Time Step and Concentration Calculation
Re: Time Step and Concentration Calculation
Dear alicec,
Thank you very much for your help.
I am using HYSPLIT to model insect dispersal. The computational particles represent insects. Insects do not loss their mass during wind dispersal process, and will eventually land on the ground.
That is why I am only interested in counting the number of particles on the ground level (deposition layer). I do not focus on counting the particles above the deposition layer.
I have looked at the output file of a simulation example from the tutorial. It has a ‘deposition’ value (mass/square meter) for each grid (square meter). Please see the attached file (the second last column: PMCH00000). I am very sorry for using a wrong word ‘concentration’ previously and making you confused. It should be ‘deposition’.
What I would like to do is to convert the ‘deposition’ value (mass/square meter) to the number of computational particles/insects (number/square meter).
I am aware that I would be able to count the number of computational particles by setting CMASS=1 (mass same as emission), as demonstrated in tutorial (section 12.1). CMASS=1 can make each particle equivalent to 1 mass unit (total number of computational particles released = total mass emitted). However, it has the ‘double counting’ issue.
For example, a ‘concentration’ value of 10 suggests a range of possibilities. One particle may have resided in the grid box for 10 time steps, or 10 particles may have been in the grid box only for one time step. In such a case, concentration calculation only results in summation of particles’ mass in each grid box, and the final result is no longer divided by the grid box volume. I guess that this ‘double counting’ issue also affects the accuracy of the number of particles in each grid (square meter) on the deposition layer.
Question 1: How am I going to calculate the number of computational particles in each grid cell (square meter) on the deposition layer without double counting?
Question 2: According to your previous reply, ‘The deposition loop with ichem=5 will set the mass to some of the particles to 0 which will flag those particles for removal from the simulation later.’. It seems that the mass on the particles is changing (from amount of mass to 0), if I understand it correctly.
I thought that the mass on the particles is not changing if I set ICHEM=5. That is why I tried to make each particle equivalent to 1 mass unit (total number of computational particles released = total mass emitted), thus it would be easy for me to count the number of particles after simulations.
I hope that it now makes sense to you. My apologies for the long reply.
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Thank you very much for your help.
I am using HYSPLIT to model insect dispersal. The computational particles represent insects. Insects do not loss their mass during wind dispersal process, and will eventually land on the ground.
That is why I am only interested in counting the number of particles on the ground level (deposition layer). I do not focus on counting the particles above the deposition layer.
I have looked at the output file of a simulation example from the tutorial. It has a ‘deposition’ value (mass/square meter) for each grid (square meter). Please see the attached file (the second last column: PMCH00000). I am very sorry for using a wrong word ‘concentration’ previously and making you confused. It should be ‘deposition’.
What I would like to do is to convert the ‘deposition’ value (mass/square meter) to the number of computational particles/insects (number/square meter).
I am aware that I would be able to count the number of computational particles by setting CMASS=1 (mass same as emission), as demonstrated in tutorial (section 12.1). CMASS=1 can make each particle equivalent to 1 mass unit (total number of computational particles released = total mass emitted). However, it has the ‘double counting’ issue.
For example, a ‘concentration’ value of 10 suggests a range of possibilities. One particle may have resided in the grid box for 10 time steps, or 10 particles may have been in the grid box only for one time step. In such a case, concentration calculation only results in summation of particles’ mass in each grid box, and the final result is no longer divided by the grid box volume. I guess that this ‘double counting’ issue also affects the accuracy of the number of particles in each grid (square meter) on the deposition layer.
Question 1: How am I going to calculate the number of computational particles in each grid cell (square meter) on the deposition layer without double counting?
Question 2: According to your previous reply, ‘The deposition loop with ichem=5 will set the mass to some of the particles to 0 which will flag those particles for removal from the simulation later.’. It seems that the mass on the particles is changing (from amount of mass to 0), if I understand it correctly.
