Time Step and Concentration Calculation
Time Step and Concentration Calculation
Dear all,
I would like to clarify something on dispersal modelling.
I am aware that time steps vary a lot during simulation. Some time steps are 6 minutes while others can be 12 minutes. The minimum time step is 1 minute.
Question1: Does HYSPLIT calculate concentration for a grid cell based on each time step? If so, how are time steps defined? Do time steps depend on meteorological data files?
According to Section 12.1 in the online tutorial, one unit of mass can be allocated to each particle released, by configuring Mass (same as emission; CMASS=1) in Advanced/Configuration Setup/Concentration menu #8 the Concentration Grid Packing Method.
The mass of particles will be summed, as they pass through the grid cells. If we define a 3hour averaging sampling period for the grid cells, a single particle may reside in the cell for multiple time steps.
For instance, a ‘concentration’ value of 10 suggests a range of possibilities. One particle may have resided in the cell for 10 time steps or 10 particles may have been in the cell only for one time step. In such a case, concentration calculation only results in summation of particles’ mass in each cell, and the final result is no longer divided by the cell volume.
Question2: Is mass calculated based on summation of particles’ mass in a cell over a certain sampling period (e.g. 24 hours) that contains multiple time steps, if CMASS=1 [Mass (same as emission)] is set? If so, does it override sampling type [averaging (type=0)] specified in the control file?
Question3: Is concentration calculated based on averaging particles’ mass in a cell over a certain sampling period (e.g. 24 hours) that contains multiple time steps, and then being divided by the cell volume, if sampling type [averaging (type=0)] and CMASS=0 [Concentration (mass/m3)] are set?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
I would like to clarify something on dispersal modelling.
I am aware that time steps vary a lot during simulation. Some time steps are 6 minutes while others can be 12 minutes. The minimum time step is 1 minute.
Question1: Does HYSPLIT calculate concentration for a grid cell based on each time step? If so, how are time steps defined? Do time steps depend on meteorological data files?
According to Section 12.1 in the online tutorial, one unit of mass can be allocated to each particle released, by configuring Mass (same as emission; CMASS=1) in Advanced/Configuration Setup/Concentration menu #8 the Concentration Grid Packing Method.
The mass of particles will be summed, as they pass through the grid cells. If we define a 3hour averaging sampling period for the grid cells, a single particle may reside in the cell for multiple time steps.
For instance, a ‘concentration’ value of 10 suggests a range of possibilities. One particle may have resided in the cell for 10 time steps or 10 particles may have been in the cell only for one time step. In such a case, concentration calculation only results in summation of particles’ mass in each cell, and the final result is no longer divided by the cell volume.
Question2: Is mass calculated based on summation of particles’ mass in a cell over a certain sampling period (e.g. 24 hours) that contains multiple time steps, if CMASS=1 [Mass (same as emission)] is set? If so, does it override sampling type [averaging (type=0)] specified in the control file?
Question3: Is concentration calculated based on averaging particles’ mass in a cell over a certain sampling period (e.g. 24 hours) that contains multiple time steps, and then being divided by the cell volume, if sampling type [averaging (type=0)] and CMASS=0 [Concentration (mass/m3)] are set?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Re: Time Step and Concentration Calculation
The simplest case is the snapshot. At the output time, the total mass of all the particles in a grid box is summed and then divided by the volume of the grid box.
To get a time average, 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.
A simple example is the following:
for 10 minute time step concentration is 100 units,
for 20 minute time step concentration is 50 units and
for 30 minute time step concentration is 200 units.
To get the hour average concentration
100 units* 10/60 + 50 units* 20/60 + 200 units* 30/60 = 133.33 units
To understand how the time step is calculated see here:
https://ready.arl.noaa.gov/hysplitusersguide/S610.htm
Note – if you don’t want the time step to vary you may set the DELT option in the SETUP.CFG file.
Simply specify DELT=5 to always get 5 minutes time steps. As you already know, the smallest time step you can specify is 1 minute.
I hope this answers all of your questions.
To get a time average, 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.
A simple example is the following:
for 10 minute time step concentration is 100 units,
for 20 minute time step concentration is 50 units and
for 30 minute time step concentration is 200 units.
To get the hour average concentration
100 units* 10/60 + 50 units* 20/60 + 200 units* 30/60 = 133.33 units
To understand how the time step is calculated see here:
https://ready.arl.noaa.gov/hysplitusersguide/S610.htm
Note – if you don’t want the time step to vary you may set the DELT option in the SETUP.CFG file.
