Perpendicular lines

[4DThinker]

While drafting the line I have the end point of a line that I now want to continue perpendicularly through/to another existing line segment. The intersection isn't defined yet, but I want it to be where my previous end point continues through the new line at 90 degrees. Your trick only works if the intersection is known and a node is placed there. I need the perpendicular snap to locate it, rather than a node locating the perpendicular line.

Yes, I can build my drawing using other drafting tricks. Like tracing the second line, )) two times to rotate it 90 degrees, then dragging an end point to where my starting line ended. Extend or trim to get it to end where I want it to be.

I know the feature I described isn't exactly promised by Vectric as being one of the new features. I was just hoping.

4D

[ger21]

I need the perpendicular snap to locate it, rather than a node locating the perpendicular line.

If I understand correctly, this would not be possible unless the start point was moved when it snapped perpendicular, correct? I've never seen a CAD program that did that.

[4DThinker]

Yes. Must be a difficult drafting trick to code for. Easy to do with conventional drafting tools (t-squares and triangles, etc.).

[ger21]

With a triangle, you are automatically starting from a point in space perpendicular to your endpoint location. Vectric is basically having you place a "virtual triangle" to select the start point.

[mtylerfl]

I guess I need a labeled picture of what you mean. From your description, I am still thinking you can do this. As long as the geometry is "awakened" on your first line end point, cursor then moved away from that and click to start the line, I don't understand why you cannot extend that through another line.

[4DThinker]

I guess I need a labeled picture of what you mean.

Attached. Now draw a line starting from the right side endpoint of "new line" toward "existing line" that snaps at 90 to "existing line".
The known endpoint is at the right end of "new line" and I want a vector from there to wherever on "existing line" would be 90 degrees from "existing line". This is easy to do using a few other steps, but would be one click magic if it worked directly.

4D

[mtylerfl]

I guess I need a labeled picture of what you mean.

Attached. Now draw a line starting from the right side endpoint of "new line" toward "existing line" that snaps at 90 to "existing line".
The known endpoint is at the right end of "new line" and I want a vector from there to wherever on "existing line" would be 90 degrees from "existing line". This is easy to do using a few other steps, but would be one click magic if it worked directly.

4D

Hi 4D,

I believe the easiest (and accurate) method would be just draw a "purposely too long" perpendicular line from the "existing" line first, then drag/snap the end of that perpendicular line to the endpoint of your "new" line. Trim any excess overlap, if you want that perpendicular line to meet exactly at the "existing" vector. Very quick, easy and simple, plus no extra tricks or drawn geometry necessary.

[4DThinker]

@mtylerfl : That's exactly the solution I've already come up with. With the new smart snap features it is at least 2 steps quicker than how I used to do it. Thanks for looking into it.

The next drafting feature I'd love is to be able make a new line and smart snap it parallel to another existing line. Something for the programmers to work on for version 9.5 perhaps?

Way back when PCs ran DOS I wrote a 3D wireframe CAD program optimized for creating perspective views. I based all the drawing tools after techniques I knew for drafting with t-square and triangles. Parallel to and perpendicular to lines were what I thought was easy to code (in BASIC) back then, but that was a younger smarter me.

4D

[FixitMike]

@mtylerfl :
The next drafting feature I'd love is to be able make a new line and smart snap it parallel to another existing line. Something for the programmers to work on for version 9.5 perhaps?

Use the Offset function to draw a parallel line.

[4DThinker]

Use the Offset function to draw a parallel line.

Works fine if you start off knowing the distance from the existing line your new line needs to be. Not so useful when you start a new line, and simply want it to be parallel to "that line over there", but not a copy and not from a point you already know the distance to(from) that "line over there". Offset also takes your cursor off the drawing screen to accomplish. None of the capabilities I'm hoping for will do something I can't already do. What they could do is simplify/shorten/speed-up the drafting process even more than the new smart snap features do.

4D

[mtylerfl]

Hi 4D,

You probably already do what I have done for that in the past...just drag an existing line while holding control-shift to make a new parallel line when the distance doesn't matter. (Or just drag while holding control if I don't care about having the same horiz or vert position as the original)

[4DThinker]

Hi 4D, You probably already do what I have done for that in the past...just drag an existing line while holding control-shift to make a new parallel line when the distance doesn't matter.

Yep. The new smart snap working in the node mode does making stretching that moved copy easy though.

[wood400]

Hi Folks. I think the new geometry and snaps are a fabulous resource . i seem to be having a bit of difficulty creating a perpendicular line from any given point on a line. If i locate the start point on an end snap or centre snap the right angle image shows perfectly. but as said if i select an non snap position and create a line i do not seem to be able to obtain the dedicated perpendicular right angle image. Can anyone give their expertise . thanks alan

[Adrian]

The perpendicular function only works from a node. There is one at the start and end of a line plus the mid-point is a virtual node. If you want to draw a line from a particular point you have to add a node there first.

[wood400]

Hi Adrian . Than you as i see this being a god send for the roofing geometry.
Cheers alan