Calculating windchill...
 

If the temperature is 21°F and I'm cycling at 12mph into a 24mph headwind, is the windchill calculated using 36mph?

Similarly, if I'm traveling at 24mph with a 24mph tailwind is windchill negated?

my guesses

yes, 21 - 36 = -15 degrees

yes

winter is coming back to my area Brrrr!

You can't subtract windspeed from temperature, can you?

Windchill and cycling ...
 

(Initially, I posted this query in the winter cycling sub-forum. But this sub-forum seems to get a lot more traffic)

If the temperature is 21°Farenheit and I'm cycling at 12mph into a 24mph headwind, is the windchill calculated using 36mph?

Similarly, if I'm traveling at 24mph with a 24mph tailwind is windchill negated?

Looks as if you have an answer in "Winter Cycling."

The only reply in that thread makes no sense because Velocity cannot subtracted from nor added to Temperature.

i.e. 21°F - 36mph ≠ -15°F

Yes, use your airspeed.

When I still had a motorcycle I used a wind chill chart like this one. Note it has the same formula mentioned above
https://www.weather.gov/safety/cold-wind-chill-chart
https://www.weather.gov/images/safety/windchill21.gif

As windchill is a relative expression of an absolute (that is, temperature) being outside @ 21ºF riding into a 24mph wind, the temperature is 21ºF. Any exposed skin will experience the sensation of temperature around 5ºF. This is above the threshold for frostbite. Riding 12mph into a 24mph headwind is absolutely not 12 + 24. You're not somehow going faster into the wind than the wind is hitting you. Windchill would be based on the wind speed, and should be essentially moot, as I don't know why anyone would be going out with bare arms and shorts in 21ºF temps to begin with.

Meanwhile, I'm sitting here thankful I can never experience such conditions here. I don't think I've ever seen it windy with the temps below 40ºF or so. Most of our wind comes from the desert, so if it's windy, it's warming.

My understanding is that this is the formula for determining windchill for stationary objects.

What I'm unsure of is if bike velocity and windspeed can be combined by simple addition to determine windchill "in motion", or if there is an exponential factor to consider as when two cars collide head on.

When two cars collide head on, each traveling at 15mph, the force of their collision is not the sum total of their combined velocities. Is it not the same when riding into a headwind?

If your relative velocity is slower than the windspeed, the wind is never going to be hitting you faster than the wind is blowing.

If the wind speed is 20mph and your speed is 10mph, the wind speed is 20mph.
If the wind speed is 10mph and your speed is 20mph, the wind speed is 20mph.

^^please reconsider. If on a still day at 40F, you set out at 60 mph (back to my motorcycling example), you are going to be very, very cold.

I can't help but think that you would rethink your reply if you'd ever ridden a bike at any speed into 25mph headwind in sub-freezing temperatures, and then compared that sensation by turning around and riding 25mph with a 25mph tailwind in the same temperature. The difference between the two experiences is immense.

Whether bare naked or buried beneath several winter layers, loss of heat is an inevitability in certain conditions, especially if one of the factors you have to consider before riding is how long you can last in the wind and sub freezing temperatures should you have to repair a flat tire far afield.

AllWeatherJeff speed is not the same as momentum

They are if my momentum is carrying me along at 15mph into a 25mph head wind.


Are you saying that windchill is the same, regardless of whether or not one is moving or standing still?

This is the essence of my initial query.

That makes sense to me. It's pretty straight forward. Windchill is the effect of wind on cooling down the body. With the wind at 0 mph, your net wind is your speed, and that's how your windchill is calculated. Years of riding in winter tells me this also.

What is your experience of the difference between riding a motorcycle 60mph into a 25mph headwind versus riding 60mph with a 25mph tailwind?

Thanks. I wasn't sure if there was some kind of exponential factor of momentum nvolved as when two cars collide head on.

Riding 10mph into a 25mph headwind, wind speed 25mph.
Riding 25mph with a 25mph tailwind, wind speed 0mph.

And I reiterate: the wind speed is the wind speed; if you were to ride 25mph into a 25mph headwind, that doesn't magically make the wind speed 50mph. That's not how air works.

according this calculator I was waaay off, should have been -1.3 degrees when I added wind speed against you with your traveling speed (36 mph wind against you). don't know why that wouldn't apply. but this is not my area of expertise. gotta explain it to me like I'm 6

https://www.weather.gov/epz/wxcalc_windchill

Maybe, maybe not. I'm trying to determine windchill factor not windspeed, per se. But I can guarantee that standing still in a 25mph wind versus traveling at 25mph into that headwind at 21°F are two entirely different experiences.


I may be mistaken, but what sounds implausible is that your response implies that at 21°F with a 25mph wind that a body will lose heat at the same rate whether standing still or in motion.

Merged duplicate thread. Please do not cross post.

Oh, for sure. I mean, traveling 25mph into a 25mph wind on a bicycle would take something between 1,700 and 2,000 watts. You'd be producing so much heat you could take your shirt off.

The windchill (relative) is as I said much earlier, around 5ºF, give or take. That's cold. If the wind chill were recalculated to say, -1ºF, would that make an earth-shattering difference?

Roger.
WilCo

Look up the windchill equation .... it is a rather hilarious exercise of "why in the world is that the case???!!!"