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Math Help

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Math Help

Postby Jeremoot » Mon Aug 08, 2011 10:37 pm

Say I have a scalar field representing n^3 points in space. If for a moment, you imagine the scalar field as a solid cube. To an onlooker, no matter where they are in space, only a portion of the points will be visible (or the closest to the onlooker in depth).

Image

To illustrate my point: the green faces, as well as the cross-hatched front face are visible to the observer. The red faces, as well as any point within the inside of the cube are not. As you rotate around the cube, the visible portion of each dimension will change. If you were to move the observer a complete 90 degrees around the cube, the amount of visible points in the x and y dimensions would flip. Knowing the position in space of both the observer and each scalar in the field, how can I flatten a 3 dimensional scalar field into a 2 dimensional square-- eliminating depth?
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Re: Math Help

Postby masterpoobaa » Mon Aug 08, 2011 11:07 pm

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Re: Math Help

Postby Astronomic » Tue Aug 09, 2011 5:23 am

If you just made the cube so that the observe was looking straight at 1 face, would it not appear as if it were a square?
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Re: Math Help

Postby Jeremoot » Tue Aug 09, 2011 8:15 am

Astronomic wrote:If you just made the cube so that the observe was looking straight at 1 face, would it not appear as if it were a square?


Yes, but if the observer is not looking directly at one face (e.g., looking at the corner of the cube), they will see portions of the other faces.
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Re: Math Help

Postby Astronomic » Tue Aug 09, 2011 8:44 am

But the qeustion was: Knowing the position in space of both the observer and each scalar in the field, how can I flatten a 3 dimensional scalar field into a 2 dimensional square-- eliminating depth? If I know where my observer is Then I turn the cube so he is facing one face... Maybe thats my relativity side comming out :P
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Re: Math Help

Postby Worldie » Tue Aug 09, 2011 9:22 am

The observer has to be perpendicular to the center of a face.
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Re: Math Help

Postby Jeremoot » Tue Aug 09, 2011 9:41 am

While that would work, haha, I'll clarify that the cube is a theoretical representation of the scalar field. We can't pick up the cube and turn it to face our observer (or vice versa for that matter :P ).

The problem lies in that depending on which of the 6 sides of the cube the observer is looking at, the depth could be in any 3 of the dimensions. Furthermore, if the observer is not perpendicular to the center of any face, they will see all 3 dimensions but only a portion of each.

If I'm still not making any sense, close one eye and hold your hand so that the palm is facing you. As you rotate your hand clockwise (or counter-clockwise if you're left-handed), the visible portion of the knife edge will become larger, while the visible portion in the other dimension becomes smaller.
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Re: Math Help

Postby Worldie » Tue Aug 09, 2011 10:02 am

You want to only end up seeing two faces on the sides with a line in the middle?
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Re: Math Help

Postby Torias » Tue Aug 09, 2011 10:04 am

I'm not entirely sure what you're asking. Are you looking for the relationship between the dimensions based on cube position and orientation, relative to the observer?
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Re: Math Help

Postby Jeremoot » Tue Aug 09, 2011 10:27 am

Torias wrote:I'm not entirely sure what you're asking. Are you looking for the relationship between the dimensions based on cube position and orientation, relative to the observer?


Yes. I want to eliminate the depth of the field based on the relationship between the cube and the observer. I'm sorry if I'm not being very clear. Here's another MS Paint diagram.

Image
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Re: Math Help

Postby Astronomic » Tue Aug 09, 2011 12:33 pm

Jeremoot wrote: Knowing the position in space of both the observer and each scalar in the field, how can I flatten a 3 dimensional scalar field into a 2 dimensional square-- eliminating depth?

So you cant move the square cube thing and you cant move the observer.......and you want to remove depth....and you cant use sledgehammers.....im confused.
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Re: Math Help

Postby halabar » Tue Aug 09, 2011 12:50 pm

Assuming the observer can't move, the only way would be to eliminate two of the dimensions... but if you can't eliminate the dimensions, I'm confused as well. If you have a scalar field that is a collection of points, the only other way would be to change the representation. A sphere perhaps?
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Re: Math Help

Postby Jeremoot » Tue Aug 09, 2011 12:54 pm

halabar wrote:Assuming the observer can't move, the only way would be to eliminate two of the dimensions... but if you can't eliminate the dimensions, I'm confused as well. If you have a scalar field that is a collection of points, the only other way would be to change the representation. A sphere perhaps?


Ah! I didn't mean to be this confusing! The observer can move, but we can't decide where they are in space to solve the problem as Astronomic was hinting at.

The answer needs to be a function of the position of both the observer and the collection of points. You need to be able to solve it wherever they are in space at any given time.
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Re: Math Help

Postby Hrobertgar » Tue Aug 09, 2011 1:09 pm

Yes. I want to eliminate the depth of the field based on the relationship between the cube and the observer. I'm sorry if I'm not being very clear. Here's another MS Paint diagram.


It sounds like you are transforming your observer from a person with two eyes who can see depth into a cyclops who cannot. Figure c is the correct answer, had trouble getting the Paint item to copy.

As for some sort of matrix or functional transormation that will convert a cube into figure c, its been a very long time since I've tankgled with that tpye of math.
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Re: Math Help

Postby Astronomic » Tue Aug 09, 2011 1:34 pm

Usually in problems like this dont even think of the observer as human. Its just a single point that can "see".
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