2:1 Diametric rotation matrix for a 2D orthographic projection
$begingroup$
I asked this question in the game dev stack exchange, but didn't get a response. This is more of a math question with an application in game development so I hope it's alright if I ask here.
I'm trying to implement a 2D isometric game, well diametric with a 2:1 pixel ratio. To do this I have a 2D image of a cube I drew with pixel editing software, where for every pixel up, there are two pixels to the right. I'd like to post an image of it but I don't have the reputation for it. It looks something like this. From what I read this is called dimetric projection, and is composed of two axis rotations, one is 45 degrees about the x axis and the other one is arctan(1/2) around the z. Here is some more detail.
I tried to calculate the rotation matrix, but when I use this to convert an (x,y,z) in Cartesian coordinates to isometric world space coordinates(x`,y`,z`) the cube images do not align correctly with each other, leading me to believe I am missing something else to this problem, or my math is wrong. I have found ways online to fudge it, but they all ignore the z axis which I require. I'm not sure if I did my rotation matrix correctly but here it is:
$$
begin{bmatrix}
1/sqrt{2} & 0 & -1/sqrt(2) \
1sqrt{10} & 2/sqrt{5} & -1/sqrt{10} \
-1sqrt{2/5} & 1/sqrt{5} & sqrt{2/5} \
end{bmatrix}
$$
Edit: I can now upload pictures of what I have working, notice how the images are offset. The blue coordinates are the Cartesian positions before the isometric transformations.
linear-algebra matrices linear-transformations rotations
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
$endgroup$
|
show 2 more comments
$begingroup$
I asked this question in the game dev stack exchange, but didn't get a response. This is more of a math question with an application in game development so I hope it's alright if I ask here.
I'm trying to implement a 2D isometric game, well diametric with a 2:1 pixel ratio. To do this I have a 2D image of a cube I drew with pixel editing software, where for every pixel up, there are two pixels to the right. I'd like to post an image of it but I don't have the reputation for it. It looks something like this. From what I read this is called dimetric projection, and is composed of two axis rotations, one is 45 degrees about the x axis and the other one is arctan(1/2) around the z. Here is some more detail.
I tried to calculate the rotation matrix, but when I use this to convert an (x,y,z) in Cartesian coordinates to isometric world space coordinates(x`,y`,z`) the cube images do not align correctly with each other, leading me to believe I am missing something else to this problem, or my math is wrong. I have found ways online to fudge it, but they all ignore the z axis which I require. I'm not sure if I did my rotation matrix correctly but here it is:
$$
begin{bmatrix}
1/sqrt{2} & 0 & -1/sqrt(2) \
1sqrt{10} & 2/sqrt{5} & -1/sqrt{10} \
-1sqrt{2/5} & 1/sqrt{5} & sqrt{2/5} \
end{bmatrix}
$$
Edit: I can now upload pictures of what I have working, notice how the images are offset. The blue coordinates are the Cartesian positions before the isometric transformations.
linear-algebra matrices linear-transformations rotations
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
$endgroup$
$begingroup$
Is your ultimate goal to compute an appropriate projection matrix?
$endgroup$
– amd
Dec 8 '18 at 1:20
$begingroup$
That is correct @amd
$endgroup$
– Ziamor
Dec 8 '18 at 1:26
$begingroup$
Can you first scale so the 2 to 1 pixel complication is removed from the problem? Then reformulate based on that? Are you wanting to rotate the cube about its center? Can you express as two successive rotations (factoring)?
$endgroup$
– John L Winters
Dec 8 '18 at 1:53
$begingroup$
Would you be able to elaborate about the scaling first part? I have enough rep now to upload images so I added a picture to help better visualize my problem. The cube also isn't truly a cube, it's an 2D image in 3d space. @JohnLWinters
$endgroup$
– Ziamor
Dec 8 '18 at 2:34
$begingroup$
Yes I voted for your question. Pretend you have a 1 to 1 pixel instead of a 2 to 1 to simplify and focus on the rotations.
$endgroup$
– John L Winters
Dec 8 '18 at 3:21
|
show 2 more comments
$begingroup$
I asked this question in the game dev stack exchange, but didn't get a response. This is more of a math question with an application in game development so I hope it's alright if I ask here.
