time, velocity and acceleration with respect to position are calculated. The total time
taken to deploy the level is also determined. From the stages of deployment shown in
Figure 3-2 to Figure 3-4, it can be concluded that the time taken to deploy the total mast
is sum of the time taken to deploy every level of the mast. Let Z, V, A, t and Av be the
values of displacement, velocity, acceleration, instantaneous time and average
acceleration. For any level (j) the force equation is
6K ((Lmax- Li) +dmin) cos0i -3Wp (j) 6Ws (j-1) -6Wtscos (90- 0d) COS i
a2(X,)
at2 (3Mp (j)+6Ms(j-1)) (3.15)
dt
As mentioned above with regards to equation (3.4) the mast is non-linear and hence
a simulation technique is adopted to determine the velocity and time at every instant of
position. The instantaneous time, velocity and acceleration for all positions is then
determined.
At position Zo = Zd
to (Zo) = 0
Vo (Zo) = 0
6K((Lax-4) +d.) cos0o -6,, cos0-0o) cos4 -3W(j)-6[( -1)
Ao (Zo) = 3M-(j)+6 (j-I)
3AM )+6H/0-1)
At position Zi = Zo + .01
SZ = 6K((L-.a-41)+d) co V-6W, cosO-0z) cosO -3W,(j)-6W3( -1)
3AM() +6(i -1)
Aiv (Zi)= .5(Ao (Zo) + A, (Zi))
V(Zo)+(Vo(Z0)2 +.04A1 (Z1))5
t (2A() -
2A,(Z,)