 # Exponentiation math operator

Hi!

Is there a possibility to get an exponentiation math operator?
Something like “^” (2^3=8).

Was going to ask the same question today before I searched the forum and found your request. I want to calculate Heat Index. The formula looks like this:

HI = -42.379 + 2.04901523T + 10.14333127RH - .22475541TRH - .00683783T^2 - .05481717RH^2 + .00122874T^2RH + .00085282TRH^2 - .00000199T^2RH^2

HI = Heat Index
F = Temperature
RH = Relative Humidity (as a percentage)

Since the exponents are all squared I can just replace them with this:

HI = -42.379 + 2.04901523T + 10.14333127RH - .22475541TRH - .00683783TT - .05481717RHRH + .00122874TTRH + .00085282TRHRH - .00000199TTRHRH

Not to figure out the actual expression on the Facer Creator…

So the calc is not working right. Got everything in and the operators and expressions work. But the value seems wrong. I found another one online for excel sheets that may work:

=ROUND(16.923+((1.85212*(10^-1))A2)+(5.37941B2)-((1.00254*(10^-1))A2B2)+
(9.41695*(10^-3)(A2^2))+(7.28898(10^-3)(B2^2))+(3.45372(10^-4)(A2^2)B2)-
(8.14971
(10^-4)A2(B2^2))+(1.02102
(10^-5)(A2^2)(B2^2))-(3.8646*(10^-5)(A2^3))+
(2.91583
(10^-5)(B2^3))+(1.42721(10^-6)(A2^3)B2)+(1.97483(10^-7)A2(B2^3))-
(2.18429
(10^-8)(A2^3)(B2^2))+(8.43296*(10^-10)(A2^2)(B2^3))-(4.81975*(10^-11)(A2^3)(B2^3)),0)

where A2 is the temperature and B2 is the relative humidity

Working on simplifying the exponents and adding in tags.

For simplification, I suggest using Wolfram Alpha: http://www.wolframalpha.com/
Make sure that your formula has all brackets and that you changed all tags into Xs and Ys and just put in into Wolfram, it should give you a simpler version if one is mathematically possible while being accurate.

I think I figured it out. With testing it appears to work so far:

(round(-42.379+(2.04901523*#WCT#)+(10.14333127*(#WCHN#))-(0.22475541*#WCT#(#WCHN#))- (0.00683783(#WCT##WCT#))-(0.05481717((#WCHN#)(#WCHN#)))+ (0.00122874(#WCT##WCT#)(#WCHN#))+(0.00085282*#WCT#((#WCHN#)(#WCHN#)))- (0.00000199*(#WCT##WCHN#))((#WCT#*(#WCHN#))))) The Relative humidity needed to be in whole numbers, not a percentage. That’s what was throwing me off.

Ok, I’m quitting for tonight. Its still giving me the wrong value by a few degrees. I used the Worlfram site and cleaned up the text a little but something is still off:

(round(-42.379+(2.04901523*#WCT#)+(10.14333127*#WCHN#)-(0.22475541*#WCT##WCHN#)- (0.00683783(#WCT##WCT#))-(0.05481717(#WCHN##WCHN#))+ (0.00122874(#WCT##WCT##WCHN#)+(0.00085282*#WCT##WCHN##WCHN#)- (0.00000199*(#WCT##WCHN##WCT#*#WCHN#)))))

This is the calculation I should be using:
http://math.info/Misc/Heat_Index/
http://www.wpc.ncep.noaa.gov/html/heatindex_equation.shtml

When I have T=85 and RH=50, I get HI=86. This is correct.
When I have T=75 and RH=94, I get HI=73. This is incorrect. Should be 77.

I’m starting to think this may be impossible. Even the government NOAA calculator produces weird results. At 120F and 94% H the HI is over 350 degrees. searched a few more websites and they state if temp is less than 80F or if RH is less than 40%, the calculation will not work. There also appears to be an upper limit as well. http://www.climateknowledge.org/heat_waves/Doc1006_Steadman_Heat_Stress_Index_JApplMet_1979.pdf
IDK, I like the idea of having it in a weather app, but i need to find a more reliable calculation.

i guess i could add a conditional to the front of this to exclude these values and redisplay the #WCT#.