Whats the difference between a wood & coal stove?

65 degrees F. minus the days average temperature (in degrees F.) = Heating Degree Days (HDD)

The easiest way to calculate it (not exact) is:

65 - [(Days High Temp + Days Low Temp)/2] = HDD

Example:

Days high temp = 20 degrees F.
Days low temp = 5 degrees F.

65 - [(20 + 5)/2] = 52.5 HDD's

I find that the calculation as done above is improved overall by multiplying by a kludge (or fudge) factor of roughly 1.06, and I believe this is so because in the depths of the winter there are fewer daylight hours than dark hours in a day, and you are trying to estimate the days average temperature from only two data points (days high and days low).

Therefore, an improvement to the above formula (with admittedly an empirical kludge, also not exact) is:

(65 - [(Days High Temp + Days Low Temp)/2]) x 1.06 = HDD

The kludge bumps up the heating degree days for the example day above to 52.5 x 1.06 = 55.7 HDD's

In my case I would expect to burn 1.57lbs/HDD x 55.7 HDD's = 87.5 lbs. of coal for the example day.

Or you can just go here: http://www.degreedays.net/

***The company I work for uses HDD's to estimate its monthly heat bill, and to keep the energy company honest.***
*** Your energy company uses monthly HDD's to compute your bill when they send you an estimated rather than an actual monthly bill. ***

Oh wow, thank you for the detailed explanation I think this winter I will use that formula to calculate coal usage.. I agree with the "Kludge" factor since HI and Low temps don't represent a true average for the day lol... Thats very interesting!

Um but the 65 degree base.... Shouldn't that be the temp you keep your house at? I run my house 75 degrees

Official HDD's are based at 65 degrees, but you are free to base as you see fit. Perhaps you will get even higher correlation of coal to HDD's in so doing?

The average temperature for the day is an integration function. The more data points, the better you can simulate this integration. 24 hourly temperature points summed and then divided by 24 would get you rather close overall. I choose to make it as simple as I can, and I reduce it to but two data points (using data that is readily available), accepting the error that this inevitably brings to the calculated HDD.