Describe the shape of projection of vertices (vector positions of a cube) onto a 2D plane from a source...
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I am having trouble with this. I can manually calculate every single projection point onto the z=0 plane from deriving vector equations to get to the z plane for each vertices. From this I can then plot all these coordinates to find the shape. Is there a better/faster method as this question shouldn't take long.
Many thanks
The question for context is here
Here is my method
linear-algebra geometry vectors mathematical-physics plane-geometry
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
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add a comment |
$begingroup$
I am having trouble with this. I can manually calculate every single projection point onto the z=0 plane from deriving vector equations to get to the z plane for each vertices. From this I can then plot all these coordinates to find the shape. Is there a better/faster method as this question shouldn't take long.
Many thanks
The question for context is here
Here is my method
linear-algebra geometry vectors mathematical-physics plane-geometry
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
$endgroup$
$begingroup$
The fact that this is a cube parallel to the projection plane gives it some simplifying properties. But you still have to compute some points. If you wrote out your own methods in detail for at least one point, perhaps someone would notice where you could reduce the effort.
$endgroup$
– David K
Dec 13 '18 at 15:11
$begingroup$
Hi David, I have updated the post with my method shown. I create vectors between the light source and each verticy and then extrapolate them to the plane z=0 to get the points of projection.
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:25
$begingroup$
I probably should have mentioned that people here tend not to like reading the math from photographs, especially if the original is handwritten. You can look at other questions that were highly rated and got good answers to see how people format questions to get those kinds of answers. You may decide it's not worth the effort. If you want to continue, step 1 is to learn some MathJax: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 13 '18 at 19:33
add a comment |
$begingroup$
I am having trouble with this. I can manually calculate every single projection point onto the z=0 plane from deriving vector equations to get to the z plane for each vertices. From this I can then plot all these coordinates to find the shape. Is there a better/faster method as this question shouldn't take long.
Many thanks
The question for context is here
Here is my method
linear-algebra geometry vectors mathematical-physics plane-geometry
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
$endgroup$
I am having trouble with this. I can manually calculate every single projection point onto the z=0 plane from deriving vector equations to get to the z plane for each vertices. From this I can then plot all these coordinates to find the shape. Is there a better/faster method as this question shouldn't take long.
Many thanks
The question for context is here
Here is my method
linear-algebra geometry vectors mathematical-physics plane-geometry
linear-algebra geometry vectors mathematical-physics plane-geometry
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
edited Dec 13 '18 at 19:23
Adam Suttle
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
asked Dec 13 '18 at 13:28
Adam SuttleAdam Suttle
11
11
$begingroup$
The fact that this is a cube parallel to the projection plane gives it some simplifying properties. But you still have to compute some points. If you wrote out your own methods in detail for at least one point, perhaps someone would notice where you could reduce the effort.
$endgroup$
– David K
Dec 13 '18 at 15:11
$begingroup$
Hi David, I have updated the post with my method shown. I create vectors between the light source and each verticy and then extrapolate them to the plane z=0 to get the points of projection.
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:25
$begingroup$
I probably should have mentioned that people here tend not to like reading the math from photographs, especially if the original is handwritten. You can look at other questions that were highly rated and got good answers to see how people format questions to get those kinds of answers. You may decide it's not worth the effort. If you want to continue, step 1 is to learn some MathJax: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 13 '18 at 19:33
add a comment |
$begingroup$
The fact that this is a cube parallel to the projection plane gives it some simplifying properties. But you still have to compute some points. If you wrote out your own methods in detail for at least one point, perhaps someone would notice where you could reduce the effort.
$endgroup$
– David K
Dec 13 '18 at 15:11
$begingroup$
Hi David, I have updated the post with my method shown. I create vectors between the light source and each verticy and then extrapolate them to the plane z=0 to get the points of projection.
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:25
$begingroup$
I probably should have mentioned that people here tend not to like reading the math from photographs, especially if the original is handwritten. You can look at other questions that were highly rated and got good answers to see how people format questions to get those kinds of answers. You may decide it's not worth the effort. If you want to continue, step 1 is to learn some MathJax: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 13 '18 at 19:33
$begingroup$
The fact that this is a cube parallel to the projection plane gives it some simplifying properties. But you still have to compute some points. If you wrote out your own methods in detail for at least one point, perhaps someone would notice where you could reduce the effort.
$endgroup$
– David K
Dec 13 '18 at 15:11
$begingroup$
The fact that this is a cube parallel to the projection plane gives it some simplifying properties. But you still have to compute some points. If you wrote out your own methods in detail for at least one point, perhaps someone would notice where you could reduce the effort.
