Isometric Drawing Tool: Converting 2D information to 3D The Next CEO of Stack...
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Isometric Drawing Tool: Converting 2D information to 3D
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$begingroup$
I was drawing with the NCTM Isometric Drawing Tool, and produced the image seen here. I also noticed that it is possible to view the isometric drawing in 2D, and was wondering if/how it is possible to generate the 3D-coordinates of the blocks by using the information given in 2D.
geometry coordinate-systems
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
$endgroup$
add a comment |
$begingroup$
I was drawing with the NCTM Isometric Drawing Tool, and produced the image seen here. I also noticed that it is possible to view the isometric drawing in 2D, and was wondering if/how it is possible to generate the 3D-coordinates of the blocks by using the information given in 2D.
geometry coordinate-systems
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
$endgroup$
add a comment |
$begingroup$
I was drawing with the NCTM Isometric Drawing Tool, and produced the image seen here. I also noticed that it is possible to view the isometric drawing in 2D, and was wondering if/how it is possible to generate the 3D-coordinates of the blocks by using the information given in 2D.
geometry coordinate-systems
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
$endgroup$
I was drawing with the NCTM Isometric Drawing Tool, and produced the image seen here. I also noticed that it is possible to view the isometric drawing in 2D, and was wondering if/how it is possible to generate the 3D-coordinates of the blocks by using the information given in 2D.
geometry coordinate-systems
geometry coordinate-systems
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
asked Mar 17 at 17:40
RuffWarriorRuffWarrior
11
11
add a comment |
add a comment |
1 Answer
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$begingroup$
Isometric projection, in this case, is the orthogonal projection on plane with equation $x+y+z=0$ (or any affine plane with equation $x+y+z=d$). Thus
A first answer is 'no' because depth information (depth in the direction of view, i.e., in the $(1,1,1)$ direction) is lost ; in the example you give, one could interpret the upper corner of the highest cube as having coordinates $(0,-1,5)$ but they could as well be
$$(0,-1,5)+t(1,1,1)=(t,t-1,t+5)$$
for any real number $t$...
A second answer is 'yes' as long as we have the absolute coordinates $(a,b,c)$ of a certain reference point $R$. Indeed in this case, one can retrieve the absolute coordinates of any point $S$ by first creating a path from (the projection) of $R$ to (the projection of) $S$ and then, while following the path, add $(pm 1,0,0), (0,pm 1,0), (0,0,pm 1) $ to the current coordinates as we are moving South-East, North-West, South-West, North-East, South or North direction resp.
Remarks :
1) all paths from $R$ to $S$ give the same final coordinates to $S$.
2) thinking to reference (affine) plane with equation $x+y+z=1$, one could work with barycentric coordinates. I can even imagine that this "Isometric Drawing Tool" has been programmed using sort of coordinates. I will try to re-create it in this way in the coming times...
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
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1 Answer
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1 Answer
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votes
$begingroup$
Isometric projection, in this case, is the orthogonal projection on plane with equation $x+y+z=0$ (or any affine plane with equation $x+y+z=d$). Thus
A first answer is 'no' because depth information (depth in the direction of view, i.e., in the $(1,1,1)$ direction) is lost ; in the example you give, one could interpret the upper corner of the highest cube as having coordinates $(0,-1,5)$ but they could as well be
$$(0,-1,5)+t(1,1,1)=(t,t-1,t+5)$$
for any real number $t$...
A second answer is 'yes' as long as we have the absolute coordinates $(a,b,c)$ of a certain reference point $R$. Indeed in this case, one can retrieve the absolute coordinates of any point $S$ by first creating a path from (the projection) of $R$ to (the projection of) $S$ and then, while following the path, add $(pm 1,0,0), (0,pm 1,0), (0,0,pm 1) $ to the current coordinates as we are moving South-East, North-West, South-West, North-East, South or North direction resp.
Remarks :
1) all paths from $R$ to $S$ give the same final coordinates to $S$.
2) thinking to reference (affine) plane with equation $x+y+z=1$, one could work with barycentric coordinates. I can even imagine that this "Isometric Drawing Tool" has been programmed using sort of coordinates. I will try to re-create it in this way in the coming times...
