I am using a 2D top view sketch to model a house. The sketch is a very rough top view drawing with some handwritten notes (no side, front or back views). The top view of the roof rafters have brought me to a halt. The 2D measurement is 20 ft. The pitch of the roof (simple shed) is 2:12 (9.462 degrees). My first thought was to use simple trig, but now I’m not so sure. Seems the base of the right triangle is 20 ft. and knowing the angle I can calculate the hypotenuse. It seems logical that the hypotenuse would represent the actual length of the top truss. But with a top view, am I looking at the hypotenuse or the base (leaning toward hypotenuse)? Can’t seem to get my head around looking at a top down view of an angled piece of geometry. Tried some simple drawings, see below. I now see the importance of a complete set of 2D drawings. Also see why a simple 3D drawing can convey much more information than a whole set of 2D drawings (also to a much wider audience).

In general when converting 2D drawings, is there an effective workflow that generates more accurate 3D representations, given incomplete/inconsistent 2D drawings?

I’ve never used the hypotenuse to draw a roof or rafter. This is new to me.

However in SketchUp like CAD and mechanical drafting you can use the protractor to find the slope of the roof. The protractor axis should be set along the wall that would be the low part of the roof. In SU you orient the protractor, click on the horizontal direction. move the direction of rotation, click, and type the pitch “2:12” and hit enter. You get a guideline that you can draw along. likewise you can use the rotate tool to rotate an edge or face that you’ve already drawn.

As a far as exactly where you want this point of rotation depends on how you are building the roof and your knowledge of construction. I could add a picture, but someone will come along and post a video soon enough.

The drawing of the top view or plan view is typically only the “run” of the roof. 20’ parallel to the ground. It is not the length of the sloped element or hypoteneuse

I did some cheating to get this. I resized the top view to be at the same scale as the side view, then resized both so that the 20’ length was correct. Then traced the lines and extruded both, and used intersect with selection to get the final piece. I don’t know how thick it’s supposed to be, so I made it an inch thick.

So, this (the run) would be equivalent to the base of the right triangle, right? If I remember my trig correctly, the hypotenuse is always greater than the base. If this is correct, then the top sketch showing 20’ would yield a truss greater than 20’ ft. in length. And since we know the angle, the truss length can be calculated. Hope I understand your thinking. Thanks.

Yeah you could or you can just use the protractor or rotate tool to begin drawing your roof slope. Then the edges on the model will give you the hypotenuse.

Yes I took it to be a general concept question, and I thought it involved reading a plan view of a building. If Claire is still in doubt, posting a picture of what she has to work with would give some idea of the approach.

" But with a top view, am I looking at the hypotenuse or the base (leaning toward hypotenuse)? "

If you are looking at typical building plan, then you are looking at the base dimension (usually length of the walls).

I don’t think this is correct. 20’ is the real length of the truss. 19’ 10 41/64" is the horizontal distance. The angle in the drawings is 6.1 degrees, which would mean that after going horizontally 19’ 10 41/64" (238.64 inches) it would have gone up 25.5 inches. The square root of (238.64 x 238.64 + 25.5 x 25.5) is exactly 20 feet.

What I got from the GIF (Box) is that when I look at a 2D top view (plan view) of a house which shows the roof rafters, I am looking at horizontal measurements. The angled top truss will be longer than the measurement on the drawing. Hope this makes sense.

Confirmed my thinking. In my sketch, the 20’ measurement was outside wall to wall, no overhangs. So, in my case the rafter measured slightly over 20’. So, now I feel more comfortable, understanding what 2D plan measurements refer to (horizontal geometry). Thanks.

It all depends which architecting school was attended - or not - by the draftsperson. Deciphering this (as well as other discrepancies) is often a guessing game with drawings I get.

Colin, sorry for the poor problem definition. Box saw through my poor question statement and hit the nail on the head, so to speak, with his GIF. [pbacot] also recognized that I was seeking guidance on not how to create the geometry but how to read 2D plans correctly.

I generated a model to confirm my understanding of the comments in this thread. See the attached images.

As can be seen the 2D top view (parallel projection) shows the 20’ base, while the 3D view shows the actual rafter length of 20’ 3 3/16". As often is the case, Box/pbacot saw through my poor explanation, and provided clear answers.

Generally speaking, plans give the length of a span (that is, the distance from one bearing member to the other). That length is generally measured on a flat horizontal plane (parallel to the ground). The cut length of the material will take into consideration the slope and the end cuts. Remember, the overall length of a material will be longer, depending on how tall the material is and what angle the ends are cut at.

Thanks for the comment. I still struggle in reading drawings of any type.

In this case the outside dimension of the building was 20’. Is the bearing member in this case the front/back wall? In that case, should the original drawing (2D top view) have shown the 20’ 3 3/16" dimension?

You mentioned “cut length” as the hands on dimension. How are “cut length” dimensions typically shown on 2D plan drawings?