Hi! Can anyone create an alpha account for me? I finally managed to build the client, but now it's asking me for a login. Looking to check out the status of the project and maybe contribute..

I uploaded a picture here: https://alphas.highfidelity.io/t/ray-trace-design/668

I suppose nobody knows from the top of their head what is the best way to find point 'I' (capital i) in the graph in the bottom-right corner?

O, D and I are in the blue plane (the rayPlane) which is parallel to the green line (axis). The blue dotted line is parallel to the green axis and 'I' is on it.

I know the distance between axis and rayPlane (N is a unit vector in that direction: its both perpendicular to axis as well as rayPlane). 'A' is a unit vector along axis, O is a point in the rayPlane whose relative position to S is known. D is a unit vector pointing from O to 'I'. Asked is to express a vector from S to 'I'.

I suppose nobody knows from the top of their head what is the best way to find point 'I' (capital i) in the graph in the bottom-right corner?

O, D and I are in the blue plane (the rayPlane) which is parallel to the green line (axis). The blue dotted line is parallel to the green axis and 'I' is on it.

I know the distance between axis and rayPlane (N is a unit vector in that direction: its both perpendicular to axis as well as rayPlane). 'A' is a unit vector along axis, O is a point in the rayPlane whose relative position to S is known. D is a unit vector pointing from O to 'I'. Asked is to express a vector from S to 'I'.

Let S be the origin of all vectors. Let R be a vector to a point on the ray, then (R - O) X D = 0.

Let P be the point directly above S in the rayPlane, thus P = d N, where d is already known.

Let B be a vector to a point on the dotted line, then (B - P) X A = 0.

Then I is the point that satisfies both: (I - O) X D = (I - P) X A = 0.

Hmm, I wonder if I can ask Wolfram Alpha to solve that for me - heheh.

Let P be the point directly above S in the rayPlane, thus P = d N, where d is already known.

Let B be a vector to a point on the dotted line, then (B - P) X A = 0.

Then I is the point that satisfies both: (I - O) X D = (I - P) X A = 0.

Hmm, I wonder if I can ask Wolfram Alpha to solve that for me - heheh.

I try to download the MD5 hash from here - http://highfidelity-public.s3-us-west-1.amazonaws.com/binaries/win/domain-server/domain-server.md5 - using Qt Network classes and I get "??9" as the reply output and if you look at the complete MD5 hash, 9 is the first character. Using the same code when I try to download - http://highfidelity-public.s3-us-west-1.amazonaws.com/binaries/mac/domain-server/domain-server.md5 - I get the correct reply output. Note that in the first case the MD5 is for Windows and in the second instance it is for Mac. What might be the problem? is there some problem with the ".md5" file on the server?

Also, the Linux version of the MD5 hash is also retrieved correctly through my code.

we really need the @ thing that pops up offering to pick a name here , back in hf

@daemeh send me an email and I can help you out. chris at Highfidelity.io

Yes! I think I have it :) ... much more beautiful than anything I could find on the net :P

Still have to put into formulas, not to mention code - but I added a new picture to https://alphas.highfidelity.io/t/ray-trace-design/668/4 with the rough description of my idea ;).