Mixing
Parallel/Offset 3D curve
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I'm trying to understand the logic (theory?) behind the offset/parallel curve for the design of a radial diffuser.
I found the equations for a 2D parametrical offset curve on wiki, where the normal unit vector to the original curve is defined as the first derivative of the original curve.
However, in this article , the normal vector is defined as a second drivative of the original curve.
What I'm trying to achieve is: I would like to define the equation of the offset curve and then play around a little bit to make it "progressive" compared to the original curve.
My original curve is:
$a_3=11.482$
$b_3=-10.629$
$c_3=0$
$r_3=81.109$
$x=r_3cos(t)+a_3$
$y=r_3sin(t)+b_3$
$z=5t(20^t)+c_3$
Edit: the offset of the curve is in the z direction
curves
asked Jul 31 at 10:34
user2882635
62
add a comment |Â
up vote
1
down vote
favorite
I'm trying to understand the logic (theory?) behind the offset/parallel curve for the design of a radial diffuser.
I found the equations for a 2D parametrical offset curve on wiki, where the normal unit vector to the original curve is defined as the first derivative of the original curve.
However, in this article , the normal vector is defined as a second drivative of the original curve.
What I'm trying to achieve is: I would like to define the equation of the offset curve and then play around a little bit to make it "progressive" compared to the original curve.
My original curve is:
$a_3=11.482$
$b_3=-10.629$
$c_3=0$
$r_3=81.109$
$x=r_3cos(t)+a_3$
$y=r_3sin(t)+b_3$
$z=5t(20^t)+c_3$
Edit: the offset of the curve is in the z direction
curves
asked Jul 31 at 10:34
user2882635
62
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 31 at 10:36
In the $mathbbR^3$ case the offset curve has not a precise meaning unless this curve is onto a surface and the offset to the curve is taken on this surface.
– Cesareo
Jul 31 at 10:55
I should be more precise, the offset of the curve is in the direction of the z component. I will update the OP.
– user2882635
Jul 31 at 11:07
after some google-fuing, I found this: researchgate.net/publication/…
– user2882635
Jul 31 at 12:19
the paper from the link in the comment above describes the my problem perfectly, but the equation they provide ( r_d(t) = r(t) + d*n(t) ), just doesn't deliver the desired offset curve to my original curve.
– user2882635
Aug 1 at 8:51
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to understand the logic (theory?) behind the offset/parallel curve for the design of a radial diffuser.
I found the equations for a 2D parametrical offset curve on wiki, where the normal unit vector to the original curve is defined as the first derivative of the original curve.
However, in this article , the normal vector is defined as a second drivative of the original curve.
What I'm trying to achieve is: I would like to define the equation of the offset curve and then play around a little bit to make it "progressive" compared to the original curve.
My original curve is:
$a_3=11.482$
$b_3=-10.629$
$c_3=0$
$r_3=81.109$
$x=r_3cos(t)+a_3$
$y=r_3sin(t)+b_3$
$z=5t(20^t)+c_3$
Edit: the offset of the curve is in the z direction
curves
asked Jul 31 at 10:34
user2882635
62
I'm trying to understand the logic (theory?) behind the offset/parallel curve for the design of a radial diffuser.
I found the equations for a 2D parametrical offset curve on wiki, where the normal unit vector to the original curve is defined as the first derivative of the original curve.
However, in this article , the normal vector is defined as a second drivative of the original curve.
What I'm trying to achieve is: I would like to define the equation of the offset curve and then play around a little bit to make it "progressive" compared to the original curve.
My original curve is:
$a_3=11.482$
$b_3=-10.629$
$c_3=0$
$r_3=81.109$
$x=r_3cos(t)+a_3$
$y=r_3sin(t)+b_3$
$z=5t(20^t)+c_3$
Edit: the offset of the curve is in the z direction
curves
asked Jul 31 at 10:34
user2882635
62
edited Jul 31 at 11:08
asked Jul 31 at 10:34
user2882635
62
asked Jul 31 at 10:34
user2882635
62
asked Jul 31 at 10:34
user2882635
62
62
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 31 at 10:36
In the $mathbbR^3$ case the offset curve has not a precise meaning unless this curve is onto a surface and the offset to the curve is taken on this surface.
– Cesareo
Jul 31 at 10:55
I should be more precise, the offset of the curve is in the direction of the z component. I will update the OP.
– user2882635
Jul 31 at 11:07
after some google-fuing, I found this: researchgate.net/publication/…
– user2882635
Jul 31 at 12:19
the paper from the link in the comment above describes the my problem perfectly, but the equation they provide ( r_d(t) = r(t) + d*n(t) ), just doesn't deliver the desired offset curve to my original curve.
– user2882635
Aug 1 at 8:51
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 31 at 10:36
In the $mathbbR^3$ case the offset curve has not a precise meaning unless this curve is onto a surface and the offset to the curve is taken on this surface.
– Cesareo
Jul 31 at 10:55
I should be more precise, the offset of the curve is in the direction of the z component. I will update the OP.
– user2882635
Jul 31 at 11:07
after some google-fuing, I found this: researchgate.net/publication/…
– user2882635
Jul 31 at 12:19
the paper from the link in the comment above describes the my problem perfectly, but the equation they provide ( r_d(t) = r(t) + d*n(t) ), just doesn't deliver the desired offset curve to my original curve.
– user2882635
Aug 1 at 8:51
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 31 at 10:36
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 31 at 10:36
In the $mathbbR^3$ case the offset curve has not a precise meaning unless this curve is onto a surface and the offset to the curve is taken on this surface.
– Cesareo
Jul 31 at 10:55
In the $mathbbR^3$ case the offset curve has not a precise meaning unless this curve is onto a surface and the offset to the curve is taken on this surface.
– Cesareo
Jul 31 at 10:55
I should be more precise, the offset of the curve is in the direction of the z component. I will update the OP.
– user2882635
Jul 31 at 11:07
I should be more precise, the offset of the curve is in the direction of the z component. I will update the OP.
– user2882635
Jul 31 at 11:07
after some google-fuing, I found this: researchgate.net/publication/…
– user2882635
Jul 31 at 12:19
after some google-fuing, I found this: researchgate.net/publication/…
– user2882635
Jul 31 at 12:19
the paper from the link in the comment above describes the my problem perfectly, but the equation they provide ( r_d(t) = r(t) + d*n(t) ), just doesn't deliver the desired offset curve to my original curve.
– user2882635
Aug 1 at 8:51
the paper from the link in the comment above describes the my problem perfectly, but the equation they provide ( r_d(t) = r(t) + d*n(t) ), just doesn't deliver the desired offset curve to my original curve.
– user2882635
Aug 1 at 8:51
add a comment |Â
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Jul 31 at 10:36
In the $mathbbR^3$ case the offset curve has not a precise meaning unless this curve is onto a surface and the offset to the curve is taken on this surface.
– Cesareo
Jul 31 at 10:55
I should be more precise, the offset of the curve is in the direction of the z component. I will update the OP.
– user2882635
Jul 31 at 11:07
after some google-fuing, I found this: researchgate.net/publication/…
– user2882635
Jul 31 at 12:19
the paper from the link in the comment above describes the my problem perfectly, but the equation they provide ( r_d(t) = r(t) + d*n(t) ), just doesn't deliver the desired offset curve to my original curve.
– user2882635
Aug 1 at 8:51