SOLUTION: The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula

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Question 1208108: The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is given below.

Let W = a piecewise function.

Top portion: t if 0 <= v < 1.79
Middle portion:
33 - [(10.45 + 10sqrt{v} - v)(33 - t)]/(22.04)
if 1.79 <= v <= 20
Bottom portion: 33 - 1.5958(33 - t) if v > 20
where v represents the wind speed (in meters per second) and t represents the air temperature (°C).
Compute the wind chill for the following:
(a) An air temperature of 10°C and a wind speed of 5 m/sec
(b) An air temperature of 10°C and a wind speed of 25 m/sec

Found 2 solutions by Boreal, math_tutor2020:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
a. 33- (10.45+10sqrt(5)-5)(23)/22.04) =33-29.02 or 3.98 C (4 C)
b. 33-1.5958(23)=-3.70 C.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Piecewise functions can be a strange beast.
I think of them as a Frankenstein's monster because we glue together various parts of other functions.

The function will change its identity based on what the input is.
The input in this case is variable v (t is the parameter and appears to be fixed to t = 10).

We have 3 scenarios
  • If 0 <= v < 1.79, then W(v) = t
  • If 1.79 <= v <= 20, then W(v) = 33 - (10.45 + 10*sqrt(v) - v)*(33-t)/22.04
  • If v > 20, then W(v) = 33 - 1.5958(33-t)
-------------------------------------------------------------------------

Part (a)

t = 10
v = 5

v = 5 fits the interval 1.79 <= v <= 20, so we use the middle piece.

W(v) = 33 - (10.45 + 10*sqrt(v) - v)*(33-t)/22.04
W(5) = 33 - (10.45 + 10*sqrt(5) - 5)*(33-10)/22.04
W(5) = 3.977965752044
W(5) = 3.98
The wind chill is approximately 3.98 degrees Celsius.

-------------------------------------------------------------------------

Part (b)

This time we have
t = 10
v = 25

v = 25 fits the interval v > 20 aka 20 < v < infinity.
So we'll use the 3rd piece of this piecewise function.

W(v) = 33 - 1.5958(33-t)
W(25) = 33 - 1.5958(33-10)
W(25) = -3.7034
The wind chill is roughly -3.7034 degrees Celsius.


Side notes:
10 degrees Celsius = 50 degrees Fahrenheit
3.98 degrees Celsius = 39.164 degrees Fahrenheit
-3.7034 degrees Celsius = 25.33388 degrees Fahrenheit
5 m/s = 18 km/hr = 11.1846815 mph
25 m/s = 90 km/hr = 55.9234073 mph
The decimal values are approximate.