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Trig/Math dudes or dudettes, need help


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Can someone figure out the straight line location to the face rotated at 5° for me on this drawing that I have highlighted?

Please and thank you.


2725_4727b188789916837a61aeeb81a43d36.png
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Hmm I'm not sure that simple trig formula finds the dimension you are looking for if I understand you correctly. It looks to me to be much more complex than that. If you are trying to get that distance normal to the 5 degree angle where it intersects that center bore at the front plane, there's a lot of other things to consider. First you need to figure out where that point at 2.375 would be at the intersection of the 5 degree angle and the .625 width. Then from there you can figure out the distance normal to the 5 degree angle. So you can start with TAN(5)= x/.625. solve for X to find the difference between the 2.375 length and what it would be at the .625 intersection point. X=0.05468. so at that point the distance up to the centerline at 5 degrees and from the .625 intersection point would be 2.32032. Now we just need to find that normal to the 5 degree angle. COS(5)=y/2.32032. y=2.31149. So your distance is 2.31149.

Another way to express this would be COS(5)*(2.375-TAN(5)*.625).

Here's the dimension done in solidworks for clarity.

trig.JPG

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