I thought that the mass on the particles is not changing if I set ICHEM=5. That is why I tried to make each particle equivalent to 1 mass unit (total number of computational particles released = total mass emitted), thus it would be easy for me to count the number of particles after simulations.
I hope that it now makes sense to you. My apologies for the long reply.
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Re: Time Step and Concentration Calculation
Questions 1 & 2 – With ichem=5 the deposition routine will transfer the entire mass of one particle to the deposition layer. So you can think of that particle as now being in the deposition layer and you are not going to be double counting it. (The fact that it stays in the particle array with 0 mass for some amount of time is irrelevant except that the number of particles written to the MESSSAGE file does include those. But they are not “seen” in the cdump output.) So if you use ichem=5 then you can get the number of particles which were deposited. Maybe you just want to use a snapshot rather than time average.
One caveat is that ichem=5 is only for dry deposition. If you have wet deposition set up then the individual particles will still lose partial mass. Although I can’t think that wet deposition is relevant for your application.
One caution  HYSPLIT was not designed to model insects. I can't really advise you here on how appropriate it is to use the dispersion and deposition routines for modeling insect behavior.
One caveat is that ichem=5 is only for dry deposition. If you have wet deposition set up then the individual particles will still lose partial mass. Although I can’t think that wet deposition is relevant for your application.
One caution  HYSPLIT was not designed to model insects. I can't really advise you here on how appropriate it is to use the dispersion and deposition routines for modeling insect behavior.
Re: Time Step and Concentration Calculation
Dear alicec,
Thank you very much for your reply and help.
I ran some simple simulations with ichem=5 yesterday before seeing your rely, and the outputs are consistent with your explanation. But I sampled the deposition value using ‘time average’.
As you suggested, ‘Maybe you just want to use a snapshot rather than time average.’
I tried it today. It could not capture the deposition value, because there was no deposition occurring in some of the last time steps within the desired time periods.
Question 1: I would just like to confirm with you. The deposition calculation (time average) in HYSPLIT is actually to transfer the entire mass of one particle to the deposition layer. Basically, it is the summation of particles’ mass on the deposition layer over the desired time period? So it is unlike concentration calculation that HYSPLIT calculates the concentration at each time step as in the snapshot and then calculates the average concentration over the desired time period by weighting each concentration by its "residence" time. Would I be correct?
It works perfectly if I only use a small number of particles (e.g. particle number = 20; total mass = 20). I would be able to work backward to get the number of particles from the deposition value on the deposition layer.
For example, in my simulation, grid spacing was 0.05*0.05 (5000m*5000m). The desired time period for sampling was 3 hours, and the total simulation duration was 12 hours. I would be able to see that there were 3 particles deposited in the first hour and in the second hour, another 3 particles had deposited again. So there were total 6 particles deposited on the ground within the first sampling interval (3 hours). After that, there was no deposition occurring. The deposition value (6 particles) can match the information shown in the MESSAGE file (14 mass units left in HYSPLIT).
However, I increased the total number of particles and total mass (e.g. particle number = 10000; total mass = 10000), The deposition value cannot exactly match the information shown in the MESSAGE file. I worked backward again to get the number of particles from the output file. I got 4410 particles, but the MESSAGE file showed there were 5800 mass units left, which means I should get 4200 (100005800) particles.
I am thinking that this could be the precision issue for the floating point numbers. When the number of particles is increased significantly in HYSPLIT, there will be a difference between values (e.g. 4410) in the output file and values (e.g. 4200) in the MESSAGE file, but they are still similar.
Question 2: Would I be correct? If not, could you please explain this phenomenon?
Please find the MESSAGE files.
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Thank you very much for your reply and help.
I ran some simple simulations with ichem=5 yesterday before seeing your rely, and the outputs are consistent with your explanation. But I sampled the deposition value using ‘time average’.
As you suggested, ‘Maybe you just want to use a snapshot rather than time average.’
I tried it today. It could not capture the deposition value, because there was no deposition occurring in some of the last time steps within the desired time periods.
