From: no@spam.net
Newsgroups: sci.math
Subject: Re: windchill factor
Date: 15 Oct 1997 21:36:14 GMT

> A student asked me if I knew the formula used to determine wind chill.
>
> I'm sure it's in some exercise in some precalculus textbook, but I have
> not the time to search it out at present.  It seemed the sort of thing
> someone on group would have immediately to hand, which is the quicker
> course of action for me to take.  Please don't waste any time searching
> for something that I can't be bothered to search for myself, but if you
> do happen to know the formula, I'd be grateful if you'd pass it on.

Let v be the maximum of the wind speed and 4 miles per hour in mph (so
if the wind speed is 2 mph, then v=4. If the wind speed is 6 mph, then
v=6).

Then the effective wind chill temperature, T_eff (for humans) is given
in Fahrenheit by:

                      T_eff = 91.4 - a(v)*(91.4-T)

or                   T_eff = (1-a(v))*91.4 + a(v)*T

where T is the (outside) temperature and a(v) is given by:

           a(v) = (10.45 + 6.68608*SQRT(v) - .44704*v)/22.034

(or, approximately,

                 1 - a(v) = (SQRT(v)-2)*(SQRT(v)-13)/49)

Note that for v>4 (and let's be reasonable... v<169), 1-a(v) is negative
(a(v)>1) and for T<91.4 (Fahrenheit),

          T_eff = (1-a(v))*91.4+a(v)*T < (1-a(v))*T+a(v)*T = T
                                    ^^^^^^^^
                            since (1-a(v))<0 and 91.4>T

so the effective wind speed temperature is smaller than the outside
temperature.

                         -=-=- Derivation -=-=-

Consider a system which loses heat to the environment at rate Q'
('=derivative with respect to time).

By Newton's Law of Cooling, for conduction/radiation, there is a term
proportional to the difference in temperatures,

                     A(environment)*(T_object - T)

which contributes to Q' (T=outside temperature, T_object being the
temperature of the object in which we are interested). A(environment) is
some constant giving the rate at which heat is transferred from the
object... Newton's Law of cooling uses T_object' instead of Q' but this
is just another constant, the specific heat, to transform from Q' to
T_object', so T_object'=(A(environment)/Specific_heat)*(T_object-T),
again, just a constant times (T_object-T) as you may have seen Newton's
Law of cooling).

There may be other terms reflecting some heat loss other than through
conduction (to an item at a different temperature), convection,
radiation... let me call that B(environment) so we get:

           Q' = A(environment)*(T_object-T) + B(environment)

In the summer, B(environment) is important, reflecting the loss of heat
due to the evaporation of sweat (the rate of sweat evaporation times the
heat of vaporization of sweat) and the rate of evaporation depends upon
humidity, whether one's skin is exposed to the free air and wind speed.

(note that if T_object<T, that is, say, the outside temperature is
 higher than the body temperature, then without the B term, Q' is
 negative, that is, the object is picking up heat, rather than radiating
 it... when the outside temperature is higher than the body temperature
 the body cannot get rid of the heat it generates without sweating... so
 no sweating causes over heating and sun stroke/ heat exhaustion).

For wind chill (winter) we ignore B(environment) (sweating) and simply
take:

                  Q' = A(environment) * (T_object - T)

A(environment) reflects insulation, wind, etc.

We take A(environment) as a product of terms, one for the effect of the
wind (a(v)) times other effects (e.g. if a(v)=1.2, then the wind is
causing heat to be lost 20% faster than with no wind) [a(v)=1 for no
wind]. Thus take, A(environment)=a(v)*c(environment) where
c(environment) reflects such things as the insulation one is wearing,
etc. (but NO wind... c(environment) is the coeficient for heat loss
without any "wind chill") and if there were no wind, one would have
a(v)=1 and would lose heat at rate c(environment)*(T_object-T).

a(v) is the actual "Wind Chill *FACTOR*" (the factor by which heat is
more rapidly lost due to a wind). If, for example, a(v)=1.25 (an
increase of heat loss of 25% due to the wind) and one wanted to wear
enough clothing to eliminate the effect of wind chill, one would want to
wear clothing which insulates one by cutting the heat loss by 20% (a
factor for insulation being 1-.2 for a reduction due to insulation of
20%)

[I believe that some clothing is rated in the percentage of heat loss
 eliminated and perhaps Canada gives the wind chill as an actual factor,
 rather than an effective wind chill temperature ?? I am not sure about
 that though.]

so that:

      c(environment)=(factor_for_insulation)*c(without_insulation)

and the rate of heat loss would be

                    a(v)*c(environment)*(T_object-T)

which equals (with these numbers)

             1.25*(1-.2)*c(without_insulation)*(T_object-T) or just
               1.25*.8*c(without_insulation)(T_object-T) or
                  c(without_insulation)*(T_object -T)

and the insulation would be just enough to cancel the effect of the wind
speed resulting in a heat loss just reflecting the object's temperature
and the outside temperature (as if there were no wind or insulation...
the effects of those two cancelling).

