I am trying to create this double angle feature, but am having trouble. I am guessing you need to make a reference plane and go from there but everything I try has failed.
What version of SolidWorks are you using?
Attach the *.sldprt file of your starting attempt here.
Did I pass the test?
Make the block with the 15° angled end.
Create a reference axis located as the dashed 25° line.
Use this axis to create a reference plane rotated 10°.
Use this plane as the sketch plane for a cut extrude.
I would mark it out with a 3D sketch first. After that, many possibilities.
Roland Schwarz wrote: I would mark it out with a 3D sketch first. After that, many possibilities.
Roland Schwarz wrote:
Since this is a compound angle you can't simply plop down the angles for the 10 degree angle as it won't give you what you want, or at least show in your drawing.
What I did was to create the reference plane for the 15 degree angle and cut that. I then created a 3d sketch and plopped points at the 1.285 on one edge and 2.25 at the other. The 2.25 is not on the surface but created by the 3d sketch and square to the world. If you just use the 10 degree angle either on the surface created by the first cut or square to the world the angle at the cut will not be 10 degrees. What I did was calculated what the short leg of the triangle was based on the top view and placed a point there.
If you actually have to make this block old school fashion, compound sine plate or sine plate on a sine plate you will have to compensate one of the angles by a formula you can find in the machinist handbook. It's something like the sine of the first angle times the tangent of the second angle or something like that. That compensation, if you calculated it out should actually leave you with the 14.52 degree angle I show in the drawing.
Yea Søren Stærk shows how there was not quite enough info in the original two views presented.
this could be the intent!!!
Let's see your 2D drawing (take the 10° out to several decimal places).
Just another way to do it.
The way I See It...
Using only the dimensions, a vertex type chamfer works.
That's a very cool solution. However your 1.285 would also have to be compensated to 2.3294 in order to get the right solution.
You would have to trig out the .3967 and the 2.3294 to get the correct end dims using this method though.
Good point Matt.
Three split lines and a Delete face with fill option works as well. I think it works out without the trig.
Split line three only requires connecting the points from the other split lines.
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