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Nov 26th, 2001, 11:26 PM
#1
Thread Starter
Addicted Member
Rad to Deg
Hello,
How can I transfert a Rad number to deg?
Thanks for any help.
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Nov 27th, 2001, 01:43 AM
#2
transcendental analytic
deg=rad*57.295779513082320876798154814105
the constant is 180/pi
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Nov 27th, 2001, 03:01 AM
#3
Hyperactive Member
Multiply it by 360/2pi.
(That's probably what Ked said, but I can't see it!)
Test: 2pi rad = 2pi * 360 / 2pi = 360. OK
pi/4 rad = pi/4 * 360 / 2pi = 360/8 = 45. OK
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Nov 27th, 2001, 03:44 AM
#4
PowerPoster
Originally posted by Jim Brown
(That's probably what Ked said, but I can't see it!)
 How do u know that he responded?? How does this ignore thing work.. dont explain here and ruin the thread ... maybe answer in ur ignore thread... i am confused... more than usual
And just so that i am not accused of chit chatting...
You are both right it is 180 / pi 
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Nov 27th, 2001, 12:26 PM
#5
Hyperactive Member
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Nov 27th, 2001, 12:37 PM
#6
transcendental analytic
OMG  I think it's rather (2*Srt(45/pi))^2 
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Nov 27th, 2001, 01:50 PM
#7
Or, more precisely:
180/SqrRoot[6*Sum{n = 1 to Infinity}(n^(-2))]
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