Calculate temperature change with elevation

It’s no secret that the temperatures high up in the mountains will be lower than that in the valleys. Have you ever set off on a warm summer hike to find the temps at the peak in the 50’s or worse?
Avoid these surprises by knowing how to calculate the temperature loss as you climb. Here are the steps involved, and a simple equation to reference.

Now, before we proceed, these are simple approximations. They’re not exact, they’re not terribly scientific, and they’re not meant to be used in such a way. Also, if you do the math using Celcius vs Fahrenheit, you will get slightly different values. Again, these are both simplified calculations designed to be done on the fly, they’re not exact.  That being said, let’s go on.

  • Look up the weather forecast

Obviously, you need a base to go on. Look up the local area forecast, and see what the high, and low temperatures are going to be.

  • Determine the elevation of reference for the forecast.

All weather forecasts are referenced to a particular elevation. Using the National Weather Service website you can a detailed forecast, and they’ll list the elevation of reference on the page. If that information isn’t available, it’s usually the same elevation as your official city elevation. Here, we’re at 2,000 feet, and our forecast are all for 2,000 feet.

  • Determine your peak elevation

Now you need to know how high up you are going to climb or descend. Reference your topo map, or find these details online. A quick Google with “Mountain (Name) elevation” will normally get you what you need.

  • Do the math

You will lose an average 3.5 degrees Fahrenheit for every 1000 feet of elevation you gain. You can also  use about 1.2 degrees Celsius per ever 1000 feet, or about 1 degree Celsius per 100 meters (source, NFW who showed me my typo on the metric conversion in the comments). Some people use 9.8 degrees Celsius per 1000 meters).

If you start out at 1000 feet, and climb to 6000 feet, that’s a 5000 foot difference (6000 – 1000 = 5000). So, since you’re gaining 5,000 feet in elevation, you’ll use a 5 in your calculation. 5,000 feet, times 3.5 degrees. Just drop the (thousand). So, ( 5 x 3.5 =  17.5 degrees). So roughly, you’ll expect to lose at least 17.5 degrees. I always round up to the nearest 5 just to factor in changes in weather that can’t be planned for, so here I will assume a 20 degree difference. Simply subtract this number form your expected low, according to the forecast, and you have the lowest expected temperature, short of some crazy weather event.

Ex: The weather man says it’s going to be 60 degrees today for the high in your city. Your city is at 6,000 feet. If you’re climbing from 6,000 feet to 14,000 feet, That’s an 8,000 foot difference. 8 times 3.5 is 28 (8 x 3.5 = 28). You can assume a 30 degree difference after rounding, so it’s only going to be 30 degrees max at the top of the mountain (A high of 60 minus your 30 difference)! Remember, that’s for the high. Always consider your low temperatures too.

Factors that affect your actual temperature

Some factors change the actual value of these calculations. Cloud cover will trap in more heat, where a clear sky will drop the temperature slightly faster. Cold fronts and air streams may also have an effect, as well as local evaporative cooling. These factors are too numerous to account for, so the equation is designed as a general scenario calculation (after rounding).


( 3.5 x Change in elevation)/1000 = temp loss due to elevation change

or (3.5ΔH=Tf)

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57 thoughts on “Calculate temperature change with elevation

    1. The 3.5 is a little complicated, but basically the average combination of variables that affect the temperature, such as cloud cover, humidity, pressure, air density, and a few other factors. The 3.5 isn’t a perfect estimator, but it works pretty well in most conditions.

    2. Mother Nature. Or … the higher you go the fewer molecules are in the air for any given volume. Heat is the interaction of molecules ( density and speed) transferring energy to your skin as they collide with it. Higher altitude equals fewer molecules banging into you which equals less energy transfer which equals lower temperature.

  1. This begins at 1000m correct? So moving from sea level to 1000m doesn’t apply to this formula. Can you explain why that is?

    1. That’s true. It comes down to how the formula works. If you’re at sea level, the you’re multiplying by 0000 feet…so 0 temperature change. You’ll always get a zero. The formula just breaks down at anything below that. You could replace 500 feet of change with a .5, or something similar to that, but near ground level humidity and pressures throw off the assumptions built into the formula anyway.
      There is some fuzziness though. If you’re looking at a forecast, your forecast may be at 2000 feet (usually the elevation of the city) or 6000 feet. So if you plan to camp at 6000 feet, but your forecast is only for 2000 feet, you would only calculate in a 4000 foot change.

