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Formula for the offset curve of an ellipsoid?
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If the parametric formula for an ellipsoid is:
$beginalign
x&=acos(theta)cos(varphi)\
y&=bcos(theta)sin(varphi)\
z&=csin(theta)endalign\$
where
$
-frac pi 2 le thetale frac pi 2
qquad
-pile varphile pi
$
And the parametric formula for an offset curve(!) is:
$x_d(t)= x(t)+fracd; y'(t)sqrt x'(t)^2+y'(t)^2$
$y_d(t)= y(t)-fracd; x'(t)sqrt x'(t)^2+y'(t)^2$
where $d$ is the distance from the curve.
Then what is the formula for the offset surface(!) of an ellipsoid? Thanks.
See:
- https://en.wikipedia.org/wiki/Ellipsoid
- https://en.wikipedia.org/wiki/Parallel_curve
(P.S. what is the derivative of the formula for an ellipsoid? Thanks.)
calculus geometry trigonometry parametric
asked Jul 20 at 1:48
posfan12
202214
add a comment |Â
up vote
0
down vote
favorite
If the parametric formula for an ellipsoid is:
$beginalign
x&=acos(theta)cos(varphi)\
y&=bcos(theta)sin(varphi)\
z&=csin(theta)endalign\$
where
$
-frac pi 2 le thetale frac pi 2
qquad
-pile varphile pi
$
And the parametric formula for an offset curve(!) is:
$x_d(t)= x(t)+fracd; y'(t)sqrt x'(t)^2+y'(t)^2$
$y_d(t)= y(t)-fracd; x'(t)sqrt x'(t)^2+y'(t)^2$
where $d$ is the distance from the curve.
Then what is the formula for the offset surface(!) of an ellipsoid? Thanks.
See:
- https://en.wikipedia.org/wiki/Ellipsoid
- https://en.wikipedia.org/wiki/Parallel_curve
(P.S. what is the derivative of the formula for an ellipsoid? Thanks.)
calculus geometry trigonometry parametric
asked Jul 20 at 1:48
posfan12
202214
For an offset curve, you move a distance $d$ along a normal to the curve. Do the same thing for the surface.
– amd
Jul 20 at 1:56
I'm in need of help with the formulas. Adding the third dimension is not easy for me.
– posfan12
Jul 20 at 2:29
Try deriving the 2-d formulas for yourself. Adding as many dimensions as you want will be easy if you understand them instead of just copying them from somewhere.
– amd
Jul 20 at 2:58
Seriously? So, if I set aside some time this coming weekend, do you think I can learn calculus?
– posfan12
Jul 20 at 7:34
You can certainly learn enough to do this. Look up how to compute the normal to a surface, which I mentioned in my first comment. The rest is simple vector addition.
– amd
Jul 20 at 18:18
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If the parametric formula for an ellipsoid is:
$beginalign
x&=acos(theta)cos(varphi)\
y&=bcos(theta)sin(varphi)\
z&=csin(theta)endalign\$
where
$
-frac pi 2 le thetale frac pi 2
qquad
-pile varphile pi
$
And the parametric formula for an offset curve(!) is:
$x_d(t)= x(t)+fracd; y'(t)sqrt x'(t)^2+y'(t)^2$
$y_d(t)= y(t)-fracd; x'(t)sqrt x'(t)^2+y'(t)^2$
where $d$ is the distance from the curve.
Then what is the formula for the offset surface(!) of an ellipsoid? Thanks.
See:
- https://en.wikipedia.org/wiki/Ellipsoid
- https://en.wikipedia.org/wiki/Parallel_curve
(P.S. what is the derivative of the formula for an ellipsoid? Thanks.)
calculus geometry trigonometry parametric
asked Jul 20 at 1:48
posfan12
202214
If the parametric formula for an ellipsoid is:
$beginalign
x&=acos(theta)cos(varphi)\
y&=bcos(theta)sin(varphi)\
z&=csin(theta)endalign\$
where
$
-frac pi 2 le thetale frac pi 2
qquad
-pile varphile pi
$
And the parametric formula for an offset curve(!) is:
$x_d(t)= x(t)+fracd; y'(t)sqrt x'(t)^2+y'(t)^2$
$y_d(t)= y(t)-fracd; x'(t)sqrt x'(t)^2+y'(t)^2$
where $d$ is the distance from the curve.
Then what is the formula for the offset surface(!) of an ellipsoid? Thanks.
