3-D Rendering

Feb 14th, 2000, 07:06 AM

Does anyone know how to do 3-D rendering in VB3? I need to know how to rotate objects, plot 3d points and lines, and calculate what to show according to the user's viewpoint.

**Darkwraith** · Jul 29th, 2003, 06:13 PM

Can DirectGraphics for VB accomplish this?

**Electroman** · Jul 29th, 2003, 06:24 PM

DirectX won't work with anything less than VB5. Only way would be to make your own 3D to 2D transformer.

**Darkwraith** · Jul 29th, 2003, 06:48 PM

I think that this might help:

http://www.exaflop.org/docs/cgafaq/cga5.html

Hope you like math!

**Electroman** · Jul 29th, 2003, 07:03 PM

Wooohoo, looks good but its all in C

How come you never said anything in this thread??

**Darkwraith** · Jul 29th, 2003, 07:08 PM

(1) just found the site
(2) you are using linear algebra which I have not had yet

(trying to learn it myself though

. )

However, I will take a look at it and see if I can add anything to the thead.

**Electroman** · Jul 29th, 2003, 07:14 PM

linear algebra

How else could you do it??
Had you already read the thread before or have you only just read it??

**Darkwraith** · Jul 29th, 2003, 07:25 PM

I saw some of the notation at a quick glance and its all vector math. I had some basic vector math in my Calculus class (cross and dot product, basic vector handling, gradients, etc.) so I kinda know what I am doing, but its takes a little while for me to understand what you are trying to convey with your mathematics.

However, it was only a quick glance (so I could be wrong) and I am playing with another vector problem at the moment. (so I have not entered the fray as of yet.

) Basically, I need to spend some quality time analyzing the problem.

http://www.vbforums.com/showthread.p...hreadid=252696

(NOTE: Just realized that the last post's smilies are off. This has been corrected.

)

**Electroman** · Jul 29th, 2003, 07:34 PM

I noticed that thread a while back

I thort it had been solved tho when I saw the first reply, just read his reply now tho and it seems he didn't take into count your resulting direction as well. I'll give it a look and see what I can do with it

.

If you read that thread I think I curse the Fact that Cos(ArcCos(x)) <> x for 0 < x < 360.

**Darkwraith** · Jul 30th, 2003, 12:28 PM

I should be coming up from with an algorithm from that thread soon. Due to the extensive help that I recieved and the fact that the only valuable part of the resulting project is that algorithm, I will post the algorithm.

(NOTE: How can that be when arccosine and cosine are inverse functions by definition?

)

**SLH** · Jul 30th, 2003, 12:42 PM

Originally posted by Darkwraith
(NOTE: How can that be when arccosine and cosine are inverse functions by definition? )

I thought cos^-1 was the inverse.

**Electroman** · Jul 30th, 2003, 01:56 PM

ArcCos is Cos^-1

And you are both right, only thing is Cos(ArcCos(x)) = x for 0 < x < 180 you were thinking along the lines I was

thats why I was cursing that it didn't work for all 360^o. try it and you'll see

.

**Darkwraith** · Jul 30th, 2003, 07:34 PM

Ok. I just started to look at the algorithm (and some of the VB code which I could understand

. ) Hopefully, I will have some results, albeit it will be slow going.

**Electroman** · Jul 30th, 2003, 08:08 PM

Does that mean you don't need help with the Physics problem thread?? Just cos I didn't see anything in that thread that helped you with the resulting direction.

**Electroman** · Jul 30th, 2003, 08:18 PM

I've only just realised that when I was on about the Cos and ArcCos stuff in this thread I put it the wrong way round, same applies tho

. it should have been:
ArcCos(Cos(x)) = x for 0 < x < 180 I'm sure you both knew what I meant

**Darkwraith** · Jul 31st, 2003, 02:49 PM

Does that mean you don't need help with the Physics problem thread??

I'm going to release the first draft of the algorithm to make sure that it makes sense and if there is an easier way to accomplish the problem.

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