3 replies to this topic

[TeaMan]

I need to draw my lot where I plan to build a shop. I am struggling with how to enter the angle using surveyor units. For example, I have a lot line that is 227.97 feet at S33degrees 16' 14"W. I typed degrees instead of the symbol since I don't know how to make the degree symbol. I suspect it starts with 227.97@. or 227.97<. I'm not sure what to put after the @ sign or < sign while the units are in surveyor mode. Can anyone help please?

Thanks
TeaMan

[Katharine Warrender]

You have to change the angular units to surveyors units under Format>drawing units. Then follow the formatting you see in the precision drop down menu eg. S33d16'14"W

[TeaMan]

Thanks Katharine. That seams to work. I had edited my post, but it created another one and for some reason looked like a duplicate. The one you replied to was the original. I had added a little information to that post, but I think you got me there without that information. What I had said is I typed the Units command which gets me to the same place you suggested. I had chosen Archatectural for length type, and Surveyors for Angle type. The part I missed was the precision drop down. It defaulted to N 0d E. I changed it to N 00d00'00" E which now lets me put the syntax I have on my drawing from the surveyors. What was confusing is that the precision drop down shows a space between the direction letter and distance and also the minutes and last direction letter. When you try to enter it that way, when you hit the space bar, it enters the command and gives a syntax error. I then tried it with no spaces and it entered properly.

I have two questions for clarity. How does the base angle work? I think it came up as a default of E for base angle and I changed it to N. I only did this when I couldn't get anything to work and thought that's why my lines were being drawn wrong. I wasn't able to enter the first angle letter such as "S" or "N", so I tried to change the base angle to "N" hoping it was what was needed to substitute for me not being able to enter it. Now that I have the proper precision, I'm wondering what the base angle does, or how it works. Right now I'm able to simply draw independent lines on my screen and move them to where the last line end was to create the proper she. This does work, but maybe there is a better way.

Second, I have two arcs. One is large and the other small. One is 143.43' and the arc is (Radius) 160.70' (ARC) 162.28' (Delta) 057d51'27" (Chord) 155.47' (Chord Bearing) S 87d22'27" W. The other one is 51.75' and the arc is (Radius) 560.12' (ARC) 308.90' (Delta) 031d35'53" (Chord) 305.00' (Chord Bearing) N 47d21'14" W. I have no idea how to draw an arc using information like this. The best I could do is approximate it, but if there is a way to enter it properly and draw it properly I'd prefer that. Can you help with how to draw an arc using this data. The data comes from a chart on the surveyors drawing with the headings for the chart being the words I have in parenthesis.

I appreciate your help. This has me much further than I was previously.

Teaman.

[TeaMan]

Katharine, I forgot to mention one thing. When I enter a line such as a line that is 158.68 feet long and angle is S 21d21'49" W, the way I enter it is, I click the line command, left click a point on my screen then type at the command line:
@158.68<S21d21'49"W
I then press Enter. This draws me a line from that point that looks very close to the surveyors drawing I have. I assume this is the way to enter the syntax. If I'm not correct with that, please help me with the proper way to enter the syntax for the line command using surveyors units.

Thanks again,
TeaMan