jtfsr Posted August 4, 2014 Share Posted August 4, 2014 I have a coordinate: 23.7dn 47.4dw How can I draw a simple line from 0,0,0 at any length to that point (23.7dn 47.4dw)? Simple huh? For the life of me I can't seem to make this work. HELP! Quote Link to comment Share on other sites More sharing options...
jtfsr Posted August 4, 2014 Author Share Posted August 4, 2014 Still can't get the command right. this is what I'm inputting: drawing a line 1' long to a coordinate.... Specify first point: Specify next point or [undo]: 1'Point or option keyword required. Is this correct? or not?... I feel like a simpleton..... Quote Link to comment Share on other sites More sharing options...
designerstuart Posted August 4, 2014 Share Posted August 4, 2014 welcome to cad tutor forum! you can draw a line from 0,0,0 by using command L then inputting 0,0,0 as your first point. the second point could be 23.7,47.4,0 but unfortunately i am unfamiliar with your units dn and dw. does this help at all? Quote Link to comment Share on other sites More sharing options...
ReMark Posted August 4, 2014 Share Posted August 4, 2014 d = degrees. n and w = north and west. Looks like you are using or want to use decimal degrees as opposed to degrees-minutes-seconds. Nor does it appear that you want to use bearings like northwest or southeast for example. Quote Link to comment Share on other sites More sharing options...
jtfsr Posted August 4, 2014 Author Share Posted August 4, 2014 Here's what I'm trying to do. I have a tiff of the Moon Crater Aristarchus. The center of the crater is at 23.7deg north by 47.4 deg west. I want to attach this tiff using the Image command and position the tiff so that the center of the crater aligns at 23.7deg north by 47.4 deg. west. I attempted to draw line from 0,0,0 to that point and attach the tiff there. Help? Quote Link to comment Share on other sites More sharing options...
ReMark Posted August 4, 2014 Share Posted August 4, 2014 What are the distances for each of those directions? Quote Link to comment Share on other sites More sharing options...
RobDraw Posted August 4, 2014 Share Posted August 4, 2014 Those are longitude and latitude coordinates. They do not work in Cartesian or Polar coordinate systems, which are the options in AutoCAD. I can't think of an easy solution ATM. This should be interesting. Quote Link to comment Share on other sites More sharing options...
eldon Posted August 4, 2014 Share Posted August 4, 2014 I think that the length of the line from 0,0,0 to your point would be 1737100000 millimetres (or so) Quote Link to comment Share on other sites More sharing options...
steven-g Posted August 4, 2014 Share Posted August 4, 2014 First, welcome to CADTutor. I don't think there is an input method to enter coordinates in that way (I could be wrong), the only way I can think of to do it directly is rather long winded, and involves using 'cal to work out the second point of the line. line (enter) specify first point - 0,0 (enter) specify second point - 'cal (enter) 100*vec1([0,0],[47.4,-27.7]) (enter) using Y for north and X for west. Failing that just draw a point that you want to use then draw the line from 0,0 and hold the cursor over that point and type in the length you need Autocad will automatically draw it in the direction of the cursor (with dynamic input on and ortho off) Quote Link to comment Share on other sites More sharing options...
RobDraw Posted August 4, 2014 Share Posted August 4, 2014 Another way would be to get a scaled map of the moon with lat./long. lines and measure from 0 lat., 0 long. using a Cartesian or Polar coordinate system. Quote Link to comment Share on other sites More sharing options...
ReMark Posted August 4, 2014 Share Posted August 4, 2014 Know anyone with Civil 3D? It could be done using the Line By Latitude/Longitude command. Quote Link to comment Share on other sites More sharing options...
RobDraw Posted August 4, 2014 Share Posted August 4, 2014 Know anyone with Civil 3D? It could be done using the Line By Latitude/Longitude command. , if it comes with an alternate system for the moon. I believe those values are different because of the size of the orbs. Quote Link to comment Share on other sites More sharing options...
jtfsr Posted August 4, 2014 Author Share Posted August 4, 2014 Thanks guys for the replys...@ ReMark- no if i did I wouldn't be here. lol... @RobDraw - why should the system or size have anything to do with it?....I'm just trying to draw a line (however long) from 0,0,0 to a point at 23.7deg N by 47.4deg W....(just trying to locate that point in space).... Quote Link to comment Share on other sites More sharing options...
ReMark Posted August 4, 2014 Share Posted August 4, 2014 Then he'll need selenographic coordinates which are comparable to latitude and longitude here on earth. Quote Link to comment Share on other sites More sharing options...
RobDraw Posted August 4, 2014 Share Posted August 4, 2014 Yeah, I already said that but didn't know what to call the system. Thanks for giving us the name, though. Quote Link to comment Share on other sites More sharing options...
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