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- How to create a curve from the equation?Does Anybo...

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Aug 13, 2010
01:38 AM

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Aug 13, 2010
01:38 AM

How to create a curve from the equation?Does Anybody know?

How to create a curve from the equation?Does Anybody know?I need this kind of information.

Thanks

12 REPLIES 12

Aug 13, 2010
02:12 AM

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Aug 13, 2010
02:12 AM

counter-questions:

where exactly is your problem?

Do you not find the menu?

Do you not know what an equasion should look like?

Aug 13, 2010
06:02 AM

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Aug 13, 2010
06:02 AM

Hi,

I know

Is this correct answer?

And seriously, here is a simple example:

Remember that "t" is changing from 0 to 1;

BR

Marek

Aug 13, 2010
06:08 AM

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Aug 13, 2010
06:08 AM

Thanks for sharing!

And interesting since I would have created something like that using the cylindrical coordinate system...

Aug 13, 2010
06:17 AM

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Aug 13, 2010
06:17 AM

Yep, equations would be easier and need less math

BLUE DUNCAN: Here you have more than simple example: http://www.meadinfo.org/2009/05/involute-curve-generation-with.html

Marek

Aug 18, 2010
11:05 AM

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Aug 18, 2010
11:05 AM

hi i m balwan & i know correct ans which type of curve u want to draw by equation

Aug 19, 2010
06:34 PM

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Aug 19, 2010
06:34 PM

You mean something like this?

Note the reversal.

Aug 19, 2010
06:35 PM

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Aug 19, 2010
06:35 PM

This maybe?

Aug 20, 2010
05:37 AM

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Aug 20, 2010
05:37 AM

I can see, that you like trigonometric functions very much 😉

Aug 23, 2010
11:53 AM

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Aug 23, 2010
11:53 AM

Actually Marek, no, I don't. I have other tricks that I use instead for geometry creation. I'm terrible at math so I have to compensate in other ways.

Aug 23, 2010
12:29 PM

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Aug 23, 2010
12:29 PM

Blue,

A little more general answer: The Curve-By-Equation technique involves the use of so-called parametric equations. (You may have had a one-day introduction and assignment dealing with them in a high school trigonometry or analytical geometry class.) The equations all depend on just one parameter which takes on all of a **finite** range of values to determine the simultaneous results for the **dependent** equations. In this case, the parameter you use must be **t** and it will take on the continuous values of 0 to 1. (The same idea is used in Variable Section Sweeps where you use **trajpar**, as opposed to **t**. Both of these techniques couldn't be more appropriate for parametric software.)

As an example, if you write the equation x=y you would create a 45 degree "climbing" straight line **going to infinity** in both directions. If, on the other hand--using the parametric technique, you write the simultaneous equations x=t and y=t you create a 45 degree line from {0,0} to {1,1} **only**. These equations never go to infinity because the independent variable stops when it "gets to" 1.

Obviously, you wouldn't use this method for something that simple. On the other hand, if you want to create some kind of interesting variable spiral or the involute for a gear tooth this technique is great. Choose your type of equations carefully. For example, if you are creating some kind of helical curve you can certainly do it with Cartesian coordinates employing sine and cosine functions for two of the 3 equations (usually x & y), but it's much easier to use Cylindrical coordinates and just type in a straightforward function for theta. As a further clarification, if you wrote theta=4*t*360, you would generate 4 complete turns in your helical geometry. (You need to multiply t--0 to 1 only--by 360 to get one complete turn).

Hope this helps!

David

Aug 23, 2010
01:22 PM

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Aug 23, 2010
01:22 PM

For example:

CS1--Cartesian:

x=2*cos(3*t*360)

y=2*sin(3*t*360)

z=3*t

CS2--Cylindrical:

r=2

theta=3*t*360

z=3*t

CS3--Spherical:

rho=2

theta=90-(2*t*45)

phi=3*t*360

CS4--Cartesian:

x=4*t-2

y=2*sin(2*t*360)

z=3*t

CS5--Cylindrical:

r=2

theta=3*t*360

z=2*(t+sin(3*t*360))

CS6--Spherical:

rho=2

theta=90-(t*180)

phi=t*360

Aug 23, 2010
04:26 PM

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Aug 23, 2010
04:26 PM

Thanks David for the detailed demo. I'm sure people will find it very useful!