Simply specify DELT=5 to always get 5 minutes time steps. As you already know, the smallest time step you can specify is 1 minute.
I hope this answers all of your questions.
Re: Time Step and Concentration Calculation
Dear alicec,
Thank you very much for your reply. It is very helpful. I have a much better understanding of concentration calculation now, but I still have two questions.
Question 1: Could you please see the example below? Would I be correct?
For example, the sampling interval is every 2 hours (120 minutes) for the volume of a grid box (25 m3). The first hour has 4 time steps while there are 3 time steps in the second hour.
Hour 1: Time step 1 (15 minutes) the total mass of all the particles is 100 mass units
Hour 1: Time step 2 (15 minutes) the total mass of all the particles is 50 mass units
Hour 1: Time step 3 (15 minutes) the total mass of all the particles is 200 mass units
Hour 1: Time step 4 (15 minutes) the total mass of all the particles is 60 mass units
Hour 2: Time step 5 (20 minutes) the total mass of all the particles is 70 mass units
Hour 2: Time step 6 (20 minutes) the total mass of all the particles is 80 mass units
Hour 2: Time step 7 (20 minutes) the total mass of all the particles is 90 mass units
If the sampling type is averaging (type=0), the concentration will be:
(100*15/120+50*15/120+200*15/120+60*15/120+70*20/120+80*20/120+90*20/120)/25 = 3.67 mass/m3
If the sampling type is now/snapshot (type=1), the concentration will be:
90/25 = 3.6 mass/m3
If the sampling type is maximum (type=2), the concentration will be:
200/25 = 8 mass/m3
Question 2: Would I be able to convert concentration to the number of particles in a grid box?
For example, I released 1000 particles and the total mass emitted was 1000 mass units. CMASS=0 [Concentration (mass/m3)] was set. I then got an average concentration value (3.67 mass/m3) for a grid box (25 m3) over a 2 hours sampling interval.
the total mass of particles = the average concentration * the volume of a grid box = 91.75 mass units
Would I be able to interpret that there were approximately 92 particles in this grid box over 2 hours, since the number of particles released and the total mass emitted were same?
Your help would be much appreciated. Thank you very much again.
Kind Regards,
Ming
Thank you very much for your reply. It is very helpful. I have a much better understanding of concentration calculation now, but I still have two questions.
Question 1: Could you please see the example below? Would I be correct?
For example, the sampling interval is every 2 hours (120 minutes) for the volume of a grid box (25 m3). The first hour has 4 time steps while there are 3 time steps in the second hour.
Hour 1: Time step 1 (15 minutes) the total mass of all the particles is 100 mass units
Hour 1: Time step 2 (15 minutes) the total mass of all the particles is 50 mass units
Hour 1: Time step 3 (15 minutes) the total mass of all the particles is 200 mass units
Hour 1: Time step 4 (15 minutes) the total mass of all the particles is 60 mass units
Hour 2: Time step 5 (20 minutes) the total mass of all the particles is 70 mass units
Hour 2: Time step 6 (20 minutes) the total mass of all the particles is 80 mass units
Hour 2: Time step 7 (20 minutes) the total mass of all the particles is 90 mass units
If the sampling type is averaging (type=0), the concentration will be:
(100*15/120+50*15/120+200*15/120+60*15/120+70*20/120+80*20/120+90*20/120)/25 = 3.67 mass/m3
If the sampling type is now/snapshot (type=1), the concentration will be:
90/25 = 3.6 mass/m3
If the sampling type is maximum (type=2), the concentration will be:
200/25 = 8 mass/m3
Question 2: Would I be able to convert concentration to the number of particles in a grid box?
For example, I released 1000 particles and the total mass emitted was 1000 mass units. CMASS=0 [Concentration (mass/m3)] was set. I then got an average concentration value (3.67 mass/m3) for a grid box (25 m3) over a 2 hours sampling interval.
the total mass of particles = the average concentration * the volume of a grid box = 91.75 mass units
Would I be able to interpret that there were approximately 92 particles in this grid box over 2 hours, since the number of particles released and the total mass emitted were same?
Your help would be much appreciated. Thank you very much again.
Kind Regards,
Ming
Re: Time Step and Concentration Calculation
Question 1 – Yes, this is correct.
Question 2 – You can work backward to get the number of particles from the concentration as long as the mass on the particles is not changing. Mass on the computational particle may change if you have wet or dry deposition turned on. (see below for more on deposition).
https://ready.arl.noaa.gov/hysplitusersguide/S314.htm
An extra note: A common question people ask is how many computational particles are needed for their simulation. One way to estimate this is to decide what is the lowest concentration that you need to resolve Cmin, the size of your concentration grid box V, and then make sure that the particle number N=Cmin * V / (mass per particle) that would produce that concentration is fairly large – probably between 10 and 100.
