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Is there an easy way to draw this?
August 20, 2017, 02:49:32 AM
I'm designing something that involves a folding A-frame in the form of a triangle, hinged at each corner. Two of the sides are rigid, but the third is divided by a hinged joint.
If the 2 rigid sides are of equal length, and the hinge on the 3rd side is in the middle, then the whole frame can be folded flat. The problem comes when the two rigid sides are of unequal length - how to place the hinge so that it still folds flat?
In the following schematic, A,B and C are the hinges at the corners, and D is the hinge on the 3rd side. The top diagram shows how 2 equal sides and a centre joint folds flat. The other 2 diagrams show what happens when the 2 rigid sides are unequal. The centre diagram shows the result of placing the hinge on the altitude bisector - this clearly doesn't fold flat.
The bottom diagram shows  the hinge placed so that the difference in lengths between the hinge arms is equal to the difference in lengths of the 2 rigid sides. This folds flat nicely, but the position of the hinge has either to be calculated and measured (very tedious), or determined by rather a lot of auxiliary drawing construction involving circles.
Is there an easy way to draw it?

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Gary Wooding
Win10 64-bit,
TC21.2 x64 Plat, Bld59
TC16.2 Plat, Bld54.0
TCC 3.5

August 20, 2017, 03:38:52 AM
I would think as long as line D-B is a bisector of angle B then the two edges AB and CB will lie against each other, obviously the lengths would be different.

The 2d Bisector tool (in the line toolbar) will easily draw this but it only works on lines not closed polylines or polygons, so if its closed then copy the triangle to a temp layer and explode it once, then bisect.

Of course I could have misunderstood the problem.

## Edit ## just thought there is a bisector construction line in later versions (checked in v21), which works on closed shapes, ## End Edit ##
« Last Edit: August 20, 2017, 03:49:39 AM by Andy H »

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August 20, 2017, 04:27:01 AM
Close, but no cigar.
This is using the angle bisector.
The only way it works is if CD-AD = CB-AB
« Last Edit: August 20, 2017, 04:29:51 AM by lemel_man »

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Gary Wooding
Win10 64-bit,
TC21.2 x64 Plat, Bld59
TC16.2 Plat, Bld54.0
TCC 3.5

August 20, 2017, 07:45:25 AM
I was right when I said "Of course I could have misunderstood the problem."   for some reason I thought the hinge was a piano hinge along the line BD, as for hinging at point D  - sorry, I don't even know what the formula is never mind how to draw it.

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August 20, 2017, 08:52:56 AM
One way (see attached screenshot):

Draw an arc through A, centered on B.
Draw a second arc, centered on C, through the point where the first arc intersects BC.
Split AC st the point where the second arc intersects AC.
Point D is the midpoint of the segment between the split point and point A.

Henry H

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August 20, 2017, 09:33:35 AM
A problem I see is the thickness of the frame material is not being accounted for.

Referencing the images in the first Post, if the frame is to be folded inward-- such that point-D gets closer to point-B when folded-- the folded-up frame will never fully fold flat.  The same can be seen to be the case in Henry's folded-up example.

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Alvin Gregorio
Intermittent TurboCAD user since yr. 2000 (ver6.5).  No formal CAD Training.
---TurboCAD: V21PP; V20.2PP; V19DL; V11.2Pro; Windows-7-Pro/64-bit; Intel-Core-i3 CPU; 2.27ghz; 4GB RAM; Intel HD Graphics (CPU based)

August 20, 2017, 11:27:16 AM
A problem I see is the thickness of the frame material is not being accounted for.

Referencing the images in the first Post, if the frame is to be folded inward-- such that point-D gets closer to point-B when folded-- the folded-up frame will never fully fold flat.  The same can be seen to be the case in Henry's folded-up example.

I suppose it's a matter of definition. If "folded flat" means that A, B, C, & D shall lie along one straight line, then it can be done by suitably shaping the various members so that they will lie above/below/beside one another when folded. Not a trivial design task, but neither is it unreasonably difficult.

Henry H

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August 20, 2017, 01:39:13 PM
...For example...

Henry H

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August 20, 2017, 08:19:53 PM
...Or...

Henry H

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August 20, 2017, 11:35:49 PM
Thanks Henry, that is a very elegant solution - much better than the one I devised.

Alvin, your observation about the frame thickness is correct in terms of how the problem was stated; I was somewhat sloppy in my definition of "flat". What I meant was that the lengths of the hinge arms would not limit the amount frame could be folded. The actual frame folds like this.

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Gary Wooding
Win10 64-bit,
TC21.2 x64 Plat, Bld59
TC16.2 Plat, Bld54.0
TCC 3.5