windseaker1
Mechanical
- Jan 7, 2009
- 34
I ran across this formula, but it is only good up to 180Degrees?
(.01743xrad)x degree of circle
And I beleive this is not true as well: given
Roll-out flat pattern:
OD = 6.375 material .375 steel
Cir. 6.375 x 3.1415 = 20.027
Formula: (.01745xR) (180 Degrees)x2 =
(.01745x3.1875) x (180)x2 =
(.055621875) x 180 x2 =
10.11937 x2 =
20.023875, which is the same as cir.??
A better formula appears to be: but for alum. only??
(I dont know where the .0078 factor cam from)
(.01745xR)-(.0078xT)x(180 Degrees)x2=
(.055621875)-(.002925)x180x2 =
(.052696875)x180x2 =
9.4854375x2 =
18.970875
But evern better "I hope" is AutoCad's factor formula:
(for mild steel: A36)
By AutoCAD: using .407 factor Times thickness
= .0763 of the 3/8
Then offset from the insideat this factor,
you get: Length was a total of 19.329”
This seems closer to a real stretch-out pattern!
Can anyone confirm??
(.01743xrad)x degree of circle
And I beleive this is not true as well: given
Roll-out flat pattern:
OD = 6.375 material .375 steel
Cir. 6.375 x 3.1415 = 20.027
Formula: (.01745xR) (180 Degrees)x2 =
(.01745x3.1875) x (180)x2 =
(.055621875) x 180 x2 =
10.11937 x2 =
20.023875, which is the same as cir.??
A better formula appears to be: but for alum. only??
(I dont know where the .0078 factor cam from)
(.01745xR)-(.0078xT)x(180 Degrees)x2=
(.055621875)-(.002925)x180x2 =
(.052696875)x180x2 =
9.4854375x2 =
18.970875
But evern better "I hope" is AutoCad's factor formula:
(for mild steel: A36)
By AutoCAD: using .407 factor Times thickness
= .0763 of the 3/8
Then offset from the insideat this factor,
you get: Length was a total of 19.329”
This seems closer to a real stretch-out pattern!
Can anyone confirm??