Question 1: I would just like to confirm with you. The deposition calculation (time average) in HYSPLIT is actually to transfer the entire mass of one particle to the deposition layer. Basically, it is the summation of particles’ mass on the deposition layer over the desired time period? So it is unlike concentration calculation that HYSPLIT calculates the concentration at each time step as in the snapshot and then calculates the average concentration over the desired time period by weighting each concentration by its "residence" time. Would I be correct?
It works perfectly if I only use a small number of particles (e.g. particle number = 20; total mass = 20). I would be able to work backward to get the number of particles from the deposition value on the deposition layer.
For example, in my simulation, grid spacing was 0.05*0.05 (5000m*5000m). The desired time period for sampling was 3 hours, and the total simulation duration was 12 hours. I would be able to see that there were 3 particles deposited in the first hour and in the second hour, another 3 particles had deposited again. So there were total 6 particles deposited on the ground within the first sampling interval (3 hours). After that, there was no deposition occurring. The deposition value (6 particles) can match the information shown in the MESSAGE file (14 mass units left in HYSPLIT).
However, I increased the total number of particles and total mass (e.g. particle number = 10000; total mass = 10000), The deposition value cannot exactly match the information shown in the MESSAGE file. I worked backward again to get the number of particles from the output file. I got 4410 particles, but the MESSAGE file showed there were 5800 mass units left, which means I should get 4200 (100005800) particles.
I am thinking that this could be the precision issue for the floating point numbers. When the number of particles is increased significantly in HYSPLIT, there will be a difference between values (e.g. 4410) in the output file and values (e.g. 4200) in the MESSAGE file, but they are still similar.
Question 2: Would I be correct? If not, could you please explain this phenomenon?
Please find the MESSAGE files.
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
 Attachments

 10000MESSAGE.txt
 (19.78 KiB) Downloaded 41 times

 20_MESSAGE.txt
 (16.78 KiB) Downloaded 40 times
Re: Time Step and Concentration Calculation
I'm editing this answer because I made a mistake in writing down the formula for the area.
So you are getting more particles that you would expect from the deposition value?
One thing to take into account is that your deposition grid area may not be square – depending on your location.
The area is calculated as
(111198.323)^2 * latspan * lonspan * cos(lat)
latspan and lonspan are the lengths (in degrees latitude, longitude) of the sides of the concentration grid.
Please confirm that you are using the correct area.
So you are getting more particles that you would expect from the deposition value?
One thing to take into account is that your deposition grid area may not be square – depending on your location.
The area is calculated as
(111198.323)^2 * latspan * lonspan * cos(lat)
latspan and lonspan are the lengths (in degrees latitude, longitude) of the sides of the concentration grid.
Please confirm that you are using the correct area.
Re: Time Step and Concentration Calculation
I was not able to reproduce your problem with the following CONTROL and SETUP.
Using 10,000 particles each with 1 unit mass.
calculating from the deposition I get 1087 particles "in" the deposition layer and
the MESSAGE file indicates 8913 particles left in the simulation.
Also  yes you are correct that the time averaging does not apply to the deposition layer.
The mass/Area indicated in the deposition layer is the total mass deposited over the averaging time divided
by the area. (Unlike the concentration which we discussed previously).
