Mixing
How to convert a 3D line into a 2D line?
$begingroup$
I wanted to know if there is an algorithm I could use to show a 3D line on a 2D screen. The line is made up of 2 points with X, Y, Z coordinates.
My end goal is to have a bunch of lines that may make up something like cubes, then to be able to move a camera around to view these cubes. The code below is what I have so far.
type Point =
{ X : int
Y : int
Z : int }
type Line =
{ A : Point
B : Point }
let line =
{ A = { X = 0; Y = 0; Z = 0 }
B = { X = 50; Y = 50; Z = 50 } }
geometry coordinate-systems
asked Jan 31 at 22:30
AlanayAlanay
1104
$endgroup$
add a comment |
$begingroup$
I wanted to know if there is an algorithm I could use to show a 3D line on a 2D screen. The line is made up of 2 points with X, Y, Z coordinates.
My end goal is to have a bunch of lines that may make up something like cubes, then to be able to move a camera around to view these cubes. The code below is what I have so far.
type Point =
{ X : int
Y : int
Z : int }
type Line =
{ A : Point
B : Point }
let line =
{ A = { X = 0; Y = 0; Z = 0 }
B = { X = 50; Y = 50; Z = 50 } }
geometry coordinate-systems
asked Jan 31 at 22:30
AlanayAlanay
1104
$endgroup$
$begingroup$
Any answer to your question would necessitate a course in 3D rendering beginning by the concept of camera model... This is almost a whole domain in itself.
$endgroup$
– Jean Marie
Feb 1 at 0:05
$begingroup$
Here is a part of such lecture notes : songho.ca/opengl/gl_projectionmatrix.html
$endgroup$
– Jean Marie
Feb 1 at 20:32
add a comment |
$begingroup$
I wanted to know if there is an algorithm I could use to show a 3D line on a 2D screen. The line is made up of 2 points with X, Y, Z coordinates.
My end goal is to have a bunch of lines that may make up something like cubes, then to be able to move a camera around to view these cubes. The code below is what I have so far.
type Point =
{ X : int
Y : int
Z : int }
type Line =
{ A : Point
B : Point }
let line =
{ A = { X = 0; Y = 0; Z = 0 }
B = { X = 50; Y = 50; Z = 50 } }
geometry coordinate-systems
asked Jan 31 at 22:30
AlanayAlanay
1104
$endgroup$
I wanted to know if there is an algorithm I could use to show a 3D line on a 2D screen. The line is made up of 2 points with X, Y, Z coordinates.
My end goal is to have a bunch of lines that may make up something like cubes, then to be able to move a camera around to view these cubes. The code below is what I have so far.
type Point =
{ X : int
Y : int
Z : int }
type Line =
{ A : Point
B : Point }
let line =
{ A = { X = 0; Y = 0; Z = 0 }
B = { X = 50; Y = 50; Z = 50 } }
geometry coordinate-systems
geometry coordinate-systems
asked Jan 31 at 22:30
AlanayAlanay
1104
asked Jan 31 at 22:30
AlanayAlanay
1104
asked Jan 31 at 22:30
AlanayAlanay
1104
asked Jan 31 at 22:30
AlanayAlanay
1104
asked Jan 31 at 22:30
AlanayAlanay
1104
1104
$begingroup$
Any answer to your question would necessitate a course in 3D rendering beginning by the concept of camera model... This is almost a whole domain in itself.
$endgroup$
– Jean Marie
Feb 1 at 0:05
$begingroup$
Here is a part of such lecture notes : songho.ca/opengl/gl_projectionmatrix.html
$endgroup$
– Jean Marie
Feb 1 at 20:32
add a comment |
$begingroup$
Any answer to your question would necessitate a course in 3D rendering beginning by the concept of camera model... This is almost a whole domain in itself.
$endgroup$
– Jean Marie
Feb 1 at 0:05
$begingroup$
Here is a part of such lecture notes : songho.ca/opengl/gl_projectionmatrix.html
$endgroup$
– Jean Marie
Feb 1 at 20:32
$begingroup$
Any answer to your question would necessitate a course in 3D rendering beginning by the concept of camera model... This is almost a whole domain in itself.
$endgroup$
– Jean Marie
Feb 1 at 0:05
$begingroup$
Any answer to your question would necessitate a course in 3D rendering beginning by the concept of camera model... This is almost a whole domain in itself.
$endgroup$
– Jean Marie
Feb 1 at 0:05
$begingroup$
Here is a part of such lecture notes : songho.ca/opengl/gl_projectionmatrix.html
$endgroup$
– Jean Marie
Feb 1 at 20:32
$begingroup$
Here is a part of such lecture notes : songho.ca/opengl/gl_projectionmatrix.html
$endgroup$
– Jean Marie
Feb 1 at 20:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Draw an isometric box with faint lines and then connect the front left-hand lower corner to the rear right-hand upper corner with a strong line.
Additional post :
I used a land survey system and calculated 2D N,E coordinates of an isometric box as
Point A, N,E of 0 , 0
Point B, N,E of -25.00 , 43.3013
Point C, N,E of 0.00 , 86.6025
Point D, N,E of 50.00 , 86.6025
Then the 2D direction and distance of the required line set in perspective is 60 degrees azimuth at a distance of 100 between Point A and Point D. The true distance of the 3D line would be the Square Root of (50^2 + 50^2 + 50^2) or 86.60 .
answered Feb 1 at 0:59
S SpringS Spring
1874
$endgroup$
add a comment |
$begingroup$
What you want from there? Projections, to take that 3-dimensional point data to a plane to put on screen. To project a line, of course, we project the two points we used to define it.
