# How To Approach a Range of Albedo in A Given Latitude Range When Estimating Temperature?

There's a lot of information about figuring in the average albedo of a planet when calculating the base temperature for a planet. What if you want to get more granular, down to each (let's say) 10 degree latitude band?

We know different things have a different albedo. Estimates vary, but for the sake of this discussion, let's say our various albedo break down like this:

• Land: .25
• Water: .06
• Land Ice: .9
• Water Ice: .65

So if we have a 10-degree band of latitude consisting of...

• .25 land
• .15 land ice
• .10 water ice
• .50 water

...when plugging albedo into our temperature equation, what number does one use?

Do we take an average of the different albedo ratings, figuring in the percentage of surface for each albedo?

Or does the strongest albedo cancel out the rest? Or the weakest?

(Caveats to keep this question from being a super-complex climate modeling rabbit-hole: don't worry about all the other variables that you and I both know go into doing a good-enough-for-worldbuilding climate model. Assume 0 elevation; assume we're looking at the average temp across the year, don't worry about coasts vs interiors, don't worry about wind or ocean currents, etc. etc.)

Links to sources appreciated -- thanks very much!

• You that the weighted average. Albedos combine linearly. (That's all I know. I have no idea why you would think that the albedo of a band of latitiude is relevant for the average temperature, or why you would think that the average tempeature is in any way useful. I live in Bucharest, where the average tempeature is about 11 °C; but this irrelevant when in July we regularly go above 40 °C, and in cold January nights we go below –15°C.) – AlexP Oct 7 '20 at 0:48
• Thanks, @AlexP. Getting a lot of input in various Worldbuilding groups that concur with your answer. As for why I'm doing this -- this is just one part of a larger process I'm doing to determine rough temperatures across the course of the year. The wide variation of temperature is figured in... elsewhere. Just needed this piece of the puzzle. – Matt Selznick Oct 7 '20 at 2:29

The weighted average. In this case, $$0.25*0.25 + 0.06*0.15 + 0.9*0.1 + 0.65*0.5 = 0.49$$.