CONTROL FILE
18 10 17 15 #STARTING YEAR MONTH DAY HOUR
2 #NUMBER OF SOURCE LOCATIONS
36.949892 94.042969 50 #SOURCE 1 LATITUDE LONGITUDE HEIGHT(magl)
36.949892 94.042969 0 #SOURCE 2 LATITUDE LONGITUDE HEIGHT
6 #TOTAL RUN TIME (hours)
0 #USE MODEL VERTICAL VELOCITY
25000 #TOP OF MODEL DOMAIN (mAGL)
3 #NUMBER OF INPUT DATA GRIDS
./gdas1/
gdas1.oct18.w2
./gdas1/
gdas1.oct18.w3
./gdas1/
gdas1.oct18.w4
1 #NUMBER OF DIFFERENT POLLUTANTS
Unit #POLLUTANT IDENTIFICATION
10000 #EMISSION RATE (per hour)
1 #HOURS OF EMISSION
00 00 00 00 00 #RELEASE START TIME:YEAR MONTH DAY HOUR MINUTE
1 #NUMBER OF CONCENTRATION GRIDS
0 0 #CONC GRID CENTER LATITUDE LONGITUDE
50 50 #CONC GRID SPACING (degrees) LATITUDE LONGITUDE
50 50 #CONC GRID SPAN (degrees) LATITUDE LONGITUDE
./
cdump.4c
2 #NUMBER OF VERTICAL CONCENTRATION LEVELS
0 100 #HEIGHT OF EACH CONCENTRATION LEVEL (magl)
00 00 00 00 00 #SAMPLING START TIME:YEAR MONTH DAY HOUR MINUTE
00 00 00 00 00 #SAMPLING STOP TIME:YEAR MONTH DAY HOUR MINUTE
0 6 0 #SAMPLING INTERVAL: TYPE HOUR MINUTE
1 #NUMBER OF DEPOSITING POLLUTANTS
10.0 2.5 1.0 #PARTICLE:DIAMETER (um), DENSITY (g/cc), SHAPE
0.0 0.0 0.0 0.0 0.0 #DEP VEL (m/s), MW (g/Mole), SFC REACT. RATIO, DIFFUSIVITY RATIO, HENRY'S CONSTANT
0.0 0.0 0.0 #WET REMOVAL: HENRY'S (Molar/atm), INCLOUD (1/s), BELOWCLOUD (1/s)
0 #RADIOACTIVE DECAY HALFLIFE (days)
0.0 #POLLUTANT RESUSPENSION (1/m)
SETUP.CFG
&SETUP
NUMPAR = 10000,
MAXPAR = 10000,
INITD = 0,
ndump = 1,
ichem=5,
ncycl = 1,
poutf = 'PARDUMP.1',
/
~
~
~
Using 10,000 particles each with 1 unit mass.
calculating from the deposition I get 1087 particles "in" the deposition layer and
the MESSAGE file indicates 8913 particles left in the simulation.
Also  yes you are correct that the time averaging does not apply to the deposition layer.
The mass/Area indicated in the deposition layer is the total mass deposited over the averaging time divided
by the area. (Unlike the concentration which we discussed previously).
CONTROL FILE
18 10 17 15 #STARTING YEAR MONTH DAY HOUR
2 #NUMBER OF SOURCE LOCATIONS
36.949892 94.042969 50 #SOURCE 1 LATITUDE LONGITUDE HEIGHT(magl)
36.949892 94.042969 0 #SOURCE 2 LATITUDE LONGITUDE HEIGHT
6 #TOTAL RUN TIME (hours)
0 #USE MODEL VERTICAL VELOCITY
25000 #TOP OF MODEL DOMAIN (mAGL)
3 #NUMBER OF INPUT DATA GRIDS
./gdas1/
gdas1.oct18.w2
./gdas1/
gdas1.oct18.w3
./gdas1/
gdas1.oct18.w4
1 #NUMBER OF DIFFERENT POLLUTANTS
Unit #POLLUTANT IDENTIFICATION
10000 #EMISSION RATE (per hour)
1 #HOURS OF EMISSION
00 00 00 00 00 #RELEASE START TIME:YEAR MONTH DAY HOUR MINUTE
1 #NUMBER OF CONCENTRATION GRIDS
0 0 #CONC GRID CENTER LATITUDE LONGITUDE
50 50 #CONC GRID SPACING (degrees) LATITUDE LONGITUDE
50 50 #CONC GRID SPAN (degrees) LATITUDE LONGITUDE
./
cdump.4c
2 #NUMBER OF VERTICAL CONCENTRATION LEVELS
0 100 #HEIGHT OF EACH CONCENTRATION LEVEL (magl)
00 00 00 00 00 #SAMPLING START TIME:YEAR MONTH DAY HOUR MINUTE
00 00 00 00 00 #SAMPLING STOP TIME:YEAR MONTH DAY HOUR MINUTE
0 6 0 #SAMPLING INTERVAL: TYPE HOUR MINUTE
1 #NUMBER OF DEPOSITING POLLUTANTS
10.0 2.5 1.0 #PARTICLE:DIAMETER (um), DENSITY (g/cc), SHAPE
0.0 0.0 0.0 0.0 0.0 #DEP VEL (m/s), MW (g/Mole), SFC REACT. RATIO, DIFFUSIVITY RATIO, HENRY'S CONSTANT
0.0 0.0 0.0 #WET REMOVAL: HENRY'S (Molar/atm), INCLOUD (1/s), BELOWCLOUD (1/s)