For a human, what is T_object? Not the internal body temperature, but
skin temperature, 91.4_Fahrenheit.

What is a(v)? I believe this is determined experimentally (formula given
above) (note that in the formula for a(v), all wind speeds lower than
4 mph give a value for v of 4 and a(4)=1 so wind speeds of 4 mph or
lower have no effect).

Now... suppose we have a wind blowing, with rate of heat loss of
a(v)*c(environment)*(T_object-T) where T_object=91.4 (in Fahrenheit). At
what temperature would be have the same rate of heat loss if there were
no wind, all else being the same? This is the effective temperature (the
temperature which would produce the same result as the actual
wind/temperature combination if there were no wind).

SO... let's write this down (for heat loss from the surface of the human
body):

Actual rate of heat loss:
                      a(v)*c(environment)*(91.4-T)
to be the same as heat loss with no wind at temperature T_eff
                     1*c(environment)*(91.4-T_eff)

This gives:

                    1*c*(91.4-T_eff) = a(v)*c*(91.4-T)

and solve for T_eff to get:

                      T_eff = 91.4 - a(v)*(91.4-T)

(one can give the effect of wind chill either by giving the actual
 factor, a(v), or the effective temperature, but that effective
 temperature depends upon the object whose rate of heat loss is being
 considered as well as the outside temperature.

 One might simply say that the wind chill effect is a 25% increase in
 heat loss (a(v)=1.25)

 rather than give the effective wind chill temperature, understood as
 referring to humans, at 32_Fahrenheit as 91.4-1.25*(91.4-32)
 or 17.15_Fahrenheit (this occurs for a wind speed of about 10.71 miles
 per hour))

                           -=-=- Finis -=-=-
==============================================================================
Date: Thu, 16 Oct 1997 12:12:33 -0600
From: gnygaard@crosby.ndak.net (Gene Nygaard)
Subject: Re: windchill factor
Newsgroups: sci.math

In article <623d0e$44b$1@news.inch.com>,
  no@spam.net wrote:
>

> Let v be the maximum of the wind speed and 4 miles per hour in mph (so
> if the wind speed is 2 mph, then v=4. If the wind speed is 6 mph, then
> v=6).
>
> Then the effective wind chill temperature, T_eff (for humans) is given
> in Fahrenheit by:
>
>                       T_eff = 91.4 - a(v)*(91.4-T)
>
> or                   T_eff = (1-a(v))*91.4 + a(v)*T
>
> where T is the (outside) temperature and a(v) is given by:
>
>            a(v) = (10.45 + 6.68608*SQRT(v) - .44704*v)/22.034
>
> (or, approximately,
>
>                  1 - a(v) = (SQRT(v)-2)*(SQRT(v)-13)/49)
>
> Note that for v>4 (and let's be reasonable... v<169), 1-a(v) is negative
> (a(v)>1) and for T<91.4 (Fahrenheit),

You'd better not use this formula with wind speeds anywhere near 169
mi/h.  It is completely invalid for any wind speed above 25 m/s = 55.9
mi/h (and not realy valid above about 20 m/s).	You can see this by
calculating the wind chill factor for wind speeds of (5 + x)^2 and (5 -
x)^2 metres per second (divide by 0.44704 to get mi/h to use in your
formula).  The formula gives the same chll factor for 36 m/s as it does
for 16 m/s, the same for 64 m/s as for 4 m/s.  At 169 mi/h it gives the
same results as at 3.83 mi/h, which is below the 4 mi/h threshhold.  If
you plug in a number for a wind speed above 100 m/s (224 mi/h) the
formula will actually tell you that you are gaining heat, instead of
losing it.

More on my web page at
http://ourworld.compuserve.com/homepages/Gene_Nygaard/


> [I believe that some clothing is rated in the percentage of heat loss
>  eliminated and perhaps Canada gives the wind chill as an actual factor,
>  rather than an effective wind chill temperature ?? I am not sure about
>  that though.]
>

Canada uses the same formula you used, converted to modern SI units of
watts per square metre.  My web page (URL above)includes these formulas
also.

Gene Nygaard
==============================================================================
From: borek@mda.ca (Michael Borek)
Newsgroups: sci.math
Subject: Re: windchill factor
Date: 28 Oct 1997 03:28:40 GMT

<3444D8F8.41C6@lie.math.missouri.edu> from Michael Wodzak (wodzak@lie.math.missouri.edu) reads:
:"A student asked me if I knew the formula used to determine wind chill.

Try http://www.tor.ec.gc.ca/comm/windchil.html

-- 
|***|
(:.:) Michael Borek          "Death before Dishonour; Beer before Lunch"
|.::| MacDonald Dettwiler     borek@mda.ca  phone:(604) 278-3411 ext 2753
|..:| Richmond, B.C., Canada  My views are my own, no one else gets credit
\___/