I'm trying to implement a 2D isometric game, well diametric with a 2:1 pixel ratio. To do this I have a 2D image of a cube I drew with pixel editing software, where for every pixel up, there are two pixels to the right. I'd like to post an image of it but I don't have the reputation for it. It looks something like this. From what I read this is called dimetric projection, and is composed of two axis rotations, one is 45 degrees about the x axis and the other one is arctan(1/2) around the z. Here is some more detail.
I tried to calculate the rotation matrix, but when I use this to convert an (x,y,z) in Cartesian coordinates to isometric world space coordinates(x`,y`,z`) the cube images do not align correctly with each other, leading me to believe I am missing something else to this problem, or my math is wrong. I have found ways online to fudge it, but they all ignore the z axis which I require. I'm not sure if I did my rotation matrix correctly but here it is:
$$
begin{bmatrix}
1/sqrt{2} & 0 & -1/sqrt(2) \
1sqrt{10} & 2/sqrt{5} & -1/sqrt{10} \
-1sqrt{2/5} & 1/sqrt{5} & sqrt{2/5} \
end{bmatrix}
$$
Edit: I can now upload pictures of what I have working, notice how the images are offset. The blue coordinates are the Cartesian positions before the isometric transformations.
linear-algebra matrices linear-transformations rotations
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
$endgroup$
I asked this question in the game dev stack exchange, but didn't get a response. This is more of a math question with an application in game development so I hope it's alright if I ask here.
I'm trying to implement a 2D isometric game, well diametric with a 2:1 pixel ratio. To do this I have a 2D image of a cube I drew with pixel editing software, where for every pixel up, there are two pixels to the right. I'd like to post an image of it but I don't have the reputation for it. It looks something like this. From what I read this is called dimetric projection, and is composed of two axis rotations, one is 45 degrees about the x axis and the other one is arctan(1/2) around the z. Here is some more detail.
I tried to calculate the rotation matrix, but when I use this to convert an (x,y,z) in Cartesian coordinates to isometric world space coordinates(x`,y`,z`) the cube images do not align correctly with each other, leading me to believe I am missing something else to this problem, or my math is wrong. I have found ways online to fudge it, but they all ignore the z axis which I require. I'm not sure if I did my rotation matrix correctly but here it is:
$$
begin{bmatrix}
1/sqrt{2} & 0 & -1/sqrt(2) \
1sqrt{10} & 2/sqrt{5} & -1/sqrt{10} \
-1sqrt{2/5} & 1/sqrt{5} & sqrt{2/5} \
end{bmatrix}
$$
Edit: I can now upload pictures of what I have working, notice how the images are offset. The blue coordinates are the Cartesian positions before the isometric transformations.
linear-algebra matrices linear-transformations rotations
linear-algebra matrices linear-transformations rotations
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
edited Dec 8 '18 at 2:31
Ziamor
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
asked Dec 8 '18 at 0:05
ZiamorZiamor
133
133
$begingroup$
Is your ultimate goal to compute an appropriate projection matrix?
$endgroup$
– amd
Dec 8 '18 at 1:20
$begingroup$
That is correct @amd
$endgroup$
– Ziamor
Dec 8 '18 at 1:26
$begingroup$
Can you first scale so the 2 to 1 pixel complication is removed from the problem? Then reformulate based on that? Are you wanting to rotate the cube about its center? Can you express as two successive rotations (factoring)?
$endgroup$
– John L Winters
Dec 8 '18 at 1:53
$begingroup$
Would you be able to elaborate about the scaling first part? I have enough rep now to upload images so I added a picture to help better visualize my problem. The cube also isn't truly a cube, it's an 2D image in 3d space. @JohnLWinters
$endgroup$
– Ziamor
Dec 8 '18 at 2:34
$begingroup$
Yes I voted for your question. Pretend you have a 1 to 1 pixel instead of a 2 to 1 to simplify and focus on the rotations.
$endgroup$
– John L Winters
Dec 8 '18 at 3:21
|
show 2 more comments
$begingroup$
Is your ultimate goal to compute an appropriate projection matrix?
