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Find angle giving highest rotated vector Z conponent (around X axis)

JBlackhole
16-Pearl

Find angle giving highest rotated vector Z conponent (around X axis)

To all

trying to remember my school days and the use of derivative (if that's the tool to use). Given the following

an X,Y,Z Cartesian co-ordinate system

a vector v with components (vx, vy, vz)

a rotation matrix around x-axis Rx(theta)

Header 1Header 2Header 3
100
0cos(theta)-sin(theta)
0sin(theta)cos(theta)

Is there a way of finding at which angle theta the z component is the highest?

Thanks

Regards

JXB

ACCEPTED SOLUTION

Accepted Solutions
StuartBruff
23-Emerald III
(To:JBlackhole)

J B wrote:

trying to remember my school days and the use of derivative (if that's the tool to use). Given the following

an X,Y,Z Cartesian co-ordinate system

a vector v with components (vx, vy, vz)

a rotation matrix around x-axis Rx(theta)

Header 1Header 2Header 3
1 0 0
0 cos(theta) -sin(theta)
0 sin(theta) cos(theta)

Is there a way of finding at which angle theta the z component is the highest?

Here's a starter for 10 ...

Stuart

View solution in original post

4 REPLIES 4

The z component of the rotated vector will be  vy*sin(theta) + vz*cos(theta)

Differentiate wrt theta and set to zero to get the maximum theta as  thetamax = atan(vy/vz)

(Check that it is a maximum not a minimum!).

Alan

.

StuartBruff
23-Emerald III
(To:JBlackhole)

J B wrote:

trying to remember my school days and the use of derivative (if that's the tool to use). Given the following

an X,Y,Z Cartesian co-ordinate system

a vector v with components (vx, vy, vz)

a rotation matrix around x-axis Rx(theta)

Header 1Header 2Header 3
1 0 0
0 cos(theta) -sin(theta)
0 sin(theta) cos(theta)

Is there a way of finding at which angle theta the z component is the highest?

Here's a starter for 10 ...

Stuart

Thanks Stuart for the input.  A lot more that I anticipated! I can see what I need I think

StuartBruff
23-Emerald III
(To:JBlackhole)

J B wrote:

Thanks Stuart for the input.  A lot more that I anticipated! I can see what I need I think

No worries, JXB.

Stuart

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