zzzyangliu-IDLresults

1. Download and read.

 There is my IDL 8.0 code: Question 2 & 3 & 4: Question 5:

Here are some helpful tips for IDL coding (By Emily)
code for i=0,143 do begin                  ;get index of 90W if (v_lon[i] eq 90+180) then v_90_lon=i endfor
 * Instead of using a loop to find longitudes: **

code

v_90_lon = where(v_lon eq 90+180)
 * You can use the where function, which is more efficient: **

;;; find array index where lon_precip = 90°W ;;; lon_precip = where(lon eq 90) ;returns -1 cause index doesn't exist! ;;; So find what index is closest to 90°W dummy = min(abs(lon_precip - (90+180)), index) precip90 = reform(precip[index, *, *]) ;;; now I've done the reform all in one step
 * Since there is not a precipitation lon exactly equal to 90, you can find the closest lon bin to 90°W using this: **

????And I'd like to know what colorbar.pro you are using. My version does not have a keyword target. But this may be a feature that IDL 8.0 has, that 7.? (the version I use) doesn't.

2. Calculate zonally averaged P and v over longitude, and plot the resulting latitude-time series.

Simply from pictures above, I think we(90W, ～ 20N) do not live in a place that generally reflects the wind field of ～ 20N. Not like the average, we may not have much wind coming from the south in summer; Besides, our precipitation is not that much similar to other places in this latitude. We have two peaks instead of one, that one in June and the other one in September.

3. Average air temperature over both lat and lon, to make a 12-month time series. Which season has the warmest global mean surface temperature? Can you understand why?

For the global, it is still in summer. From the picture above, the warmest month is in July, when the north hemisphere gets the most solar radiation from the sun. Because land mass difference is obvious between both hemisphere, and the heat capacity of land is much less than of water, therefore, temperature arises more in north hemisphere summer. As a result, as same as north hemisphere, the global average temperature gets a peak in July.

4. Make a map of the temporal (i.e. seasonal) standard deviation of precipitation, expressed as a percentage of the annual mean precipitation. This might be one definition of a "monsoonal" climate.

5. What is the space-time standard deviation of 'air' (temperature)? Best to build this step by step: find the global annual mean of air^2, the global annual mean of air, compute var = meansquare - mean^2, then stdev = sqrt(var).

Space-time standard deviation is 21.7762. But, we should consider about the latitude weight, which will generate a deviation of 15.2159. This number is smaller than the standard one, because we considered the pole region weights less than the tropics, which is true because our earth is a sphere.