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STAND AND FIGHT!
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Discussion Starter #1
Anybody else not like the new chill factor chart?

http://webmaster10.com/ldr/windchill.html

here's the new values



here's the older values



by the old chart, 38 degrees at 60 mph is 8 degrees chill factor, on the new chart it is 20 degrees. Both are bogus, IMHO, because both glove combos I got cold in would be good almost indefinitely in either 8 or 20 degrees in still air.

The old chart is what I keep in mind dressing overall for the ride to work, 50 degrees at 50 mph feels like 26, the new chart says 38, which isn't painfully cold, IMHO 50 at 50 w/o proper clothing is painfully cold.
 

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STAND AND FIGHT!
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Discussion Starter #3
I would think so too, seems like you can feel the difference, cold with high humidity seems to draw the heat out of you worse, I think.
 

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I definitely agree. For evidence of this, just think about when you ride over a river etc during the fall at night, just at the threshold of when you would start to wear a jacket. You can definitely tell when you are over bodies of water by the sudden temperature drop while passing over the body of water, then suddenly warm up again once you have passed over it.
 

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I totally agree with you guys. When I got my bike back in march people thought i was crazy, but i rode to work when it was 26 f. Thankfully it was only 16 blocks.lol.
 

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SNAFU organizer
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They Govment charts???,,,, They probably both lying.
 

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Something doesn't make sense with that "new" formula....

Windchill (ºF) = 35.74 + 0.6215T - 35.75(V^0.16) + 0.4275T(V^0.16)

Suppose both temp and wind are both zero to make the equation simple. Since 0 raised to any power is still zero we have :

Windchill (0ºF) = 35.74 + 0.6215(0) - 35.75(0^0.16) + 0.4275*(0)*(0^0.16)
Windchill (0ºF) = 35.74 + 0 - 35.75(0) + 0.4275*(0)*(0)
Windchill (0ºF) = 35.74 + 0 - 0 + 0 = 35.74

So they're trying to tell us that at 0ºF and no wind, the windchill will make it feel like ~36ºF? :confused:


To take it further, lets use 1 MPH for the windspeed and leave temp = 0ºF, still an easy equation since 1 raised to any power is still 1.

Windchill (0ºF) = 35.74 + 0.6215(0) - 35.75(1^0.16) + 0.4275*(0)*(1^0.16)
Windchill (0ºF) = 35.74 + (0) - 35.75(1) + 0.4275*(0)*(1)
Windchill (0ºF) = 35.74 + (0) - 35.75 + (0) = -0.01 ºF

So a 1 MPH difference in windspeed makes a 36ºF difference in windchill?
:confused:


:stac:
 

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STAND AND FIGHT!
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Discussion Starter #9
I like the chart going in the wrong direction, 100 degrees at 60mph has a "chill" effect of 105

It would seem likely, even inevitable that anyone riding in extreme heat, like Arizona in 120 degrees without insulating gear would get cooked, and suffer heat prostration or heat stroke.

I've ridden some in 106, maybe 108, but getting stuck in traffic was a bigger problem than overheating due to wind chill. There definitely needs to be some Mylar covered heat reflective gear for summer riding, if the infrared heat was rejected, the air temp might be less debilitating.

I had decided this year if it got that hot I was going to spray my mesh riding suit with aluminim paint but it never seemed to get than hot. Maybe I was still recovering from my broken ankle.
 

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STAND AND FIGHT!
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Discussion Starter #10
Must have been desperate to get the two wheeled fix. Saw 17F at a bank... at 60mph...
Haven't seen temps that cold for awhile, but last time I did, the bike ran so much better when the air is that cold and dry and dense, it makes it worth it...
 

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I break stuff.
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Cold is... Cold.

Although it seems a bit more tolerable on a sunny day.
 

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What?
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I went on a short 50 mile ride with a buddy the other night. It was funny. I didn't need gloves on top of the hills, but boy when I dipped down I wish I had them. Of course it was for five seconds at a time.

brr.....

ahhhh....

brr....

ahhh....

brr....

ahhh....

:D
 
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