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Narutokun
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In Topic: Distance to a pixel
Today, 01:07 PM
Looking over the output results, it appears that the x and y values of the vector after being multiplied by the inverse projection matrix are the angles to the pixel in radians (x converted to degrees is roughly 60 which fits with my horizontal FOV of 120 deg, and y is roughly 3.75 deg which fits my vertical FOV of 7.5 deg). I'm guessing the key is in what relationship the z and w values have with my far and near clip planes?
In Topic: Distance to a pixel
Today, 11:13 AM
Here is the geometry shader I am working with. The basic idea, is that the X axis of the texture (depth buffer) I am sampling from will correlate with the X axis of the final image, and the distance to the pixel calculated using the depth buffer will correspond to the Y axis of the final image (in otherwords, a modified Bscope radar display where the horizontal corresponds to the azimuth of the return and the vertical corrosponds to the slant range).
Here I have the camera even further from the target.
[maxvertexcount(80)] void GS( point GS_INPUT sprite[1], inout TriangleStream<PS_INPUT> triStream ) { PS_INPUT v; uint ux = sprite[0].id % 256; uint uy = sprite[0].id / 256; float x = float(ux) / 256.0f; float y = float(uy) / 75.0f; float x1 = x + (1.0f / 256.0f); float y1 = y + (1.0f / 75.0f); float4 fdepth = float4( depth.SampleLevel(SampleType, float2(x, y1), 0), depth.SampleLevel(SampleType, float2(x, y), 0), depth.SampleLevel(SampleType, float2(x1, y1), 0), depth.SampleLevel(SampleType, float2(x1, y), 0) ); float4 v1 = float4((2.0f*x)  1.0f, 1.0f  (2.0f*y1), fdepth.x, 1.0f); float4 v2 = float4((2.0f*x)  1.0f, 1.0f  (2.0f*y), fdepth.y, 1.0f); float4 v3 = float4((2.0f*x1)  1.0f, 1.0f  (2.0f*y1), fdepth.z, 1.0f); float4 v4 = float4((2.0f*x1)  1.0f, 1.0f  (2.0f*y), fdepth.w, 1.0f); float4 m1 = mul(v1, inprj); float4 m2 = mul(v2, inprj); float4 m3 = mul(v3, inprj); float4 m4 = mul(v4, inprj); float l1 = length(m1.xyz / m1.w); float l2 = length(m2.xyz / m2.w); float l3 = length(m3.xyz / m3.w); float l4 = length(m4.xyz / m4.w); v.m = 0; v.p.x = v1.x; v.p.y = l1  1.0f; v.p.zw = float2(0, 1.0f); v.t = float2(x, y1); triStream.Append(v); v.p.x = v2.x; v.p.y = l2  1.0f; v.p.zw = float2(0, 1.0f); v.t = float2(x, y); triStream.Append(v); v.p.x = v3.x; v.p.y = l3  1.0f; v.p.zw = float2(0, 1.0f); v.t = float2(x1, y1); triStream.Append(v); v.p.x = v4.x; v.p.y = l4  1.0f; v.p.zw = float2(0, 1.0f); v.t = float2(x1, y); triStream.Append(v); triStream.RestartStrip(); }Here is the first sample, using the top left corner of the texture source image (dimensions are 256x75). The screenshot below gives you an idea of where the camera is looking. The actual texture source image would only contain a thin strip from the center of the screenshot.