See:
- https://en.wikipedia.org/wiki/Ellipsoid
- https://en.wikipedia.org/wiki/Parallel_curve
(P.S. what is the derivative of the formula for an ellipsoid? Thanks.)
calculus geometry trigonometry parametric
asked Jul 20 at 1:48
posfan12
202214
edited Jul 20 at 4:34
asked Jul 20 at 1:48
posfan12
202214
asked Jul 20 at 1:48
posfan12
202214
asked Jul 20 at 1:48
posfan12
202214
202214
For an offset curve, you move a distance $d$ along a normal to the curve. Do the same thing for the surface.
– amd
Jul 20 at 1:56
I'm in need of help with the formulas. Adding the third dimension is not easy for me.
– posfan12
Jul 20 at 2:29
Try deriving the 2-d formulas for yourself. Adding as many dimensions as you want will be easy if you understand them instead of just copying them from somewhere.
– amd
Jul 20 at 2:58
Seriously? So, if I set aside some time this coming weekend, do you think I can learn calculus?
– posfan12
Jul 20 at 7:34
You can certainly learn enough to do this. Look up how to compute the normal to a surface, which I mentioned in my first comment. The rest is simple vector addition.
– amd
Jul 20 at 18:18
add a comment |Â
For an offset curve, you move a distance $d$ along a normal to the curve. Do the same thing for the surface.
– amd
Jul 20 at 1:56
I'm in need of help with the formulas. Adding the third dimension is not easy for me.
– posfan12
Jul 20 at 2:29
Try deriving the 2-d formulas for yourself. Adding as many dimensions as you want will be easy if you understand them instead of just copying them from somewhere.
– amd
Jul 20 at 2:58
Seriously? So, if I set aside some time this coming weekend, do you think I can learn calculus?
– posfan12
Jul 20 at 7:34
You can certainly learn enough to do this. Look up how to compute the normal to a surface, which I mentioned in my first comment. The rest is simple vector addition.
– amd
Jul 20 at 18:18
For an offset curve, you move a distance $d$ along a normal to the curve. Do the same thing for the surface.
– amd
Jul 20 at 1:56
For an offset curve, you move a distance $d$ along a normal to the curve. Do the same thing for the surface.
– amd
Jul 20 at 1:56
I'm in need of help with the formulas. Adding the third dimension is not easy for me.
– posfan12
Jul 20 at 2:29
I'm in need of help with the formulas. Adding the third dimension is not easy for me.
– posfan12
Jul 20 at 2:29
Try deriving the 2-d formulas for yourself. Adding as many dimensions as you want will be easy if you understand them instead of just copying them from somewhere.
– amd
Jul 20 at 2:58
Try deriving the 2-d formulas for yourself. Adding as many dimensions as you want will be easy if you understand them instead of just copying them from somewhere.
– amd
Jul 20 at 2:58
Seriously? So, if I set aside some time this coming weekend, do you think I can learn calculus?
– posfan12
Jul 20 at 7:34
Seriously? So, if I set aside some time this coming weekend, do you think I can learn calculus?
– posfan12
Jul 20 at 7:34
You can certainly learn enough to do this. Look up how to compute the normal to a surface, which I mentioned in my first comment. The rest is simple vector addition.
– amd
Jul 20 at 18:18
You can certainly learn enough to do this. Look up how to compute the normal to a surface, which I mentioned in my first comment. The rest is simple vector addition.
– amd
Jul 20 at 18:18
add a comment |Â
1 Answer
1
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oldest
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up vote
0
down vote
The formula that you need is just a couple of lines up from the ones that you’ve quoted from the Parallel Curve Wikipedia article, namely,
• $vecx_d(t) = vecx(t)+dvecn(t)$ with the unit normal
$vecn(t)$.
This generalizes to any number of parameters. You’ll have to decide which of the two possible unit normals is appropriate, but you have the same decision to make in two dimensions. (The formulas in the Wikipedia article use the normal that’s 90° clockwise from the tangent vector.)
All that’s left for you to do is to figure out how to compute $vecn$ for the parameterization that you’re using. Typically, for a surface in 3D defined by two parameters, this is computed by taking the cross product of the partial derivatives with respect to each parameter, normalized, but this is problematic for your parameterization since this cross product vanishes at the poles. Fortunately, you can use the gradient of $(x/a)^2+(y/b)^2+(z/c)^2$ instead. This gives you an outward-pointing normal, which is probably what you want, anyway.
answered Jul 21 at 19:17
amd
25.9k2943
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The formula that you need is just a couple of lines up from the ones that you’ve quoted from the Parallel Curve Wikipedia article, namely,
• $vecx_d(t) = vecx(t)+dvecn(t)$ with the unit normal
$vecn(t)$.
This generalizes to any number of parameters. You’ll have to decide which of the two possible unit normals is appropriate, but you have the same decision to make in two dimensions. (The formulas in the Wikipedia article use the normal that’s 90° clockwise from the tangent vector.)