  2. on the same latitude the temperature at X is 25 degree Celsius what is the temperature at altitude 1000 metres above sea level

    1. Assuming 1000 meters is above your starting point, you would loose about 10 degrees Fahrenheit (or about 1.6 celcius if my conversion is correct). Some fluctuation due to humidity and cloud cover.

  3. if i go up 1 km high , what is the temp variation on the 1st kilo meter’s end.
    what is the temperature variance

  4. can you help to solve this qn
    temperature at zimbambwe village 950m is 24c.What will the temperature be at Zimbambwe town 5895m above the sea level

    1. Hi there!
      Sure thing. Although, if this is for a class, keep in mind they may use an alternative formula with slightly different results (this is a simplified formula).
      950m is 3116.8 feet (my local measurement standard and the standard the formula was designed for) while 5895m is 19340.55 feet. 24C is 75.2F
      So, 19,340.55 feet – 5,895 feet = 13,445.55 change in elevation. Multiply that by 3.5, and divide by 1000, that’s 47.059 degrees change in temperature. 75.2F – 47.059 (initial temperature minus your change) and we get 28.141 F. Convert that back to C for your local usage, and that’s -2.144 C. Nice and chilly!
      Check my math, but that sounds about right. Let me know if any of that doesn’t make any sense.

      1. Errm, sir i think you are wrong this time. the Normal Lapse Rate(NLR) says: at every 1000m temp. decreases by 6.5 C.
        To solve Franks question, first we find difference in height, initial height was 950m then it changes to 5895m, therefore the difference in Height is 4945m. now we 4945m divided by 1000m times 6.5 =32.1425
        initial temp at 950m (which is 24 C minus 32.1425= -8.1425 C) therefore temperature at 5895m = -8.1425 C

      2. It’s quite expected that we get different numbers. I’ve not checked my math (or yours), but the biggest factor here is that I’m using a different equation. The one I’m using is designed as a worse case scenario (other factors other than elevation come into play here) for backpacking purposes. It errors on the high side, and purposefully so. Bravo for knowing about lapse rate. When incorporated with other factors it can be very accurate.

  5. Nice,I feel like I learn something from you.
    Question:if the air temperature at sea level is 35 degree celisious,what would be the air temperature at 4000 meter above mean sea level?

    1. I’m happy to hear I’m providing some new knowledge!
      First, we need to convert to Fahrenheit for my formula. 35C=95F. 4000 meters = about 13,000 feet. Sea level = 0 feet.
      3.5 x 13 = 45.5 degrees change. So, 95F-45.5=49.5F, or 9.7C. Brrrrr!

      1. Question : calculate the temperature at the base of the mountain if it is 11000m high and temperature is -20 C.

  6. Question : calculate the temperature at the base of the mountain if it is 11000m high and temperature is -20 C.

  7. Question: how this equation is reliable while calculating the daily temperature of past historical data in India. I’m calculating the daily temperature of different blocks of district in India….and I have daily temperature data of only one block. from this equation, can I calculate the daily temperatures for other blocks ???

    1. Hi, I wouldn’t use this in any scientific fashion. It’s a rough estimate that’s designed to be done off hand. There are other equations that would be more accurate, but require more information. This is just an estimation.

      1. Hi.., Basically I m calculating daily temperatures of 14 blocks of the district….with reference to your equation… I m applying your logic for estimating the daily temperatures… I have a question…. Do you have any kinda research work related to this equation….or any kinda equations 😊😊😊

      2. That’s awesome! I’m glad to see some research going on out there.
        I don’t have any published research on this particular equation. I really just made this so I could quickly figure out, roughly, what temperatures I could expect if I decided to climb a mountain and camp on the top. It’s come in quite handy!

  8. I live in Europe where we don’t use the english system. Is it possible to calculate this change when the mountain elevations are in meters?

    1. Hmm. I’m sure there is a way. It’s as simple at 3 feet per meter, so I conversion should be possible. One option is to convert your meters to feet, use the equation and go back. Let me see if I can come up with a meters version that works.