$endgroup$
– David K
Dec 13 '18 at 15:11
$begingroup$
Hi David, I have updated the post with my method shown. I create vectors between the light source and each verticy and then extrapolate them to the plane z=0 to get the points of projection.
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:25
$begingroup$
Hi David, I have updated the post with my method shown. I create vectors between the light source and each verticy and then extrapolate them to the plane z=0 to get the points of projection.
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:25
$begingroup$
I probably should have mentioned that people here tend not to like reading the math from photographs, especially if the original is handwritten. You can look at other questions that were highly rated and got good answers to see how people format questions to get those kinds of answers. You may decide it's not worth the effort. If you want to continue, step 1 is to learn some MathJax: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 13 '18 at 19:33
$begingroup$
I probably should have mentioned that people here tend not to like reading the math from photographs, especially if the original is handwritten. You can look at other questions that were highly rated and got good answers to see how people format questions to get those kinds of answers. You may decide it's not worth the effort. If you want to continue, step 1 is to learn some MathJax: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 13 '18 at 19:33
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
You should get a hexagon a rectangle or a square.
$(frac 14, frac 54), (frac 14, frac 94),(frac 54, frac 54),(frac 54, frac 94), (-frac 12, frac 32),(-frac 12, frac 52),(frac 32, frac 52),(frac 12, frac 32)$
How do you get these.
$P_n - lambda L$ and find $lambda$ such that the $z$ component equals $0.$
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should get a hexagon a rectangle or a square.
$(frac 14, frac 54), (frac 14, frac 94),(frac 54, frac 54),(frac 54, frac 94), (-frac 12, frac 32),(-frac 12, frac 52),(frac 32, frac 52),(frac 12, frac 32)$
How do you get these.
$P_n - lambda L$ and find $lambda$ such that the $z$ component equals $0.$
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
add a comment |
$begingroup$
You should get a hexagon a rectangle or a square.
$(frac 14, frac 54), (frac 14, frac 94),(frac 54, frac 54),(frac 54, frac 94), (-frac 12, frac 32),(-frac 12, frac 52),(frac 32, frac 52),(frac 12, frac 32)$
How do you get these.
$P_n - lambda L$ and find $lambda$ such that the $z$ component equals $0.$
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
$endgroup$
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
add a comment |
$begingroup$
You should get a hexagon a rectangle or a square.
$(frac 14, frac 54), (frac 14, frac 94),(frac 54, frac 54),(frac 54, frac 94), (-frac 12, frac 32),(-frac 12, frac 52),(frac 32, frac 52),(frac 12, frac 32)$
How do you get these.
$P_n - lambda L$ and find $lambda$ such that the $z$ component equals $0.$
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
$endgroup$
You should get a hexagon a rectangle or a square.
$(frac 14, frac 54), (frac 14, frac 94),(frac 54, frac 54),(frac 54, frac 94), (-frac 12, frac 32),(-frac 12, frac 52),(frac 32, frac 52),(frac 12, frac 32)$
How do you get these.
$P_n - lambda L$ and find $lambda$ such that the $z$ component equals $0.$
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
edited Dec 13 '18 at 19:55
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
answered Dec 13 '18 at 19:31
Doug MDoug M
44.8k31854
44.8k31854
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
add a comment |
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
I ended up with a pattern of imgur.com/a/XG2rsX9
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:37
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
$begingroup$
That doesn't look like a cube.
$endgroup$
– Doug M
Dec 13 '18 at 19:50
add a comment |
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$begingroup$
The fact that this is a cube parallel to the projection plane gives it some simplifying properties. But you still have to compute some points. If you wrote out your own methods in detail for at least one point, perhaps someone would notice where you could reduce the effort.
$endgroup$
– David K
Dec 13 '18 at 15:11
$begingroup$
Hi David, I have updated the post with my method shown. I create vectors between the light source and each verticy and then extrapolate them to the plane z=0 to get the points of projection.
$endgroup$
– Adam Suttle
Dec 13 '18 at 19:25
$begingroup$
I probably should have mentioned that people here tend not to like reading the math from photographs, especially if the original is handwritten. You can look at other questions that were highly rated and got good answers to see how people format questions to get those kinds of answers. You may decide it's not worth the effort. If you want to continue, step 1 is to learn some MathJax: math.stackexchange.com/help/notation
$endgroup$
– David K
Dec 13 '18 at 19:33