt bsin Β Ω t
(8)
We now differentiate once to obtain the parametric equations for velocity ...
v x BΒ sin Βt b Β Ω sin Β Ω t
v y BΒ cos Βt b Β Ω cos Β Ω t
... and again to obtain those for acceleration :
a x BΒ 2 cos Βt b Β Ω 2 cos Β Ω t
a y BΒ 2 sin Βt b Β Ω 2 sin Β Ω t
We can now find an expression for a
a
2
2
by squaring and adding a x and a y :
BΒ 2 cos 2 Βt BΒ 2 sin 2 Βt b Β Ω 2 cos 2 Β Ω t b Β Ω 2 sin 2 Β Ω t
2
2
2
2 BΒ 2 b Β Ω 2 cos Βt cos Β Ω t sin 2 Β t sin Β Ω t
a
2
BΒ 2 b Β Ω 2 2 BΒ 2 b Β Ω 2
2
2
(9)
Now we know that the minimum value of a , which occurs when the cosine term is negative,
must be greater than or equal to zero for the curve to be convex,
therefore we can establish the following inequality :
BΒ 2 b Β Ω 2 2 BΒ 2 b Β Ω 2 0
2
2
BΒ 2 b Β Ω 2 0
2
BΒ 2 b Β Ω 2 0
BΒ 2 b Β Ω 2
(10)
BΒ 2 and b Β Ω 2 are the acceleration vectors of the primary and secondary rotors respectively, therefore
from (10) we know that the Octopus will generate convex curves when a p a s . Therefore, if we set Β
1, b 0.5 and Ω 3, the curves generated will start being convex for values of B 8. Figures 49-51 show that
this principle, indeed, works:
a s a s a s a p a s a p
figure 49: concave figure 50: just convex figure 51: convex
For the curves to be convex, the net acceleration a must always point toward the origin, (hence a p
122
a s ).