  9. This confuses me because the conversion at the very beginning is incorrect.
    F = 9/5C+32.
    3.5F/1000ft x 3.2808ft/m = 3.5F/304.8m
    3.5F/304.8m x 5/9C/F = 1.94C/304.8m
    1.94C/304.8m x = .0064C/m
    .0064C/m x 100m = 0.64C/100m

    If 3.5F/1000ft is the correct value, then it should be 0.64C/100m not 1.2C/100m.
    If the 1.2C/100m is correct then it would become 6.6F/1000ft.

    Which is it?

    1. Hi,

      These are both approximations and there were created by two different people. The 3.5F/1000 feet errors on the caution side for low temperatures. Otherwise, the 9.8C/1000 meters is constructed under slightly different assumptions. You will definitely get different values for each. They’re both rough approximations, and for a more precise number you’ll need to factor in more variables than I know how to integrate. Neither of these should be considered precise, nor interchangeable. It’s simply a quick on the fly calculation.

  10. Thank you very much..i just found the appropriate material for my teaching reinforecement lesson..simplified and concise to my convenience.much appreciated.

    1. That’s a tricky one. There, as far as I’m aware, is no real predictable correlation between elevation and water temperature in a well. Factors that would affect it include dept, source of the water, and probably to the least extent, outside air temperature. But, I only have a couple classes under my geology belt so I’m not super useful on that one sadly.

  11. “You can also use about 1.2 degrees Celsius per ever 1000 feet, or about 2 degrees Celsius per 100 meters ”

    I’m not sure *what* went wrong in the calculations here, but if you consider that 1000 feet is nearly 305 meters it’s obvious that something did!

    1. It’s meant to be quick and easy to calculate on the fly, not really for accuracy. It’s a ballpark estimate. Wouldn’t bet my life on it but it gives me an idea of what I can expect. +/- a bit.

  12. You say

    “You will lose an average 3.5 degrees Fahrenheit for every 1000 feet of elevation you gain. You can also use about 1.2 degrees Celsius per ever 1000 feet, or about 2 degrees Celsius per 100 meters (source for the Celcius calculations). Some people use 9.8 degrees Celsius per 1000 meters).”

    Your conversion of 2 degrees C per 100m must be wrong. I’d suggest it was half that. And that is of course reduced in say cloud or snow.

    So I agree with all you say, but just wanted to correct you on your Celsius calculation, which I would suggest should be 10c per 1000m or 1c per 100m.


  13. hi help me at once if the initial altitude is given that is 1500masl then what is the level of temperature for the following elevation 1 when altitude increases to 1591.5m. (36-degree celicieos)

  14. The conversion between Fahrenheit and Celsius is exactly 9 (F) to 5 (C). Nearly everyone above has got that wrong. Pure water at sea level freezes/boils at 32 / 212 degrees F and 0 / 100 degrees C . The respective differences are 180 and 100, which gives an exact ratio of 9 to 5. The 32 is added or subtracted (from the F figure) before or after the conversion is made, depending on which way the conversion goes. Try converting 37 degrees C to degrees F. The answer should be 98.4, which is a typical but arbitrary figure for normal blood temperature in general population of the U.S. The equivalent in almost all of the rest of the world is 37 degrees C. For those who want simplicity, – 40 degrees is the same in both systems,

  15. I live in Mexico City and in general the weather is not accurate due to old equipment, Mexico city is the capital of Mexico and is a valley in a mountain 2240meters over the sea level is chilly cold to freezing all year long summer is overcast rainy and cold, but weather channel always says we’re at 22°C Lol which is not accurate, my home termometer says different but with this simple explanation I understan it more. I had many problems at university explainig that mexico city is cold, this explanation make it easier. Thank you 🙂

    1. I’m so happy to have helped!
      Weather stations generally do their forecasts for at the bottom of the mountain, in my experience, so knowing how much it can change as you go up is vital!
      Congrats for living in Mexico City. That place looks amazing!

  16. No this is wrong, these values are not what we use, for example in aviation. They correspond to dry air and so they are never true.

    The temperature decreases by 6.5 °C per 1000 m of altitude, or by 1.98 °C per 1000 ft in a standard atmosphere, that’s what people use as reference. The real value depends on the humidity along the way, so it’s more complex, but it varies around that rate (called the lapse rate).

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