Of course, the best solution? Find a standard library that implements this stuff already. In Asymptote (a language specializing in graphics, that I use when I want to create a mathematical diagram), that's the "three" module. Over 3000 lines of code, over 150 instances of "projection".
answered Feb 1 at 0:13
jmerryjmerry
17k11633
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Draw an isometric box with faint lines and then connect the front left-hand lower corner to the rear right-hand upper corner with a strong line.
Additional post :
I used a land survey system and calculated 2D N,E coordinates of an isometric box as
Point A, N,E of 0 , 0
Point B, N,E of -25.00 , 43.3013
Point C, N,E of 0.00 , 86.6025
Point D, N,E of 50.00 , 86.6025
Then the 2D direction and distance of the required line set in perspective is 60 degrees azimuth at a distance of 100 between Point A and Point D. The true distance of the 3D line would be the Square Root of (50^2 + 50^2 + 50^2) or 86.60 .
answered Feb 1 at 0:59
S SpringS Spring
1874
$endgroup$
add a comment |
$begingroup$
Draw an isometric box with faint lines and then connect the front left-hand lower corner to the rear right-hand upper corner with a strong line.
Additional post :
I used a land survey system and calculated 2D N,E coordinates of an isometric box as
Point A, N,E of 0 , 0
Point B, N,E of -25.00 , 43.3013
Point C, N,E of 0.00 , 86.6025
Point D, N,E of 50.00 , 86.6025
Then the 2D direction and distance of the required line set in perspective is 60 degrees azimuth at a distance of 100 between Point A and Point D. The true distance of the 3D line would be the Square Root of (50^2 + 50^2 + 50^2) or 86.60 .
answered Feb 1 at 0:59
S SpringS Spring
1874
$endgroup$
add a comment |
$begingroup$
Draw an isometric box with faint lines and then connect the front left-hand lower corner to the rear right-hand upper corner with a strong line.
Additional post :
I used a land survey system and calculated 2D N,E coordinates of an isometric box as
Point A, N,E of 0 , 0
Point B, N,E of -25.00 , 43.3013
Point C, N,E of 0.00 , 86.6025
Point D, N,E of 50.00 , 86.6025
Then the 2D direction and distance of the required line set in perspective is 60 degrees azimuth at a distance of 100 between Point A and Point D. The true distance of the 3D line would be the Square Root of (50^2 + 50^2 + 50^2) or 86.60 .
answered Feb 1 at 0:59
S SpringS Spring
1874
$endgroup$
Draw an isometric box with faint lines and then connect the front left-hand lower corner to the rear right-hand upper corner with a strong line.
Additional post :
I used a land survey system and calculated 2D N,E coordinates of an isometric box as
Point A, N,E of 0 , 0
Point B, N,E of -25.00 , 43.3013
Point C, N,E of 0.00 , 86.6025
Point D, N,E of 50.00 , 86.6025
Then the 2D direction and distance of the required line set in perspective is 60 degrees azimuth at a distance of 100 between Point A and Point D. The true distance of the 3D line would be the Square Root of (50^2 + 50^2 + 50^2) or 86.60 .
answered Feb 1 at 0:59
S SpringS Spring
1874
edited Feb 1 at 21:06
answered Feb 1 at 0:59
S SpringS Spring
1874
answered Feb 1 at 0:59
S SpringS Spring
1874
answered Feb 1 at 0:59
S SpringS Spring
1874
1874
add a comment |
add a comment |
$begingroup$
What you want from there? Projections, to take that 3-dimensional point data to a plane to put on screen. To project a line, of course, we project the two points we used to define it.
Of course, the best solution? Find a standard library that implements this stuff already. In Asymptote (a language specializing in graphics, that I use when I want to create a mathematical diagram), that's the "three" module. Over 3000 lines of code, over 150 instances of "projection".
answered Feb 1 at 0:13
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
What you want from there? Projections, to take that 3-dimensional point data to a plane to put on screen. To project a line, of course, we project the two points we used to define it.
Of course, the best solution? Find a standard library that implements this stuff already. In Asymptote (a language specializing in graphics, that I use when I want to create a mathematical diagram), that's the "three" module. Over 3000 lines of code, over 150 instances of "projection".
answered Feb 1 at 0:13
jmerryjmerry
17k11633
$endgroup$
add a comment |
$begingroup$
What you want from there? Projections, to take that 3-dimensional point data to a plane to put on screen. To project a line, of course, we project the two points we used to define it.
Of course, the best solution? Find a standard library that implements this stuff already. In Asymptote (a language specializing in graphics, that I use when I want to create a mathematical diagram), that's the "three" module. Over 3000 lines of code, over 150 instances of "projection".
answered Feb 1 at 0:13
jmerryjmerry
17k11633
$endgroup$
What you want from there? Projections, to take that 3-dimensional point data to a plane to put on screen. To project a line, of course, we project the two points we used to define it.
Of course, the best solution? Find a standard library that implements this stuff already. In Asymptote (a language specializing in graphics, that I use when I want to create a mathematical diagram), that's the "three" module. Over 3000 lines of code, over 150 instances of "projection".
answered Feb 1 at 0:13
jmerryjmerry
17k11633
answered Feb 1 at 0:13
jmerryjmerry
17k11633
answered Feb 1 at 0:13
jmerryjmerry
17k11633
answered Feb 1 at 0:13
jmerryjmerry
17k11633
17k11633
add a comment |
add a comment |
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$begingroup$
Any answer to your question would necessitate a course in 3D rendering beginning by the concept of camera model... This is almost a whole domain in itself.
$endgroup$
– Jean Marie
Feb 1 at 0:05
$begingroup$
Here is a part of such lecture notes : songho.ca/opengl/gl_projectionmatrix.html
$endgroup$
– Jean Marie
Feb 1 at 20:32