- Joined
- Nov 9, 2017
- Messages
- 174
Sure looks like 0.500" in the X direction to me. No trig tables handy, but y=0.375+0.5*sin(0.5/0.625)?
This is a little more challenging than that.
I would not insult you guys with a simple problem
Sure looks like 0.500" in the X direction to me. No trig tables handy, but y=0.375+0.5*sin(0.5/0.625)?
You know you are old school if you know how to use one of these.
View attachment 226519
I have one mounted on my toolbox.
You know you are old school if you know how to use one of these.
View attachment 226519
I have one mounted on my toolbox.
Can i use my slide rule?
Well, if you find the angle at 0,0 in the right triangle (1.0 long, .375 high) then you should have the three sides of the oblique triangle (hyp of earlier tri, [.625/2]+ .625, .625) and so you can get the angle of it at the 0,0 point added to the first tri totaled will give you the angle which with the side ([.625/2]+ .625) you can derive the length of the side in X.
Maybe..
I could lay it out in qcad or define the geometry in apt360, or do the math but I don't know where I left my log tables (logarithm table).
I'll leave it for the youngsters to figure out, I am up to my trig tables in trying to configure a iptables/shorewall firewall computer before the hard drive in the old on pukes.....