HumbleChief Posted May 27, 2020 Share Posted May 27, 2020 Difficult to ask this question clearly but am working with an ACAD user who will give me cabinet designs and elevations that produce in 3D for our clients. The 2D elevations are produced at a 90 degree angle to each wall but the cabinets that create the corners cannot remain at their 2D dimensions as they are truncated at the angle in the corner, 22.5 in this case. Curious about a formula that might recreate each remaining length of cabinet after that truncation? Hope the video can clarify the question. Got to be some geometry/trig brains out their that have worked this out? Mark? Thanks Link to comment Share on other sites More sharing options...
robdyck Posted May 27, 2020 Share Posted May 27, 2020 5:12 my friend. Reduce cabinets by 10" to meet at the frame, reduce by 10.5" to meet at the doors/drawers and reduce by at least 11" if you'd like the doors and drawers to open! For a bay wall like you've shown, a simpler method is to know the finished wall length, and subtract 12" at each 135degree corner. 1 Link to comment Share on other sites More sharing options...
robdyck Posted May 27, 2020 Share Posted May 27, 2020 A good idea would be to give yourself some reference lines at 24" and 26" away from the drywall. In your example, you'd keep things centered, especially if there's a 135 degree corner at each end. 1 Link to comment Share on other sites More sharing options...
HumbleChief Posted May 27, 2020 Author Share Posted May 27, 2020 23 minutes ago, robdyck said: 5:12 my friend. Reduce cabinets by 10" to meet at the frame, reduce by 10.5" to meet at the doors/drawers and reduce by at least 11" if you'd like the doors and drawers to open! For a bay wall like you've shown, a simpler method is to know the finished wall length, and subtract 12" at each 135degree corner. Not sure I understand. A 24" cabinet across the back wall is reduced to approx. 14" across the front. An 18" cabinet is reduced to approx. 8". Is there a 5:12 ratio in there somewhere? When I tried a ratio of different cabinet front lengths it was not consistent across all cab dims. There's another formula I'm looking for. EDIT the answer is here, just didn't get it. Thanks Link to comment Share on other sites More sharing options...
robdyck Posted May 27, 2020 Share Posted May 27, 2020 4 minutes ago, HumbleChief said: Is there a 5:12 ratio in there somewhere? At every corner...that is 135 degrees 1 Link to comment Share on other sites More sharing options...
HumbleChief Posted May 27, 2020 Author Share Posted May 27, 2020 1 minute ago, robdyck said: At every corner...that is 135 degrees GOT IT - nice and thank you Link to comment Share on other sites More sharing options...
HumbleChief Posted May 27, 2020 Author Share Posted May 27, 2020 10 minutes ago, robdyck said: 13 minutes ago, robdyck said: 5:12 my friend. Reduce cabinets by 10" to meet at the frame, reduce by 10.5" to meet at the doors/drawers and reduce by at least 11" if you'd like the doors and drawers to open! For a bay wall like you've shown, a simpler method is to know the finished wall length, and subtract 12" at each 135degree corner. ...and that 12" gives enough breathing room for frames and doors and drawers etc.? Link to comment Share on other sites More sharing options...
robdyck Posted May 27, 2020 Share Posted May 27, 2020 No Problem. Every good framer knows that's how you derive the length of sub-fascia at bays. Here's an overview with cabinet boxes at the corners. Notice the doors overlap. 1 Link to comment Share on other sites More sharing options...
robdyck Posted May 27, 2020 Share Posted May 27, 2020 And 11" away from the corner...a tidy inconspicuous bit of filler is all that's required. 1 Link to comment Share on other sites More sharing options...
robdyck Posted May 27, 2020 Share Posted May 27, 2020 3 minutes ago, HumbleChief said: ...and that 12" gives enough breathing room for frames and doors and drawers etc.? 11" is enough as long as the kids don't try to open them both at the same time! Link to comment Share on other sites More sharing options...
HumbleChief Posted May 27, 2020 Author Share Posted May 27, 2020 Nice and a very simple solution for my ACAD designer. Link to comment Share on other sites More sharing options...
BrownTiger Posted May 27, 2020 Share Posted May 27, 2020 Technically speaking it is: x = 24"/tan(angle / 2) ; So if your corner angle is 135; tan(135/2) = tan(67.5) = 2.4; So inside corners walls should be reduced by 24"/2.4 = 10"... For practical purposes there is a small gap between cabinets... 10 1/2" - 11". Lol trigonometry 101? Link to comment Share on other sites More sharing options...
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