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How to define a Sinus function in 0 - 2pi
3

How to define a Sinus function in 0 - 2pi

How to define a Sinus function in 0 - 2pi

(OP)
Hello All

I need to create a sinus profile ( function in x-y plane ) to create a surface , what I do is in the space I make a line between two nodes , that the length I select is 1 m , but I doubt if I have to select sth else . then I create the law , sin(x) , that I will have my curve from 0 to 1 m !?

then I correct it in function , by making a dependent variable between the axis and the curve , but still the problem is that If I want a complete sine curve , I have to enter a weird number like 16000 or 6000 , cause CATIA is working on units , how can I solve it for a sine function ?

RE: How to define a Sinus function in 0 - 2pi

(OP)
is there anything un clear ?

RE: How to define a Sinus function in 0 - 2pi

fog in CATIA will work on [0,1] so when you say y=sin(x) with x defined as angle , then x will not represent [0,2PI] but [0,1]

but if you do y=sin(x*2*PI) then you should have the result you're looking for.

Now if you want your fog to work with the length of the curve you have to bring that in your function.

y=sin ( length(curve.1) /1m * 1 rad ) or something like that...

Have fun.

Eric N.
indocti discant et ament meminisse periti

RE: How to define a Sinus function in 0 - 2pi

(OP)
Thank you , I also did this , but doesn't work , still I have the same problem

RE: How to define a Sinus function in 0 - 2pi

(OP)
The *@*PI has a negative affect on the amount of y of this function !! I don't know how I can get a normal sinus !!!

RE: How to define a Sinus function in 0 - 2pi

(OP)
I am trying so many different conditions , in a way that with knowing the wavelength of the sine and the length that we want to have the wave on it : I made this function but doesn't work , despite this fact that : y=sin(x) with x and y both of the type length is dimensionless , so we should multiply it as a y= Asin(2*PI/Lamda*x) that lamda and A are both of the dimension length , but it doesn't work in CATIA , what's the problem , also in HELP of CATIA , Link if I change the length of the line it doesn't look like a sine anymore , I attach mmy file as well , I appreciate your help , thank you ver much

RE: How to define a Sinus function in 0 - 2pi

(OP)
Anyone any Idea ???

RE: How to define a Sinus function in 0 - 2pi

I am sailing in BVI, will look at it next week when I am back

Eric N.
indocti discant et ament meminisse periti

RE: How to define a Sinus function in 0 - 2pi

I am sailing in BVI, will look at it next week when I am back

Eric N.
indocti discant et ament meminisse periti

RE: How to define a Sinus function in 0 - 2pi

(OP)
Thanks a lot , have a great time Eric

RE: How to define a Sinus function in 0 - 2pi

strange...

I have some also some weird behavior with V5R20. Let me check with V6 R2013x where curve by equation is a bit different.

Eric N.
indocti discant et ament meminisse periti

RE: How to define a Sinus function in 0 - 2pi

sorry but V6 2013x has same problem with curve by equation... i suggest you open a PMR

Eric N.
indocti discant et ament meminisse periti

RE: How to define a Sinus function in 0 - 2pi

(OP)
Thank you so much for trying , but what is PMR ?

RE: How to define a Sinus function in 0 - 2pi

it is a bug report you fill on DS website in order to get this fixed. Talk to your catia admin or VAR for more info.

Eric N.
indocti discant et ament meminisse periti

RE: How to define a Sinus function in 0 - 2pi

frztrb,

try something like this in law:

FormalLength.1=sin(50*2*PI*FormalReal.1 /10*1 rad)*50 mm

FormalLength.1-Y coordinates of Sinus Function in fog law (type of Length)
FormalReal.1 - parameter t with values [0,1] varied on required length, 1000mm for example (type of Real)

Playing with k=50, n=10 and d=50 you can get pretty nice sinus functions,
if you look at Sinus Function/ Law as y= sin (k*2*Pi*t/n *1 rad) * d mm

Just don't forget units ( multiplying expressions with 1rad, 1mm, etc)

RE: How to define a Sinus function in 0 - 2pi

(OP)
Thank you all , specially donotaskme

while I have one doubt , the formula in it's general definition is : y = A sin( 2*Pi/L*x) , so when you devide the L ( lambda the wavelength ) into k and n , first , what is the type of n as parameter ? Real ? and how the parameter t can have only the limit [ 0,1 ] ? and multiplying in 1 rad , changes the unit of the argument from real to angle or plays another role ?

thank you so much for your help

RE: How to define a Sinus function in 0 - 2pi

(OP)
the answer to limit of t already was given I had forgotten

RE: How to define a Sinus function in 0 - 2pi

Hello frztrb. I have an unusal and strange but usefull method to créate analitacal funtions to use as curve profiles.
-1.First you need define a parameter type lenght, and you need give that parameter the y axis length that you need to plot. In your case 2*PI. For example define de parameter x0=2*3.1516
-2. Define a law. Click on "law" button. Define de paramater "x" type length and "y" tipe length. Define de function as:
y=sin(x*x0*1rad/1m)*1m
-3. Start "Generative Shape design" workbench.
- Create a line with lenght x0, on "y" direction
- Use "Paralel curve" button to define de sinusoidal curve.
-Paralel curve definition:
-Curve: The line you created with length x0
-Support: YZ plane
- Law: mark "advanced" and click on law you created (You need to have lays visibility active on tree)

And it is done. Now if you change x0 length you change thw y fo the plct curve. If you use x0=4*3.1416 you have two complete peridos, if use pi you have one semiperiod....

RE: How to define a Sinus function in 0 - 2pi

(OP)
thanks a lot NachoAB

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