8 Replies Latest reply: Feb 11, 2017 9:53 PM by Joe Easley

Square Root and Wind Chill

Hello - i need to calculate the wind chill in my application.  i have the formula pasted below but i don't know how to write this in the application.  i am fine setting it up as a variable or whatever.  i have Temperature in my data as [Temp] and windspeed as [Wind], in Fahrenheit and mph respectively.  Thanks in advance for your help!Q!

Wind chill temperature = 35.74 + 0.6215T - 35.75V (**0.16) + 0.4275TV(**0.16)

In the formula, V is in the wind speed in statute miles per hour, and T is the temperature in degrees Fahrenheit.

Note: In the formula, ** means the following term is an exponent (i.e. 10**(0.5 ) means 10 to the 0.5 power, or the square root of V), - means to subtract, + means to add. A letter next to a number means to multiply that quantity represented by the letter by the number. The standard rules of algebra apply.

• Re: Square Root and Wind Chill

May be like this

Wind,

35.74 + (0.6215 * Temp) - pow(35.75 * Wind, 0.16) + pow(0.4275 * Temp * Wind, 0.16) as [Wind Chill]

• Re: Square Root and Wind Chill

Thanks for the fast response.  the calculation is going the wrong way, wind chill should always be less than temp, but see below for a few of my results:

• Re: Square Root and Wind Chill

I think there was a slight misunderstanding of the expression. I checked online and this seems to work:

Table:

35.74 + (0.6215 * Temp) - (35.75 * pow(Wind, 0.16)) + (0.4275 * Temp * pow(Wind, 0.16)) as [Wind Chill];

Temp, Wind

37, 5

37, 8

43, 11

46, 4

46, 6

52, 2

];

• Re: Square Root and Wind Chill

so that worked perfectly.  and then i ran into another problem.  it turns out there is a point at which the wind chill is replaced with the heat index.  the logic is if the temp is greater than 80 and relative humidity is greater than 40%, then use heat index for apparent temp, else use windchill for apparent temp.

so what i would like to do is figure out how to have individual windchill (solved) and heat index variables, as well as an apparent temp variable that factors the logic.

heat index formula below (wikipedia entry).  you have been SO helpful, and i appreciate anything else youre willing to help with.

i have temp in the model as well as humidity (CmpHumidity)

• Re: Square Root and Wind Chill

The expression in QlikView should be like this:

-42.379 +

(2.04901523 * Temp) +

(10.14333127 * CmpHumidity) +

(-0.22475541 * Temp * CmpHumidity) +

(-683783 * pow(10,-3) * pow(Temp, 2)) +

(-5.481717 * pow(10, -2) * pow(CmpHumidity, 2)) +

(1.22874 * pow(10, -3) * pow(Temp, 2) * CmpHumidity) +

(8.5282 * pow(10, -4) * Temp * pow(CmpHumidity, 2)) +

(-1.99 * pow(10, -6) * pow(Temp, 2) * pow(CmpHumidity, 2))

Not sure how this is related to Wind Chill... Can you elaborate on that part?

• Re: Square Root and Wind Chill

when measuring "apparent temperature" for analysis, sometimes you need windchill, sometimes you need heat index.  i learned this as i was getting deeper into the app.

• Re: Square Root and Wind Chill

So this information will reside in one field with flags or are you looking to create two fields? One for windchill and one for heat index?

• Re: Square Root and Wind Chill

if

(CmpTemp >= 70,(\$(vHeatindex)),

if(CmpTemp <= 60,(\$(vWindchill)),CmpTemp))

i can now grap HeatIndex OR WIndchill independently OR run one field as apparent temp.

you have been such a tremendous help!!