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

    Square Root and Wind Chill

    Joe Easley

      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
          Sunny Talwar

          May be like this

           

          LOAD Temp,

                    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
              Joe Easley

              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:

               

              Capture.PNG

                • Re: Square Root and Wind Chill
                  Sunny Talwar

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

                   

                  Table:

                  LOAD *,

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

                  LOAD * INLINE [

                      Temp, Wind

                      37, 5

                      37, 8

                      43, 11

                      46, 4

                      46, 6

                      52, 2

                  ];

                   

                  Capture.PNG

                    • Re: Square Root and Wind Chill
                      Joe Easley

                      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)

                       

                      heat index.PNG