All that’s left for you to do is to figure out how to compute $vecn$ for the parameterization that you’re using. Typically, for a surface in 3D defined by two parameters, this is computed by taking the cross product of the partial derivatives with respect to each parameter, normalized, but this is problematic for your parameterization since this cross product vanishes at the poles. Fortunately, you can use the gradient of $(x/a)^2+(y/b)^2+(z/c)^2$ instead. This gives you an outward-pointing normal, which is probably what you want, anyway.
answered Jul 21 at 19:17
amd
25.9k2943
add a comment |Â
up vote
0
down vote
The formula that you need is just a couple of lines up from the ones that you’ve quoted from the Parallel Curve Wikipedia article, namely,
• $vecx_d(t) = vecx(t)+dvecn(t)$ with the unit normal
$vecn(t)$.
This generalizes to any number of parameters. You’ll have to decide which of the two possible unit normals is appropriate, but you have the same decision to make in two dimensions. (The formulas in the Wikipedia article use the normal that’s 90° clockwise from the tangent vector.)
All that’s left for you to do is to figure out how to compute $vecn$ for the parameterization that you’re using. Typically, for a surface in 3D defined by two parameters, this is computed by taking the cross product of the partial derivatives with respect to each parameter, normalized, but this is problematic for your parameterization since this cross product vanishes at the poles. Fortunately, you can use the gradient of $(x/a)^2+(y/b)^2+(z/c)^2$ instead. This gives you an outward-pointing normal, which is probably what you want, anyway.
answered Jul 21 at 19:17
amd
25.9k2943
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The formula that you need is just a couple of lines up from the ones that you’ve quoted from the Parallel Curve Wikipedia article, namely,
• $vecx_d(t) = vecx(t)+dvecn(t)$ with the unit normal
$vecn(t)$.
This generalizes to any number of parameters. You’ll have to decide which of the two possible unit normals is appropriate, but you have the same decision to make in two dimensions. (The formulas in the Wikipedia article use the normal that’s 90° clockwise from the tangent vector.)
All that’s left for you to do is to figure out how to compute $vecn$ for the parameterization that you’re using. Typically, for a surface in 3D defined by two parameters, this is computed by taking the cross product of the partial derivatives with respect to each parameter, normalized, but this is problematic for your parameterization since this cross product vanishes at the poles. Fortunately, you can use the gradient of $(x/a)^2+(y/b)^2+(z/c)^2$ instead. This gives you an outward-pointing normal, which is probably what you want, anyway.
answered Jul 21 at 19:17
amd
25.9k2943
The formula that you need is just a couple of lines up from the ones that you’ve quoted from the Parallel Curve Wikipedia article, namely,
• $vecx_d(t) = vecx(t)+dvecn(t)$ with the unit normal
$vecn(t)$.
This generalizes to any number of parameters. You’ll have to decide which of the two possible unit normals is appropriate, but you have the same decision to make in two dimensions. (The formulas in the Wikipedia article use the normal that’s 90° clockwise from the tangent vector.)
All that’s left for you to do is to figure out how to compute $vecn$ for the parameterization that you’re using. Typically, for a surface in 3D defined by two parameters, this is computed by taking the cross product of the partial derivatives with respect to each parameter, normalized, but this is problematic for your parameterization since this cross product vanishes at the poles. Fortunately, you can use the gradient of $(x/a)^2+(y/b)^2+(z/c)^2$ instead. This gives you an outward-pointing normal, which is probably what you want, anyway.
answered Jul 21 at 19:17
amd
25.9k2943
edited Jul 21 at 19:45
answered Jul 21 at 19:17
amd
25.9k2943
answered Jul 21 at 19:17
amd
25.9k2943
answered Jul 21 at 19:17
amd
25.9k2943
25.9k2943
add a comment |Â
add a comment |Â
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For an offset curve, you move a distance $d$ along a normal to the curve. Do the same thing for the surface.
– amd
Jul 20 at 1:56
I'm in need of help with the formulas. Adding the third dimension is not easy for me.
– posfan12
Jul 20 at 2:29
Try deriving the 2-d formulas for yourself. Adding as many dimensions as you want will be easy if you understand them instead of just copying them from somewhere.
– amd
Jul 20 at 2:58
Seriously? So, if I set aside some time this coming weekend, do you think I can learn calculus?
– posfan12
Jul 20 at 7:34
You can certainly learn enough to do this. Look up how to compute the normal to a surface, which I mentioned in my first comment. The rest is simple vector addition.
– amd
Jul 20 at 18:18