July CET

[Roger J Smith]

The reason why Michael is getting a higher avarage may not be related to his instrument exposure at all. From what he says, he's taking an hourly average. In the summer months, this will be higher than the average based on daily max and min, which is the way most monthly averages are calculated. The reason is fairly obvious -- July has 16 hours of daylight and 8 hours of night. Only the hours around dawn are usually close to the daily minimum, and many hours during the daytime are almost as warm as the maximum. Michael, if you have a few minutes, just list your daily highs and lows from that hourly data, and see what the average of those happen to be. Betcha it will be lower than the average you have for the hourly data.

[The PIT]

The reason why Michael is getting a higher avarage may not be related to his instrument exposure at all. From what he says, he's taking an hourly average. In the summer months, this will be higher than the average based on daily max and min, which is the way most monthly averages are calculated. The reason is fairly obvious -- July has 16 hours of daylight and 8 hours of night. Only the hours around dawn are usually close to the daily minimum, and many hours during the daytime are almost as warm as the maximum. Michael, if you have a few minutes, just list your daily highs and lows from that hourly data, and see what the average of those happen to be. Betcha it will be lower than the average you have for the hourly data.

Which is one of the reasons why my Davis differs from my Stephson screen as it takes the averages for the full day not just the Max and Min.

[johnholmes]

yet another day of above 28C, the last of this spell.

My mean is 21.1C way ahead of any other month for any records in this area.

11 over 30C or more

17 of 28C or more

23 of 25C or more.

some record.

to comment about the values shown by Michael.

The mean of the month is calculated by taking the max each day the min each day and usually getting a mean for the day.

Or

simply adding up all the max temps all the min temps, divide each by the number of days in the month, add the result and divide by 2.

As has been already commented you cannot take the hourly means and do it for the reason already given

John

Which is one of the reasons why my Davis differs from my Stephson screen as it takes the averages for the full day not just the Max and Min.

exactly

Edited July 29, 2006 by johnholmes

[shuggee]

The mean of the month is calculated by taking the max each day the min each day and usually getting a mean for the day.

Or

simply adding up all the max temps all the min temps, divide each by the number of days in the month, add the result and divide by 2.

But which is the standard/accepted way of doing it? As those two methods will produce different results.

[johnholmes]

back in a tic

it makes no difference, assuming you show each temp, max, min, mean to the same decimal point, which should be to 1 decimal place in all temp readings and averages so derived.

Thus avge min=13.9C

avge max=28.3C

Mean=21.1C

mean each day and the average taken=21.1C

John

Edited July 29, 2006 by johnholmes

[Methuselah]

But which is the standard/accepted way of doing it? As those two methods will produce different results.

[johnholmes]

see my post above it makes no difference

jh

[Methuselah]

see my post above it makes no difference

jh

Sorry John. I'm on the vodders, so I see things a tad oddly!

I was taking Shuggs's post entirely out of context!

[The PIT]

Problem is it doesn't really reflect the day Question. Say 9:00 it's 30C ten minutes a cold front goes through the rest of the day is at 15C. If you measured hourly you'd get a lot lower average than you would, which would reflect more accuratly the true feel of the day. The days average would be 15.6C while the other method it would be 22.5C An extreme example of course but most people would remember the 15.6C average.

[johnholmes]

bit early for that mate!

[The PIT]

Sorry John. I'm on the vodders, so I see things a tad oddly!

I was taking Shuggs's post entirely out of context!

Must have been a bad day.

[johnholmes]

Problem is it doesn't really reflect the day Question. Say 9:00 it's 30C ten minutes a cold front goes through the rest of the day is at 15C. If you measured hourly you'd get a lot lower average than you would, which would reflect more accuratly the true feel of the day. The days average would be 15.6C while the other method it would be 22.5C An extreme example of course but most people would remember the 15.6C average.

one way of looking at it Pit. However the accepted very long term way, long before we were able to have our own AWS set up, is the way I've described. For those who are really keen that event would be logged in the daily weather diary for anyone to read, I often make comments to cover that kind of thing.

Maybe those with recording kit like the Davis should show both, the old way and the Davis way.

jh

this is the Davis explanation

Mean Temperature

The mean temperature for the day. At the bottom of the column, the mean temperature for the month is displayed.

If "Calculate using integration method" is checked in the Degree-Day section of the NOAA setup dialog window, then the mean temperature is calculated by adding up all the temperature measurements for that day and then dividing by the number of samples. If "Calculate using integration method" is not checked in the NOAA setup, the mean temperature is the average of the daily high and low temperatures.

