If we have the satellite state vector and the station coordinates and the modified julian date, how can we find topocentric right ascension and declination, azimuth and elevation.

Sketch the isometric and draw the orthographic view with neat arrangement ,it's better to do in catia V5 ( no hand sketch ,only catia v5 )

What would be the Direction Cosine Matrix if there was a rotation about q for 120 degrees. I know that q would be lambda and theta = 120 degrees. But we don't have any measure numbers for lambda, so I can't create a DCM using Principle Rotation Parameters. Would you do it using the Euler Angles method? If so, how?

Re-derive analytically with the intermediate steps and motivate with pictograms the transfor-mations from geocentric right ascension and declination to topocentric right ascension anddeclination and to topocentric azimuth and elevation. Show the hour angle and the siderealtime explicitly in your drawings. Define all terms and variables.  Please draw by hand

I’m having trouble in my statics course. We are covering moments and I am confused about (1) what to use for the distance (2) when to use rFsin(theta), r(cross)F, or M=Fd. I  keep using the wrong method and I really just need to know the way I approach the problem. For instance, on the problem below, I used rxF and the solutions said instead to simply use rFsin(theta)

GIVEN FRONT AND SIDE VIEW, SKETCH THE MISSING VIEW ALSO DRAW THE ISOMETRICVIEW USING AUTO CAD -16 MATL - SAE 1050 Ø8 3-101 R 10 R 25 FØ16.1 R 25 R 28 Ø 20. R 10 Ø 12.5 R 13 Connector.

A pictorial view of a machinebracket is shown in the figure. Draw using the first angle projection method front view, top view and right end side view. 30 80 050 100 D60 20% BOSS 30 15 Ø40 70 R40 LOBE Note : sketch the figure úsing CATIA V5 drafting no hand writing

Which datum has size? Please refer to image attached.

Copy of Based on the kinematic system below If we rotate (A-B) 90 degrees counter-clockwise, and (B-P) 90 degrees clockwise where will be the point P y (2,4) Kinematic skeleton 42) O (4,4) (4,6) (6,4) O (6,6)

I had a theoretical question about attitude determination. In the attached images, I gave two axis and angles. The coefficient of the axes are the same and the angles are the same. The only difference is the vector basis. Lets say there is a rotation going from n hat to b hat. Then, you introduce a intermediate rotation s hat. So, I want to know if the DCM produced from both axis and angles will be the same or not. Does the vector basis affect the numerical value of the DCM? The DCM formula only cares about the coefficient of the axis and the angle. So, they should be the same right?