Octagon Cut
Octagon Cut
Octagon Cut
I am trying to figure the angles needed to cut/weld tube steel into an oblong 1'x7' octagon.?
this one was kinda tough ... I assume that an oblong 1'x7'
octagon is what you get when you "tilt' a stop-sign shaped
thing at an angle to where the width is 7 feet and the
"perceived" height is 1 foot.
If so, cut :
2 pieces = 2.899495 feet
2 pieces = 0.414213 feet
4 pieces = 2.071068 feet
the angles (between the pieces) are 163.7398 degrees
and 106.2602 degrees.
The angles (at the 'ends' of the pieces) are:
4 pieces = 2.071068 feet ("square-off": ie "zero degrees)
2 pieces = 2.899495 feet (16.3 deg 'off-normal')
2 pieces = 0.414213 feet (73.7 deg 'off-normal')
By "off-normal". I mean that a 'square-off' cut is zero
degrees off-normal ... the two longer pieces are fairly
"blunt" and the two short pieces are pretty "sharp", in
that if you put these cuts at the end of a spear, they
would form a sharp point.
if you put the pieces on the floor, you will know
where the angles go.
good luck ...
here's the BASIC program to determine these values:
/*************************/
pii = 355 / 113
a = 1 + 2 ^ .5
PRINT "a = "; a
b = 1 / a
PRINT "1/a = "; 1 / a
b7 = 7 * b
PRINT "b7 = "; b7
c = (7 - b7) / 2
d = (1 - b) / 2
PRINT "a,b,c,d="; a; b; c; d
thet = ATN(d / c)
PRINT "theta = "; thet
dthet = 360 * (thet / pii)
PRINT "dthet="; dthet
alpha = 180 - dthet
PRINT "alpha = "; alpha
psi = 90 - dthet
PRINT "psi = "; psi
beta = 180 - psi
PRINT "beta ="; beta
e = (c ^ 2 + d ^ 2) ^ .5
PRINT "e = "; e
Beautiful Simulated Diamond Octagon cut