$endgroup$
– amd
Dec 8 '18 at 1:20
$begingroup$
That is correct @amd
$endgroup$
– Ziamor
Dec 8 '18 at 1:26
$begingroup$
Can you first scale so the 2 to 1 pixel complication is removed from the problem? Then reformulate based on that? Are you wanting to rotate the cube about its center? Can you express as two successive rotations (factoring)?
$endgroup$
– John L Winters
Dec 8 '18 at 1:53
$begingroup$
Would you be able to elaborate about the scaling first part? I have enough rep now to upload images so I added a picture to help better visualize my problem. The cube also isn't truly a cube, it's an 2D image in 3d space. @JohnLWinters
$endgroup$
– Ziamor
Dec 8 '18 at 2:34
$begingroup$
Yes I voted for your question. Pretend you have a 1 to 1 pixel instead of a 2 to 1 to simplify and focus on the rotations.
$endgroup$
– John L Winters
Dec 8 '18 at 3:21
$begingroup$
Is your ultimate goal to compute an appropriate projection matrix?
$endgroup$
– amd
Dec 8 '18 at 1:20
$begingroup$
Is your ultimate goal to compute an appropriate projection matrix?
$endgroup$
– amd
Dec 8 '18 at 1:20
$begingroup$
That is correct @amd
$endgroup$
– Ziamor
Dec 8 '18 at 1:26
$begingroup$
That is correct @amd
$endgroup$
– Ziamor
Dec 8 '18 at 1:26
$begingroup$
Can you first scale so the 2 to 1 pixel complication is removed from the problem? Then reformulate based on that? Are you wanting to rotate the cube about its center? Can you express as two successive rotations (factoring)?
$endgroup$
– John L Winters
Dec 8 '18 at 1:53
$begingroup$
Can you first scale so the 2 to 1 pixel complication is removed from the problem? Then reformulate based on that? Are you wanting to rotate the cube about its center? Can you express as two successive rotations (factoring)?
$endgroup$
– John L Winters
Dec 8 '18 at 1:53
$begingroup$
Would you be able to elaborate about the scaling first part? I have enough rep now to upload images so I added a picture to help better visualize my problem. The cube also isn't truly a cube, it's an 2D image in 3d space. @JohnLWinters
$endgroup$
– Ziamor
Dec 8 '18 at 2:34
$begingroup$
Would you be able to elaborate about the scaling first part? I have enough rep now to upload images so I added a picture to help better visualize my problem. The cube also isn't truly a cube, it's an 2D image in 3d space. @JohnLWinters
$endgroup$
– Ziamor
Dec 8 '18 at 2:34
$begingroup$
Yes I voted for your question. Pretend you have a 1 to 1 pixel instead of a 2 to 1 to simplify and focus on the rotations.
$endgroup$
– John L Winters
Dec 8 '18 at 3:21
$begingroup$
Yes I voted for your question. Pretend you have a 1 to 1 pixel instead of a 2 to 1 to simplify and focus on the rotations.
$endgroup$
– John L Winters
Dec 8 '18 at 3:21
|
show 2 more comments
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$begingroup$
Is your ultimate goal to compute an appropriate projection matrix?
$endgroup$
– amd
Dec 8 '18 at 1:20
$begingroup$
That is correct @amd
$endgroup$
– Ziamor
Dec 8 '18 at 1:26
$begingroup$
Can you first scale so the 2 to 1 pixel complication is removed from the problem? Then reformulate based on that? Are you wanting to rotate the cube about its center? Can you express as two successive rotations (factoring)?
$endgroup$
– John L Winters
Dec 8 '18 at 1:53
$begingroup$
Would you be able to elaborate about the scaling first part? I have enough rep now to upload images so I added a picture to help better visualize my problem. The cube also isn't truly a cube, it's an 2D image in 3d space. @JohnLWinters
$endgroup$
– Ziamor
Dec 8 '18 at 2:34
$begingroup$
Yes I voted for your question. Pretend you have a 1 to 1 pixel instead of a 2 to 1 to simplify and focus on the rotations.
$endgroup$
– John L Winters
Dec 8 '18 at 3:21