General Question Can I change docking port angles.

[Kendo]

I have a ship which I,ve added a couple of docking ports. The problem is, on one of them, due to the shape of the ship, I need to change the angle slightly from the standard 90%, so it fits level with the ship. Can this be done using Spacecraft3. Many thanks, Ken.

 

[ADSWNJ]

Oh interesting! I was looking for a ship with irregular angled docking ports to test out my docking port rotation matrices.

One thing though ... will you also have a way to gimbal the thrusters in docking mode, so e.g. Pitch Up and Yaw Right act at right angles to the docking alignment? Otherwise it will be a bear to fly.

 

[Zatnikitelman]

Yes, you will need to alter the DIR and ROT parameters in the .ini file. The resulting vector must be normalized (sqrt(X^2+Y^2+Z^2)=1) and ROT must be a right angle to the DIR vector. If you're only rotating in one dimension, you could use simple trigonometry to find the new values, otherwise, you'd need either rotation matrices, or normalize a vector that describes the DIR and ROT.

 

[Kendo]

I,m lost lol, this is one line I have. What figures do I need to change to alter the alignment.
For example 5 degrees to the right. The first three figures are the exact central position. I need to angle it slightly to the right.
-18.9 -22.11 -8.12 -1 0 0 0 1 0 023

 

[N_Molson]

Where does that 023 comes from ?

 

[Kendo]

ISD freq

 

[N_Molson]

All is in that sequence of numbers :

-1 0 0 0 1 0

Which hide the two DIR and ROT vectors described above, each expressed in [x ; y ; z] :

DIR V_[-1 ; 0 ; 0]
ROT V_[0 ; 1 ; 0]

 

[C3PO]

I need to angle it slightly to the right.
-18.9 -22.11 -8.12 -1 0 0 0 1 0 023

Not sure what you mean by "to the right". The DIR vector -1 0 0 points straight out the left side.
If you want it to point 5° forward the entry would be:
-18.9 -22.11 -8.12 -0.996195 0 0.087156 0 1 0 023
And 5° aft would be:
-18.9 -22.11 -8.12 -0.996195 0 -0.087156 0 1 0 023

 

[Kendo]

Thanks guys.
C3PO, if I change the numbers on just one side,(6th block of figures) does anything get distorted visually. It lines up better with these figures.
-8.32 -3.78 11.705 -0.996195 0 -0.067156 0 1 0 023

Or do both sets of figures have to add up to something. something?
Also how did you work out these figures?. Many thanks.

 

[meson800]

C3PO just changed the DIR_V vector for the docking port using trigonometry
Your initial vector for "straight left" is -1 0 0. This is because the normal equation for the vector is sin(theta); 0; cos(theta); "Straight left" is -90 degrees, and sin(-90) = -1 and cos(-90) = 0

To get 5 degrees forward, C3PO did the sin and cos of -85 degrees, to get the vector -0.996195 0 0.087156

5 degrees aft is the sin and cos of -95 degrees, resulting in the vector -0.996195 0 -0.087156

 

[C3PO]

:hesaid:

And the vector should have a length of 1.0.

---------- Post added at 03:07 PM ---------- Previous post was at 02:56 PM ----------

I still have a very old post-it on my desk.

I drew it when I first started out tinkering with Orbiter, and I still use it to check the directions.

 

[Kendo]

Thanks guys, you have been a great help.
Is there an easy way to get the figures,ie a converter, or do I have to work it out in my head?

 

[N_Molson]

I like that page for it's simplicity :

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

Else you can always try with Wolfram Alpha, which can be a bit tricky to use but gives excellent results