ux 0 uint uy 0 uint x 0.000000000 float y 0.000000000 float x1 0.003900000 float y1 0.013300000 float fdepth x = 0.968700000, y = 0.964700000, z = 0.968600000, w = 0.964700000 float4 m1 x = 1.048700000, y = 0.063800000, z = 0.099900000, w = 1.068700000 float4 m2 x = 1.048700000, y = 0.065500000, z = 0.099900000, w = 1.064700000 float4 m3 x = 1.040500000, y = 0.063800000, z = 0.099900000, w = 1.068600000 float4 m4 x = 1.040500000, y = 0.065500000, z = 0.099900000, w = 1.064700000 float4 l1 0.987600000 float l2 0.991300000 float l3 0.980000000 float l4 0.983700000 float //Inverse projection matrix inprj[0] x = 1.048700000, y = 0.000000000, z = 0.000000000, w = 0.000000000 float4 inprj[1] x = 0.000000000, y = 0.065500000, z = 0.000000000, w = 0.000000000 float4 inprj[2] x = 0.000000000, y = 0.000000000, z = 0.000000000, w = 1.000000000 float4 inprj[3] x = 0.000000000, y = 0.000000000, z = 0.099900000, w = 0.100000000 float4Here is what the final output looks like. A bit confusing to say the least.
Here I have the camera even further from the target.
ux 0 uint uy 0 uint x 0.000000000 float y 0.000000000 float x1 0.003900000 float y1 0.013300000 float fdepth x = 0.999300000, y = 0.999200000, z = 0.999300000, w = 0.999200000 float4 m1 x = 1.048700000, y = 0.063800000, z = 0.099900000, w = 1.099300000 float4 m2 x = 1.048700000, y = 0.065500000, z = 0.099900000, w = 1.099200000 float4 m3 x = 1.040500000, y = 0.063800000, z = 0.099900000, w = 1.099300000 float4 m4 x = 1.040500000, y = 0.065500000, z = 0.099900000, w = 1.099200000 float4 l1 0.960000000 float l2 0.960300000 float l3 0.952600000 float l4 0.952900000 float //Inverse projection matrix inprj[0] x = 1.048700000, y = 0.000000000, z = 0.000000000, w = 0.000000000 float4 inprj[1] x = 0.000000000, y = 0.065500000, z = 0.000000000, w = 0.000000000 float4 inprj[2] x = 0.000000000, y = 0.000000000, z = 0.000000000, w = 1.000000000 float4 inprj[3] x = 0.000000000, y = 0.000000000, z = 0.099900000, w = 0.100000000 float4I hope this helps to give a better idea of what I'm trying to do.
In Topic: Distance to a pixel
Today, 09:16 AM
You should be able to figure this one out yourself. You can either read the documentation carefully to see what the values mean, or you can reverse engineer it: Set up a few test situations where you know how far the pixel is and look at what values you get for those cases. You can probably figure out a formula easily from a few examples.
Unfortunately this one isn't quite possible because the program I am working with has no way to provide me with the distance to a pixel so I would be working blind (I am working through an API, not source of the program). I only have a rough guess that I can work with which is anything but reliable.
Make sure the coordinates are in NDC: 1 to 1 for x/y (and z in GL) and 0 to 1 for z in D3D.I tried making an inverse matrix of the projection matrix and multiplying the screen and depth coordinates of the pixel from it but I didn't get the expected results.
Then after the matrix multiply, divide xyz by w to get a views pace position, and calculate its length to get distance from the camera.
I did a shader debug run using VS 2013's shader debugging library. The screen coordinates I calculate using texture coordinates from the area of the screen I am sampling (I convert them to the proper +1.0f ranges), and I can confirm the Z values are 0 to 1. After multiplying with the inverse projection matrix I divide the xyz by the w and put it through the length function. The value still comes out quite low (like around 1.2 to 1.8) and my final image looks a bit like a hyperbola.
I will post some screenshots shortly of the program which may help you visualise things a bit more effectively.
In Topic: Converting world rotation to body rotation
Today, 05:32 AM
I do have the plane orientation data. I'm just wondering how I should build the rotation matrix from it (inverse or something?) which I can then use to get the camera rotation values relative to the plane rather than the world.
In Topic: Converting world rotation to body rotation
22 August 2016  06:07 AM
I think you misunderstood me a bit.... I didn't design the system. The camera data is literally the raw view matrix. It gets rotated by the aircraft rotation as well as eyepoint rotations. I am trying to reverse the aircraft rotation of the camera view direction so I can get view direction relative to what is inside the aircraft for the purposes of a custom 3D sound environment.