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
$endgroup$
add a comment |
$begingroup$
Isometric projection, in this case, is the orthogonal projection on plane with equation $x+y+z=0$ (or any affine plane with equation $x+y+z=d$). Thus
A first answer is 'no' because depth information (depth in the direction of view, i.e., in the $(1,1,1)$ direction) is lost ; in the example you give, one could interpret the upper corner of the highest cube as having coordinates $(0,-1,5)$ but they could as well be
$$(0,-1,5)+t(1,1,1)=(t,t-1,t+5)$$
for any real number $t$...
A second answer is 'yes' as long as we have the absolute coordinates $(a,b,c)$ of a certain reference point $R$. Indeed in this case, one can retrieve the absolute coordinates of any point $S$ by first creating a path from (the projection) of $R$ to (the projection of) $S$ and then, while following the path, add $(pm 1,0,0), (0,pm 1,0), (0,0,pm 1) $ to the current coordinates as we are moving South-East, North-West, South-West, North-East, South or North direction resp.
Remarks :
1) all paths from $R$ to $S$ give the same final coordinates to $S$.
2) thinking to reference (affine) plane with equation $x+y+z=1$, one could work with barycentric coordinates. I can even imagine that this "Isometric Drawing Tool" has been programmed using sort of coordinates. I will try to re-create it in this way in the coming times...
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
$endgroup$
add a comment |
$begingroup$
Isometric projection, in this case, is the orthogonal projection on plane with equation $x+y+z=0$ (or any affine plane with equation $x+y+z=d$). Thus
A first answer is 'no' because depth information (depth in the direction of view, i.e., in the $(1,1,1)$ direction) is lost ; in the example you give, one could interpret the upper corner of the highest cube as having coordinates $(0,-1,5)$ but they could as well be
$$(0,-1,5)+t(1,1,1)=(t,t-1,t+5)$$
for any real number $t$...
A second answer is 'yes' as long as we have the absolute coordinates $(a,b,c)$ of a certain reference point $R$. Indeed in this case, one can retrieve the absolute coordinates of any point $S$ by first creating a path from (the projection) of $R$ to (the projection of) $S$ and then, while following the path, add $(pm 1,0,0), (0,pm 1,0), (0,0,pm 1) $ to the current coordinates as we are moving South-East, North-West, South-West, North-East, South or North direction resp.
Remarks :
1) all paths from $R$ to $S$ give the same final coordinates to $S$.
2) thinking to reference (affine) plane with equation $x+y+z=1$, one could work with barycentric coordinates. I can even imagine that this "Isometric Drawing Tool" has been programmed using sort of coordinates. I will try to re-create it in this way in the coming times...
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
$endgroup$
Isometric projection, in this case, is the orthogonal projection on plane with equation $x+y+z=0$ (or any affine plane with equation $x+y+z=d$). Thus
A first answer is 'no' because depth information (depth in the direction of view, i.e., in the $(1,1,1)$ direction) is lost ; in the example you give, one could interpret the upper corner of the highest cube as having coordinates $(0,-1,5)$ but they could as well be
$$(0,-1,5)+t(1,1,1)=(t,t-1,t+5)$$
for any real number $t$...
A second answer is 'yes' as long as we have the absolute coordinates $(a,b,c)$ of a certain reference point $R$. Indeed in this case, one can retrieve the absolute coordinates of any point $S$ by first creating a path from (the projection) of $R$ to (the projection of) $S$ and then, while following the path, add $(pm 1,0,0), (0,pm 1,0), (0,0,pm 1) $ to the current coordinates as we are moving South-East, North-West, South-West, North-East, South or North direction resp.
Remarks :
1) all paths from $R$ to $S$ give the same final coordinates to $S$.
2) thinking to reference (affine) plane with equation $x+y+z=1$, one could work with barycentric coordinates. I can even imagine that this "Isometric Drawing Tool" has been programmed using sort of coordinates. I will try to re-create it in this way in the coming times...
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
edited Mar 19 at 21:37
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
answered Mar 19 at 9:18
Jean MarieJean Marie
31.1k42255
31.1k42255
add a comment |
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