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- How to create a Hyperbola curve using Equation Curve (Autodesk Inventor 2013)

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Message 1 of 9

Anonymous

5623 Views, 8 Replies

02-19-2013
10:34 PM

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Not applicable

02-19-2013
10:34 PM

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I want to draw hyperbola curve whose equation is x^2/a^2 - y^2/b^2=1. May I use equation curve tool to make the above hyperbola. If it is possible please help me to do so.

I have also made a hyperbola curve of aforesaid equation through conventional method on mathematical ground in the past. I am posting the file of the same to make the question clearer.

Regards

SORJM

Solved! Go to Solution.

02-19-2013
10:34 PM

I want to draw hyperbola curve whose equation is x^2/a^2 - y^2/b^2=1. May I use equation curve tool to make the above hyperbola. If it is possible please help me to do so.

I have also made a hyperbola curve of aforesaid equation through conventional method on mathematical ground in the past. I am posting the file of the same to make the question clearer.

Regards

SORJM

Solved! Go to Solution.

Solved by karthur1. Go to Solution.

8 REPLIES 8

Message 2 of 9

02-20-2013
06:01 AM

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Accepted solution

First thing, write your equation in terms of Y=SQRT(b^2((x/a)^2-1)). For the equation curve, set x=t

Kirk A.

Windows 7 x64 -12 GB Ram

Intel i7-930 @ 3.60ghz

nVidia GTS 250 -1GB (Driver 301.42)

INV Pro R2013, SP1.1 (update1)

Vault Basic 2013

02-20-2013
06:01 AM

First thing, write your equation in terms of Y=SQRT(b^2((x/a)^2-1)). For the equation curve, set x=t

Kirk A.

Windows 7 x64 -12 GB Ram

Intel i7-930 @ 3.60ghz

nVidia GTS 250 -1GB (Driver 301.42)

INV Pro R2013, SP1.1 (update1)

Vault Basic 2013

Message 3 of 9

02-20-2013
08:26 AM

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Thanks for the quick and valuable response. It shows your excellent command over the software.

Regards

SORJM

02-20-2013
08:26 AM

Thanks for the quick and valuable response. It shows your excellent command over the software.

Regards

SORJM

Message 4 of 9

10-02-2017
02:42 PM

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Here is another example of an hyperbola with these parameters.

Center (420 , 0) mm

a = 300mm

b = 80mm

So the equation is:

((x-420)/ 300)^2 - (y/80)^2 = 1

x(t) =t

y(t)=sqrt(( ( ( ( t - 420 ul ) / 300 ul ) ) ^ 2 ul - 1 ul ) * ( ( 80 ul ) ^ 2 ul ))

See attached part

10-02-2017
02:42 PM

Here is another example of an hyperbola with these parameters.

Center (420 , 0) mm

a = 300mm

b = 80mm

So the equation is:

((x-420)/ 300)^2 - (y/80)^2 = 1

x(t) =t

y(t)=sqrt(( ( ( ( t - 420 ul ) / 300 ul ) ) ^ 2 ul - 1 ul ) * ( ( 80 ul ) ^ 2 ul ))

See attached part

Message 5 of 9

05-16-2019
06:02 AM

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Is it possible to draw a hyperbola curve from a sheet of metal by the given values?

a=1.9

b=1.46

c=2.4

and the following equation is x^{2}/a^{2} – y^{2}/b^{2} = 1

Regards,

Amjad

@karthur1 wrote:

Here is another example of an hyperbola with these parameters.

Center (420 , 0) mm

a = 300mm

b = 80mm

So the equation is:

((x-420)/ 300)^2 - (y/80)^2 = 1

x(t) =t

y(t)=sqrt(( ( ( ( t - 420 ul ) / 300 ul ) ) ^ 2 ul - 1 ul ) * ( ( 80 ul ) ^ 2 ul ))

See attached part

05-16-2019
06:02 AM

Is it possible to draw a hyperbola curve from a sheet of metal by the given values?

a=1.9

b=1.46

c=2.4

and the following equation is x^{2}/a^{2} – y^{2}/b^{2} = 1

Regards,

Amjad

@karthur1 wrote:

Here is another example of an hyperbola with these parameters.

Center (420 , 0) mm

a = 300mm

b = 80mm

So the equation is:

((x-420)/ 300)^2 - (y/80)^2 = 1

x(t) =t

y(t)=sqrt(( ( ( ( t - 420 ul ) / 300 ul ) ) ^ 2 ul - 1 ul ) * ( ( 80 ul ) ^ 2 ul ))

See attached part

Message 6 of 9

05-16-2019
06:12 AM

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Your equation is in the form X^2-Y^2=1 (X^2/a^2) = (X/a)^2.... which is like the example in message 5 I posted.

If you still cant get it, post your ipt and I will see what I can do.

05-16-2019
06:12 AM

Your equation is in the form X^2-Y^2=1 (X^2/a^2) = (X/a)^2.... which is like the example in message 5 I posted.

If you still cant get it, post your ipt and I will see what I can do.

Message 7 of 9

05-16-2019
07:39 AM

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@karthur1 wrote:

Your equation is in the form X^2-Y^2=1 (X^2/a^2) = (X/a)^2.... which is like the example in message 5 I posted.

If you still cant get it, post your ipt and I will see what I can do.

05-16-2019
07:39 AM

@karthur1 wrote:

If you still cant get it, post your ipt and I will see what I can do.

Message 8 of 9

05-16-2019
11:08 AM

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Here is the equation for what you asked for. a=1.9, b=1.46

Not sure where to put the "C" value you gave.

Here is the results after I mirrored the initial curve a couple times. See attached file (Inv Version 2018)

05-16-2019
11:08 AM

Here is the equation for what you asked for. a=1.9, b=1.46

Not sure where to put the "C" value you gave.

Here is the results after I mirrored the initial curve a couple times. See attached file (Inv Version 2018)

Message 9 of 9

05-17-2019
02:09 PM

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Since the equation that I worked out if for a vertical hyperbola, I had to rotate the axis for your problem. To do that, I had to change some things around

Where a=1.90 and b=1.46, I then have this for the equation curve.

See attached file of the part.

05-17-2019
02:09 PM

Since the equation that I worked out if for a vertical hyperbola, I had to rotate the axis for your problem. To do that, I had to change some things around

Where a=1.90 and b=1.46, I then have this for the equation curve.

See attached file of the part.

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