Wind chill calculation debate

I am in Canada, Montreal and yesterday’s with a temperature of + 1Celsius and wind in the range of 10 km/hr we should see a Temp feel different but it is always the same as the outside temperature. So the Temp feel doesn’t show the wind shill or the humides when it is very hot and humid. Any solutions coming soon ???l

Indeed, during the cold seasons, when the temperature is at or below 10°C (50°F) and wind speed is at or above 4.8 kilometres per hour (3.0 mph) then the wind chill factor should be provided on the same line as the temperature/dew point/humidity.

But, during the hot/warm seasons, when the temperature is at or above 21°C and the humidity is at or above 20% then the humidex factor should be provided on the same line as the temperature/dew point/humidity.

The humidex (which only used in Canada) has different results than the heat index which is used in other countries.

Thank you but you said Indeed !!!

Indeed it doesn’t work as I said ?

Yesterday it was 1C and wind at 10km/h and no Wind Chill in the device while the wind chill was -3C.

So INDEED it doesn’t work !!!


Envoyé de mon iPhone

Wind chill and Heat index work as designed,

// Wind Chill Temperature //
// e.Observations.prototype._getWindChill
WeatherCalc.calcWindChill = function(T, W) {
	// (air temp, wind speed)
	var F = _convertCToF(T);
	var V = WeatherCalc.convertWindSpeed(W, 'mph');
	if (V < 5 || F > 50) return parseFloat(T);
	var WC = 35.74 + (0.6215 * F) - ( 35.75 * Math.pow(V, .16) ) + ( 0.4275 * F * Math.pow(V, .16) );
	return _convertFToC(WC);
// Heat Index Temperature //
// e.Observations.prototype._getHeatIndex
WeatherCalc.calcHeatIndex = function(T, RH) {
	// (air temp in C, humidity)
    var F = _convertCToF(T);
    RH = parseInt(RH, 10);
    if (RH < 40 || F < 80) return T;
    var r = 61 + 1.2 * (F - 68) + .094 * RH,
        s = .5 * (F + r),
        n = 0;
    if (s > 79) {
        if (n = 2.04901523 * F - 42.379 + 10.14333127 * RH - .22475541 * F * RH - 6.83783 * Math.pow(10, -3) * Math.pow(F, 2) - 5.481717 * Math.pow(10, -2) * Math.pow(RH, 2) + 1.22874 * Math.pow(10, -3) * Math.pow(F, 2) * RH + 8.5282 * Math.pow(10, -4) * F * Math.pow(RH, 2) - 1.99 * Math.pow(10, -6) * Math.pow(F, 2) * Math.pow(i, 2), RH <= 13 && F >= 80 && F < 112) {
            var o = (13 - i) / 4,
                l = Math.sqrt((17 - Math.abs(F - 95)) / 17);
            n -= o * l;
        } else if (RH > 85 && F >= 80 && F <= 87) {
            var c = (RH - 85) / 10,
                u = (87 - F) / 5;
            n += c * u;
    } else n = s;
    return _convertFToC(n);
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are you sure? I ( and thought that windchill is not depended on humidity, so it would make sense to show windchill even when humidity <40 (but of course only at low temperature (<50F) and some wind (>5mph))

I didn’t mention that it is dependent on humidity, just how it is designed to work.

I’m also not stating to correctness.

I will state that, I believe there may be changes in the future to allow for calculations for different geopolitical areas.

I tried to formulate it carefully, and not claiming that you stated that it was depending on humidity. I stated that I think that it is not depending on humidity and therefore it would make sense to show windchill even when humidity <40. (provided the other conditions are met)
if windspeed > 5mph and temperature <50F = windchill
would be more likely. So that’s why I’m wondering if you are sure? Where did you find this?

This is what WeatherFlow is using.

Thus: Wind chill and Heat index are working as designed.

You can read more here:

And another discussion here:

And a statement from David here:

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I don’t believe the humidity is being used. It doesn’t mathematically contribute anything to the windchill.

The only thing I can think of, that people might be getting confused with is, ‘apparent temperature’ humidity is needed along with other measurements to get a figure.

The WF link that you provided doesn’t mention humidity being linked with wind chill. lol

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Well. I don’t know what else I can convey to you. I have spent many months studying every line of the code.

You are welcome to read it yourself. LOL LOL LOL.

I don’t need to read anything. You posted WeatherFlow’s own FAQ regarding how the windchill is calculated, which rightly shows no mention of humidity being used. So, I’m not sure where you plucked that line from?

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You are correct. You don’t have to read anything.

It was quite a straightforward question in my previous post. Where did you pluck line from that shows humidity needing to be above 40% to show the windchill? The humidity only applies to the heat index. Sheesh.

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Editing your answer and replacing
If humidity >= 40 AND temperature < 50°F = Wind Chill
with the actual formula was a smart thing to do. Now I have no problems, no test for humidity at all :wink:

Note that @ledoux_mi stated that at 1°C and wind speed of 10km/h he noticed that windchill is not correctly seen. Not sure if that was in the app or in the webbrowser. The code for the webbrowser seems ok though and when I looked on the map for a cold and windy station, the wind chill was calculated correctly.

As an extra remark, Weatherflow points to for the formula. But uses a limit of 3mph instead of 5mph, so there might be a small implementation problem with the formula.


Oh, that… Yes, that was an accidental from a cut and paste. Is that what some are getting riled up about? So I’m not allowed to make typing errors…

but no worries… I won’t meltdown in public.


I wasn’t getting riled about a typing error. I was stating that humidity does not apply to calculating windchill.

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That is interesting. Here is what I find…

The wind chill calculator only works for temperatures at or below 50 ° F and wind speeds above 3 mph.

So WeatherFlow and has a 1 mph difference in calculating.

Do you have a reference at that states 3 mph or above?

That’s exactly what I found as well. “The wind chill calculator only works for temperatures at or below 50 ° F and wind speeds above 3 mph.”
or here

But the limit that is used in weatherflow’s code you just posted is 5 mph.

If it is less than 5 it does not calculate. So it has to be greater than 4.9999999999. So yes. Technically 5.

@WFstaff Will you please comment on this.