I use the latter so they agree on my site

Edited July 29, 2006 by johnholmes

[The PIT]

JohnH perhaps I should change my method but then it does give another view of things I suppose.

[shuggee]

back in a tic

it makes no difference, assuming you show each temp, max, min, mean to the same decimal point, which should be to 1 decimal place in all temp readings and averages so derived.

Thus avge min=13.9C

avge max=28.3C

Mean=21.1C

mean each day and the average taken=21.1C

John

It does make a difference John:

1. Add together the daily min and max to get the daily mean - add these all toghether and then divide by 31 (number. of days In July).

vs

2. Add together all 31 maxes and divide by 31 to get average max. Add together all 31 minimums and divide by 31 to get average min.. Then add them together and divide by 2.

For instance (here's a four day example):

Day 1 - Min 10 - Max 20

Day 2 - Min 12 - Max 22

Day 3 - Min 14 - Max 23

Day 4 - Min 14 - Max 24

Using Method 1. You get [(10+20)/2] + [(12 + 22)/2] + [(14 + 23)/2] + [(14 +24)/2] = [15 + 17 + 18.5 + 19] / 4 = 17.375

Using Method 2. You get [(10 + 12 + 14 + 14)] / 2 + [(20 + 22 + 23 + 24)] / 2 = (24 + 44.5) = 68.5 / 4 = 17.125

So which one is it?!?!

(Btw - I need a cider now )

[Roger J Smith]

They're actually the same -- check your addition in "method 2" part 1 ... it should come to 50/2 = 25, not 48/2 = 24 as you show.

From that point on, the methods give the same results.

If you look at each method, what you are essentially doing is adding up the total of max and min and dividing by twice the number of days. The two methods are just different in the steps used to do this.

[Nick H]

It does make a difference John:

1. Add together the daily min and max to get the daily mean - add these all toghether and then divide by 31 (number. of days In July).

vs

2. Add together all 31 maxes and divide by 31 to get average max. Add together all 31 minimums and divide by 31 to get average min.. Then add them together and divide by 2.

For instance (here's a four day example):

Day 1 - Min 10 - Max 20

Day 2 - Min 12 - Max 22

Day 3 - Min 14 - Max 23

Day 4 - Min 14 - Max 24

Using Method 1. You get [(10+20)/2] + [(12 + 22)/2] + [(14 + 23)/2] + [(14 +24)/2] = [15 + 17 + 18.5 + 19] / 4 = 17.375

Using Method 2. You get [(10 + 12 + 14 + 14)] / 2 + [(20 + 22 + 23 + 24)] / 2 = (24 + 44.5) = 68.5 / 4 = 17.125

So which one is it?!?!

(Btw - I need a cider now )

Roger, you are right!

Edited July 29, 2006 by Nick H

[shuggee]

They're actually the same -- check your addition in "method 2" part 1 ... it should come to 50/2 = 25, not 48/2 = 24 as you show.

From that point on, the methods give the same results.

If you look at each method, what you are essentially doing is adding up the total of max and min and dividing by twice the number of days. The two methods are just different in the steps used to do this.

LOL Git.

Yes - they do come out the same Roger

But I'm 99.9% certain that the two methods do result in a difference over a longer period. Help!!

Can someone be bothered adding up a whole month using the two methods and seeing if that's true - I can't right now!

[The PIT]

(Btw - I need a cider now )

Oi get of my Cider

[johnholmes]

It does make a difference John:

1. Add together the daily min and max to get the daily mean - add these all together and then divide by 31 (number. of days In July).

vs

2. Add together all 31 maxes and divide by 31 to get average max. Add together all 31 minimums and divide by 31 to get average min.. Then add them together and divide by 2.

For instance (here's a four day example):

Day 1 - Min 10 - Max 20

Day 2 - Min 12 - Max 22

Day 3 - Min 14 - Max 23

Day 4 - Min 14 - Max 24

Using Method 1. You get [(10+20)/2] + [(12 + 22)/2] + [(14 + 23)/2] + [(14 +24)/2] = [15 + 17 + 18.5 + 19] / 4 = 17.375

Using Method 2. You get [(10 + 12 + 14 + 14)] / 2 + [(20 + 22 + 23 + 24)] / 2 = (24 + 44.5) = 68.5 / 4 = 17.125

So which one is it?!?!

