# [AIRCRAFT] FUSELAGE VIEW CAMERA

• Jan, can you explain the way these new Fuselage views are set up because I've been trying to change them for my aircraft and the results aren't always what I expect, so I think I haven't fully grasped how they are set up. I want the camera to look more toward the tail of the aircraft so the whole things is in view... and I've been playing about with the middle direction figure. If I change it from -0.8 to -0.5 that would be 45 degree backwards Yes?

KR Matt

Code
``````            <[camera][CameraLeftFuselageView][]
<[string8][Body][LeftWing]>
<[tmvector3d][R0][ -0.6 6.9 0.6 ]>
<[tmvector3d][X0][ 0.0 -1.0 0.0 ]>
<[tmvector3d][Direction][ 0.1 -0.8 -0.1 ]>
<[float64][Cone][0.0]>
<[bool][InCockpit][false]>
<[string8][Tags][fuselage wing left]>
>``````
• Official Post

Hi Matt,

a 45 degree backwards from left side to the right would be -1.0 -1.0 0.0

or if you want to normalize the vector it's -0.707107 -0.707107 0.0.

-0.866 -0.5 0.0 would be 30 degrees from the rear towards the right side.

What you probably want is -0.5 -0.866 0.0 which is mostly to the right and 30 degrees back.

Keep in mind that the 3d vector for the Direction and X0 parameters are:

positive x = forward, positive y = left, positive z = up.

Regards,

Jan

• So Basically the camera direction is a vector in radians? I have a job working out the values I need... tried -0.5 -0.866 0.0 [/tt]

that gave me...

using the hat switch to turn the camera I wanted more like this...

After some trial and error I used this code....

Which Looks like this... left

right...

KR Matt

• Interpreting that for left view as example <[tmvector3d][Direction][ -0.5 -2.2 0.0 ]>

2.2 radians is 126 degrees and its negative because it looking right rather than left along the width axis

0.5 is 28.6 degrees, being negative it means the camera is tilted down toward the front, along the length axis, by that angle, from the origin point

and 0.0 means the vertical axis is perpendicular

Have I got that right?

Can you explain what X0 is? I assume R0 is the 3D xyz point the camera is looking from

KR Matt

• Official Post

Hi Matt,

it's a unit vector in a sphere of radius 1.

The three components are the Cartesian coordinates of the tip of the vector and each value ranges from -1.0 to 1.0.

On my personal webpage I have several tools, including a rotation matrix generator. If you plug in the Euler rotation angles in degrees it spits out the X0, Y0 and Z0 vectors (X0 could be your Direction parameter) as well as the B0 matrix for rigidbodies and controls.tmd clickspots.

Software Products – Aerofly-Sim.de

The unit calculator also has conversions for RPM, angles, masses, forces, etc.

Regards,

Jan

• Clear as mud 😁 I had to look up what a Euler rotation was. Understood about 10% of what I read. Not sure I understand why you need more than a fixed co-ordinate framework xyz and a point location to specify these cameras... I don't follow but maybe I don't need to. What I've done above works. I'll try using that gismo of yours and see what it spits out in the range of -1 to 1. If I can't get this to work can I leave it as is?... That won't cause any problems for AF4?

KR Matt

• Official Post

Hi Matt, no, the above code should work, but it means the same as <[tmvector3d][Direction][ -0.227 -1.0 0.0 ]>.

You can think of the values as pointing 100% in negative y direction and 27% aft.

Regards,

Jan

• ## MDIveyMay 25, 2023 at 8:26 PM

Changed the title of the thread from “AIRCRAFT: FUSELAGE VIEW CAMERA” to “[AIRCRAFT] FUSELAGE VIEW CAMERA”.