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).

**Equation **

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

**or (3.5ΔH=Tf)**

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Sweet. I feel like I learn something from you every day.

Love this!

Thanks! I hope it helped / made sense :p

Keep up the good work on your blog. I follow it closely. You’re doing great.

This is great! Where does the “3.5” comes from?

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.

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.

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

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.

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

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.

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

or

what is the temperature variance

You can expect a (roughly) 6 degree F difference per km that you climb.

Do you have a wife

Not currently. Although I do have a partner. 🙂

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

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.

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

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.

Thank you for your enlightenment. I’m grateful.

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?

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!

if 900 m altitude has 30c degree temperature what is the temperature in 6000m altitude?please

You can expect about -2.5 C from that big of a climb (or 27 F).

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

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

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 ???

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.

do you have an other equations for calculating daily temperature….

Sorry, I do not. What kind of calculations are you trying to do?

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 😊😊😊

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!

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?

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.

So, rough calculation. For Celsius and meters, the change would be about 9.8 degrees C per every 1000 meters you climb. Or about 1.16 degree per every 100 meters.

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?

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.

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

what will you say about well water elevation and temperature changes?

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.

“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!

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.

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.

Thanks

NFW

I’m inclined to agree. Looking it over, that does seem to be more accurate (although I see other estimations out there too).