(Btw - I need a cider now )

no idea where you get your result, I get exactly the same whichever way.

You have to remember to work to the same decimal point, one as I said.

LOL Git.

Yes - they do come out the same Roger

But I'm 99.9% certain that the two methods do result in a difference over a longer period. Help!!

Can someone be bothered adding up a whole month using the two methods and seeing if that's true - I can't right now!

lay off the cider Shuggee they do NOT no matter how long you check it for.

they are the same, IF you use the same number of decimal points for each method.

John

LOL Git.

Yes - they do come out the same Roger

But I'm 99.9% certain that the two methods do result in a difference over a longer period. Help!!

Can someone be bothered adding up a whole month using the two methods and seeing if that's true - I can't right now!

that is what my original reply did, well to today=29 days, adding 2 more days will make sod all difference.

what any of us have to remember is that Hadley, etc etc, Met Office, and any observer linked into the observing network, will ALL work out their figures the way I have described. Using max and min totaling them up, dividing each by the number of days in the month, adding that result and dividing by 2. OR a similar method. So if we want to compare our data with CET, long term averages then that is how it MUST be done, otherwise it all becomes pretty meaningless.

John

Edited July 29, 2006 by johnholmes

[shuggee]

Aye - fair enough John. Good to get clarity

[johnholmes]

okay mate

[paul tall]

The reason why Michael is getting a higher avarage may not be related to his instrument exposure at all. From what he says, he's taking an hourly average. In the summer months, this will be higher than the average based on daily max and min, which is the way most monthly averages are calculated. The reason is fairly obvious -- July has 16 hours of daylight and 8 hours of night. Only the hours around dawn are usually close to the daily minimum, and many hours during the daytime are almost as warm as the maximum. Michael, if you have a few minutes, just list your daily highs and lows from that hourly data, and see what the average of those happen to be. Betcha it will be lower than the average you have for the hourly data.

As I said earlier, the reason his average is so high is because he gets temperatures several degrees higher than surrounding stations. His maxes clearly, are over exposed, innacurate, call it what you want. It doesn't matter whether you take a sample every hour, or whether ave max+ave min/2, the basic fact of inaccuarately high temperatures skews his average artificially well upwards. End of story.

[paul tall]

I'd also like to add that a computer generated mean quite often brings a lower mean than the traditional method.

Which is actually the opposite of your theory Roger.

For example, my average max up to yesterday was 24.9º. The average min was 13.3º. Therefore a mean of 19.1º.

But the virtual weather station software which I use was giving a mean of 18.8º. As we all know, these software packages take an average of every sample, in my case its every 10 minutes.

If you have a day here where the min is 13 and the max is 25, often the temperature will not rise above 21 until about 11am, and will have dipped below 21 by say 9pm. It might not get above 18 until 9am, and be below 18 by 11pm. There are slightly more cool readings captured than warm ones, which drags the mean down by a notch or two

[johnholmes]

I'd also like to add that a computer generated mean quite often brings a lower mean than the traditional method.

Which is actually the opposite of your theory Roger.

For example, my average max up to yesterday was 24.9º. The average min was 13.3º. Therefore a mean of 19.1º.

But the virtual weather station software which I use was giving a mean of 18.8º. As we all know, these software packages take an average of every sample, in my case its every 10 minutes.

If you have a day here where the min is 13 and the max is 25, often the temperature will not rise above 21 until about 11am, and will have dipped below 21 by say 9pm. It might not get above 18 until 9am, and be below 18 by 11pm. There are slightly more cool readings captured than warm ones, which drags the mean down by a notch or two

paul see my input about Davis as to how to get it to give the average of the max and min not throughout the day. I assume its a similar thing with other auto wx stns

John

here is the post

this is the Davis explanation

Mean Temperature

The mean temperature for the day. At the bottom of the column, the mean temperature for the month is displayed.

If "Calculate using integration method" is checked in the Degree-Day section of the NOAA setup dialog window, then the mean temperature is calculated by adding up all the temperature measurements for that day and then dividing by the number of samples. If "Calculate using integration method" is not checked in the NOAA setup, the mean temperature is the average of the daily high and low temperatures.

I use the latter so they agree on my site

jh

Edited July 29, 2006 by johnholmes

[paul tall]

Are you saying that you can sort of programme it to average out the maxes and mins, rather than averaging every sample John?

If this can be done, I'll have to look into it.