how to make parametric equation of cube
$begingroup$
this is my hyperbolic equation
$y = x^2$
then convert to paramteric equation, like this:
$x = u$
$y = u^2$
so i insert the equation into x and y Axis Generator
x: u
y: u
z: 0
the output is: 2D hyperbolic object
when i insert v parameter into z Axis
the output is: 3D hyperbolic object
my question is how to make 3D cube using parametric equation and how to insert it into the Generator.
any help thank youu :-) I LOVE YOU
another exemple:
x: cos(u)
y: sin(u)
z: v
the output is: pipe
where:
u minimal: 0
u maximal: 6.28
v minimal: 0
v maximal: 1
parametric
asked Dec 14 '18 at 7:35
vaianavaiana
32
$endgroup$
add a comment |
$begingroup$
this is my hyperbolic equation
$y = x^2$
then convert to paramteric equation, like this:
$x = u$
$y = u^2$
so i insert the equation into x and y Axis Generator
x: u
y: u
z: 0
the output is: 2D hyperbolic object
when i insert v parameter into z Axis
the output is: 3D hyperbolic object
my question is how to make 3D cube using parametric equation and how to insert it into the Generator.
any help thank youu :-) I LOVE YOU
another exemple:
x: cos(u)
y: sin(u)
z: v
the output is: pipe
where:
u minimal: 0
u maximal: 6.28
v minimal: 0
v maximal: 1
parametric
asked Dec 14 '18 at 7:35
vaianavaiana
32
$endgroup$
$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52
$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00
$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation withoutuandvparameter. @MohammadZuhairKhan
$endgroup$
– vaiana
Dec 14 '18 at 8:16
add a comment |
$begingroup$
this is my hyperbolic equation
$y = x^2$
then convert to paramteric equation, like this:
$x = u$
$y = u^2$
so i insert the equation into x and y Axis Generator
x: u
y: u
z: 0
the output is: 2D hyperbolic object
when i insert v parameter into z Axis
the output is: 3D hyperbolic object
my question is how to make 3D cube using parametric equation and how to insert it into the Generator.
any help thank youu :-) I LOVE YOU
another exemple:
x: cos(u)
y: sin(u)
z: v
the output is: pipe
where:
u minimal: 0
u maximal: 6.28
v minimal: 0
v maximal: 1
parametric
asked Dec 14 '18 at 7:35
vaianavaiana
32
$endgroup$
this is my hyperbolic equation
$y = x^2$
then convert to paramteric equation, like this:
$x = u$
$y = u^2$
so i insert the equation into x and y Axis Generator
x: u
y: u
z: 0
the output is: 2D hyperbolic object
when i insert v parameter into z Axis
the output is: 3D hyperbolic object
my question is how to make 3D cube using parametric equation and how to insert it into the Generator.
any help thank youu :-) I LOVE YOU
another exemple:
x: cos(u)
y: sin(u)
z: v
the output is: pipe
where:
u minimal: 0
u maximal: 6.28
v minimal: 0
v maximal: 1
parametric
parametric
asked Dec 14 '18 at 7:35
vaianavaiana
32
asked Dec 14 '18 at 7:35
vaianavaiana
32
edited Dec 14 '18 at 7:47
vaiana
asked Dec 14 '18 at 7:35
vaianavaiana
32
asked Dec 14 '18 at 7:35
vaianavaiana
32
asked Dec 14 '18 at 7:35
vaianavaiana
32
32
$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52
$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00
$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation withoutuandvparameter. @MohammadZuhairKhan
$endgroup$
– vaiana
Dec 14 '18 at 8:16
add a comment |
$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52
$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00
$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation withoutuandvparameter. @MohammadZuhairKhan
$endgroup$
– vaiana
Dec 14 '18 at 8:16
$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52
$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52
$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00
$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00
$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation without
u and v parameter. @MohammadZuhairKhan$endgroup$
– vaiana
Dec 14 '18 at 8:16
$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation without
u and v parameter. @MohammadZuhairKhan$endgroup$
– vaiana
Dec 14 '18 at 8:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:
$|x|le a/2\|y|le a/2\|z|le a/2$
Parameterize it as:
$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
$endgroup$
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
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
$begingroup$
The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:
$|x|le a/2\|y|le a/2\|z|le a/2$
Parameterize it as:
$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
$endgroup$
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
add a comment |
$begingroup$
The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:
$|x|le a/2\|y|le a/2\|z|le a/2$
Parameterize it as:
$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
$endgroup$
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
add a comment |
$begingroup$
The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:
$|x|le a/2\|y|le a/2\|z|le a/2$
Parameterize it as:
$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
$endgroup$
The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:
$|x|le a/2\|y|le a/2\|z|le a/2$
Parameterize it as:
$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
edited Dec 24 '18 at 10:33
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
answered Dec 14 '18 at 8:07
Shubham JohriShubham Johri
5,192717
5,192717
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
add a comment |
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13
add a comment |
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$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52
$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00
$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation without
uandvparameter. @MohammadZuhairKhan$endgroup$
– vaiana
Dec 14 '18 at 8:16