Does N E 1 know how 2 calculate wind chill?
359998
Formula used to calculate wind chill\n\nThe term "wind chill" goes back to the Antarctic explorer Paul Siple, who coined it a 1939 dissertation, "Adaptation of the Explorer to the Climate of Antarctica." During the 1940s, Siple and Charles Passel conducted experiments on the time needed to freeze water in a plastic cylinder that was exposed to the elements. They found that the time depended on how warm the water was, the outside temperature and the wind speed.    More information   \n   \nNew wind chill chart.\nWhy formula was changed.\nWind chill applies only to living things.\n \n \n \n\nThe formulas used to calculate wind chill were based on those experiments. \n\nIn the fall of 2001, the U.S. National Weather Service and the Canadian weather replaced the formulas with new ones (one for Fahrenheit temperatures and one for Celsius readings). The new formulas are based on greater scientific knowledge and on experiments that tested how fast the faces of volunteers cooled in a wind tunnel with various combinations of wind and temperature. \n\nThe new formula for winds in mph and Fahrenheit temperatures is: \n\nWind chill temperature = 35.74 + 0.6215T - 35.75V (**0.16) + 0.4275TV(**0.16) \n\nIn the formula, V is in the wind speed in statute miles per hour, and T is the temperature in degrees Fahrenheit. \n\nNote: In the formula, ** means the following term is an exponent (i.e. 10**(0.5 ) means 10 to the 0.5 power, or the square root of V), - means to subtract, + means to add. A letter next to a number means to multiply that quantity represented by the letter by the number. The standard rules of algebra apply. \n\nFor reference, the old wind chill formula was: \n\nT(wc) = 0.0817(3.71V**0.5 + 5.81 -0.25V)(T - 91.4) + 91.4 \n\nSource for both formulas: The National Weather Service \n\n\nhttp://www.usatoday.com/weather/resources/basics/windchill/wind-chill-formulas.htm
756155
Celsius to fahrenheit:\n(C*9) / 5 +32\nmultiply it by 9, then divide it by 5, then add 32.\n\nbackwards:\n((F) - 32  ) /9 *5\n\nthere are some converters on the web if you don't want to use the formula:\nwww.wbuf.noaa/gov/tempfc.htm
129713
MeghnarMajhi is right.\n\nTemperature <==> Heat
562439
Sin's answer makes sense, but in terms of the environment I'd have thought it would be the highest temperature which combined with wind chill would take the felt temperature down to 0C/32F. But don't trust me - I'm not a scientist.
111333
Begin by subtracting 32 from the Fahrenheit number. \nDivide the answer by 9. \nThen multiply that answer by 5.
32289
One answer would be to convert 0C into 273 Kelvin or 32 F and then do some multiplication.\n\nBut "cold" is really a subjective measure based on physiology. I suspect if you did a test with actual humans you would find that around -10.6 C feels twice as cold as 0 C.\n\nThat is what this wind chill calculator finds, if wind speed is always zero.
921196
If you run x1 meters against wind, the work you produce is A1=F*x1+f*x1; F is the force of your muscles to overcome the resistance of air with no wind, f is the force of the wind in your face. Let x1 be covered in time t1, then P1=A1/t1 is your power against the wind. \nIf you run x2 meters with the wind, the work you produce is A2=F*x2-f*x2; F & f being the same.\nLet x2 be covered in time t2, then P2=A2/t2 is your power with the wind. They say that dP=P1-P2=50W or dP=(F+f)*(x1/t1)-(F-f)*(x2/t2). As x1/t1=x2/t2=v, then dP=2*f*v. Thus f=dP/(2*v)=50/7=7.14N  is the force of the wind, m=80kg is excessive datum!
600668
If it is 0 deg Centigrade i.e. 32 deg Farrenheit and double the cold i.e. 16 deg F mening -8.9 deg Centigrade
559800
ask at the national weather service database.  all reporting stations have their data recorded, if there is one there, they know the answer.
178900
I can't figure it out, my brain is slow tonight.\nHave x=air plane speed\nhave y=wind speed\nx+0=225\nso x=225\n225+y=300\n225-y=210\nFrom there I'm lost
246263
wind chill isn't actually a measure of how cold the air feels. a wind chill temperature is, rather, a calculation of how cold it would have to be to cause the same rate of heat loss from your skin if there were no wind blowing. wind chill approximates how cold it feels because the rate of heat loss corresponds with skin temperature, and skin temperature is what our nerves sense.\n\nthe chilling effect of wind comes from two sources:\n\n1. the disturbance of the insulating boundary layer of warm air over the skin\n\n2. the evaporative cooling effect of moisture loss from the skin.\n\nin practice, wind can also cause drafts within clothing and hence heat loss, but this effect is not considered in the chill factor calculation because of its obvious variability. \n\nnote: you cannot cool to a temperature much lower than that of the wind, but you can cool faster to that temperature. depending on the humidity, it is possible to drop below the air temperature by a few degrees.\n\nclick on link by source for more info, you'll also find a wind chill formula on their page.
241018
= wind speed measurement over shore and water surface.\n\nclick link to read more...
945879
The first method would be to compare you utility bills to see if there is a drastic change (or any) between last year at this time and the most recent bill with the insulation on.  \n\nFor an exact calculation, you need the following:\n\nThe R value of the insulation\nLength of pipe insulated\nTemperature of the water inside the pipe\n\nThen plug them into this equation:\n\nQ=UA(Tinside - Toutside)\nwhere Q is the amout of Btus \nU is the thermal insulation (or lack there of) expressed as =1/R\nTinside = water temperature\nToutside = air temperature (average) where the pipe is installed\nA = the surface area of the pipe\n\nOnce you get that for the insulation, do it once with out the insulation.  For the second R value without the insulation, use a R-value of 0.5.  The two different Q's you get will be the amount of Btu's you saved.  \n\nTo convert that to $, you need to convert the Btu's (Q) to kW.  Divide the Btu's by 3.412 to get watts.  That will be the amount watts you saved at your hot water heater/boiler or whatever is heating the water (assuming it is electric).  You then look at your electric bill to determine the cost $/W or $/kWh.  Take the number you got for watts and convert to kilo Watts by dividing by 1000.  Then multiply it by the cost off of you utility bill for $/kWh.  \n\nThis is all for a one hour period.  You need to calculate that for a month, day, or year (720, 24, 8765 hours respectively).  You now have what your savings for this project was.
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
280680
I can't do the Math, maybe you can't either.  So for future conversions, try this:\nFahrenheit to Celsius Converter @\nhttp://www.wbuf.noaa.gov/tempfc.htm
911974
Go with a line graph, that way you can also find out (or estimate) temperatures at times when you didnt measure it.
565339
For all practical purposes, the answer is zero, but obviously they want something a little more sophisticated.  Refer to Newton's Law of Cooling:\n\nT(t)=T0+|T1-T0|e^(-kt)\n\nFirst solve for k, using the given info:\n\n6=0+|100-0|e^(-k*6)\n6=100e^(-6k)\n.06=e^(-6k)\nln(.06)=-6k\nk=.469\n\nThen, substitute k back into Newton's Law of Cooling equation, and use the new time of 2 hours, making sure to use 120 minutes.  I believe your answer will be choice (b), but check my math!