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Offline Topaz  
#1 Posted : 11 April 2018 16:16:36(UTC)
Topaz

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Hi,

i calculated the Reynoldsnumber where i need the revolutions per minute. What struck me was the conversion from rpm to 1/s -> 60rpm = 1 rps = 1 s^-1. According to smath 60rpm = 6,28 s^-1.

Is there a bug or are i'm doing the math wrong?



thanks a lot!

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Offline Davide Carpi  
#2 Posted : 11 April 2018 16:38:51(UTC)
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please read here (SS-2395) (you can vote and comment it), here or here (and other...)

base unit of a revolution is coded as 2pi radians (not 1), hence the multiplication factor.

2018-04-11 15_33_50-SMath Studio - [Page1_].png
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thanks 1 user thanked Davide Carpi for this useful post.
on 11/04/2018(UTC)
Offline alyles  
#3 Posted : 11 April 2018 16:41:00(UTC)
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This is because the standard unit in SMath for rotation is radian = 2*pi which is ~6.28. If you look at the definition of rpm it is pi/30 1/s or 2*pi/60.

Untitled.png

In order to get the right result you need to divide by 2*pi. What i would suggest is creating your own unit for RPMs as done in the attached.

rpm.sm (3kb) downloaded 43 time(s).



Edit: Davide beat me to it.
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Offline Jean Giraud  
#4 Posted : 11 April 2018 17:42:37(UTC)
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Originally Posted by: alyles Go to Quoted Post
This is because the standard unit in SMath for rotation is radian = 2*π

... and this is because trig functions are numerically approximated in radian.
Those functions are native in Microsoft, used by math software as is.

Offline Topaz  
#5 Posted : 12 April 2018 10:24:30(UTC)
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thanks a lot for your quick reply, good to know Good
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