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View Full Version : V 2004---use a reference to constrain a line to X or Y



ajs
26th Oct 2007, 08:29 pm
I'm a newbie to Autocad. For years, I've been using a program called FastCad. FastCad has a modifier that I use frequentl. that I can't seem to find an AutoCad equivalent for:

What I would like to do is to be able to do is draw a line and then pick the endpoint of another (reference) line to set the "X" or "Y" position.

For example, in FastCad, what I would do is start my line and then go to:

Modifier > Same X

What this would do is allow me to choose the endpoint of another line that occupied a different Y position but because of the "Same X" modifier, the line that I am drawing will not snap to the end of the reference line but will instead match the "X" endpoint of the reference.

Example:


Reference line: _______________________ "I use this endpoint as a reference for the new line"



New Line Start : ______________________ New line end

New line ends at same "X" point as reference line but ignores reference line "Y" if I choose "Same X" and use the endpoint of the reference line.

Thanks in advance!

---AJ

eldon
26th Oct 2007, 08:38 pm
You probably need Point Filters :)

lpseifert
26th Oct 2007, 08:51 pm
If I understand your question, I think you want to use Point Filters. For example, start the line, when it prompts you for next point type .y. Then pick a point, continue with the line and it will be constrained to the y value that you previously picked. Other options are .x, .z, .xy, .xz, .yz.

Raggi_Thor
26th Oct 2007, 10:29 pm
Or turn on OTRACK and OSNAP, you also need POLAR or ORTHO:

Strix
26th Oct 2007, 11:59 pm
so FROM wouldn't work this way?

bookmarking for FAQ too

ajs
27th Oct 2007, 12:33 am
Thank you for the responses. the point filter accomplishes what I need.

Raggi_Thor
27th Oct 2007, 07:25 pm
With FROM I think you have to type @x,y afterwords.

Alan Cullen
28th Oct 2007, 01:15 am
With FROM you can type @x,y (relative coords), or x,y, (actual coords) afterwords. You can also type @250.5<135d30'15" (relative distance, bearing). :lol:

I know what your intention was there, Raggi. I just added that for future reference for any searches.