0 #RADIOACTIVE DECAY HALFLIFE (days)
0.0 #POLLUTANT RESUSPENSION (1/m)
SETUP.CFG
&SETUP
NUMPAR = 10000,
MAXPAR = 10000,
INITD = 0,
ndump = 1,
ichem=5,
ncycl = 1,
poutf = 'PARDUMP.1',
/
~
~
~
Re: Time Step and Concentration Calculation
Dear alicec,
Thanks for your reply.
I am not clear about the formula for calculating the area.
The area is calculated as
(111198.323)^2 * lat * lon * cos(lat)
The grid area in my simulation is 5000 m * 5000 m, because the grid spacing is 0.05. How is your formula related to the grid spacing (0.05)?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Thanks for your reply.
I am not clear about the formula for calculating the area.
The area is calculated as
(111198.323)^2 * lat * lon * cos(lat)
The grid area in my simulation is 5000 m * 5000 m, because the grid spacing is 0.05. How is your formula related to the grid spacing (0.05)?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Re: Time Step and Concentration Calculation
Dear alicec,
Thank you very much for your reply and help. I really appreciate your help.
I used your area calculation formula to calculate the area. I did the calculation in an Excel file. Please see the attached output file.
It seems not right. That would be great if you could explain your formula in detail.
I have also attached CONTROL and SETUP files for my simulation. Could you please run them on your computer to see whether you could get the same output?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Thank you very much for your reply and help. I really appreciate your help.
I used your area calculation formula to calculate the area. I did the calculation in an Excel file. Please see the attached output file.
It seems not right. That would be great if you could explain your formula in detail.
I have also attached CONTROL and SETUP files for my simulation. Could you please run them on your computer to see whether you could get the same output?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
 Attachments

 10000_setup.txt
 (547 Bytes) Downloaded 41 times

 10000_control.txt
 (294 Bytes) Downloaded 42 times

 output.csv
 (24.84 KiB) Downloaded 43 times
Re: Time Step and Concentration Calculation
I made an error when writing down the formula. Please see the edited answer.
The area is calculated as
(111198.323)^2 * latspan * lonspan * cos(lat)
latspan and lonspan are the lengths (in degrees latitude, longitude) of the sides of the concentration grid.
The area is calculated as
(111198.323)^2 * latspan * lonspan * cos(lat)
latspan and lonspan are the lengths (in degrees latitude, longitude) of the sides of the concentration grid.
Re: Time Step and Concentration Calculation
Thank you very much, alicec.
I still could not get the correct numbers using your formula [e.g. (111198.323)^2 * 0.05 * 0.05 * COS(E2)]. I got negative values. Please see the attached file. Thanks.
I still could not get the correct numbers using your formula [e.g. (111198.323)^2 * 0.05 * 0.05 * COS(E2)]. I got negative values. Please see the attached file. Thanks.
 Attachments

 output.csv
 (21.65 KiB) Downloaded 42 times
Re: Time Step and Concentration Calculation
Did you convert degrees to radians when using the cosine function in excel?