Question 2 – You can work backward to get the number of particles from the concentration as long as the mass on the particles is not changing. Mass on the computational particle may change if you have wet or dry deposition turned on. (see below for more on deposition).
https://ready.arl.noaa.gov/hysplitusersguide/S314.htm
An extra note: A common question people ask is how many computational particles are needed for their simulation. One way to estimate this is to decide what is the lowest concentration that you need to resolve Cmin, the size of your concentration grid box V, and then make sure that the particle number N=Cmin * V / (mass per particle) that would produce that concentration is fairly large – probably between 10 and 100.
Re: Time Step and Concentration Calculation
Dear alicec,
I actually would like to include dry deposition for my dispersal simulation, and will only sample concentration in a grid cell (m2) on the ground level (0 m).
How do I ensure that the mass on the particles is not changing?
According to the link that you sent, 'In the special case where the dry deposition velocity is set to a value less than zero, the absolute value will be used to compute gravitational settling but with no mass removal.'
Do you think that the mass on the particles will not be changing if I set Deposition velocity (m/s) to a negative value?
According to the tutorial, the particles will stick to the surface/ground, if ICHEM=5 (Deposit particles rather than reducing their mass) is set.
Do you think that I would be able to sample concentration in a grid cell (m2) on the ground level (0 m), if ICHEM=5 is set?
Thank you very much, again.
Kind Regards,
Ming
I actually would like to include dry deposition for my dispersal simulation, and will only sample concentration in a grid cell (m2) on the ground level (0 m).
How do I ensure that the mass on the particles is not changing?
According to the link that you sent, 'In the special case where the dry deposition velocity is set to a value less than zero, the absolute value will be used to compute gravitational settling but with no mass removal.'
Do you think that the mass on the particles will not be changing if I set Deposition velocity (m/s) to a negative value?
According to the tutorial, the particles will stick to the surface/ground, if ICHEM=5 (Deposit particles rather than reducing their mass) is set.
Do you think that I would be able to sample concentration in a grid cell (m2) on the ground level (0 m), if ICHEM=5 is set?
Thank you very much, again.
Kind Regards,
Ming
Re: Time Step and Concentration Calculation
If you use the option to not have any deposition, then the particles will not change mass, but you also will not see any mass in the
deposition layer (0 m).
You could use ichem=5 if you want to calculate deposition without the particles losing mass.
deposition layer (0 m).
You could use ichem=5 if you want to calculate deposition without the particles losing mass.
Re: Time Step and Concentration Calculation
Dear alicec,
Thank you very much for your help. I really appreciate it. Now, I have a much better understanding of how HYSPLIT works.
I would just like to confirm some of my final thoughts. Please see the example below and the attached MESSAGE file.
Number particle released: 10,000
Emission rate: 3333.333333 mass units/hour
Emission duration: 3 hours
Total mass emitted (10,000 = 3333.333333 * 3) is equivalent to the total number of particles released (10,000).
1 particle = 1 mass unit
ICHEM=5 (Deposit particles rather than reducing their mass)
Total run time (simulation duration): 25 hours
After the simulation, I looked at the MESSAGE file. It showed how many particles and how much mass were in the atmosphere.
Hour 1: 3335 (particles) vs. 3000.531 (mass units)
Hour 2: 6336 (particles) vs. 5826.308 (mass units)
Hour 3: 9162 (particles) vs. 8442.341 (mass units)
Hour 25: 4781 (particles) vs. 4709.621 (mass units)
Question 1: The total number of particles did not match the total mass units exactly, but they were similar. Does this actually how HYSPLIT works during simulations?
Question 2: After Hour 3, there should be 10,000 rather than 9162 in the MESSAGE file. It seems that the particles started depositing on the ground during the first three hours, although they were still in emission duration. Would I be correct?
Question 3: In this simulation example, there is for instance a concentration value (3.67 mass/m2) for a grid cell (25 m2) on the ground level.
the total mass of particles = the average concentration * the size of a grid cell = 91.75 mass units
Since I made 1 particle = 1 mass unit, would I be able to interpret that there were approximately 92 particles in this grid cell?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Thank you very much for your help. I really appreciate it. Now, I have a much better understanding of how HYSPLIT works.
I would just like to confirm some of my final thoughts. Please see the example below and the attached MESSAGE file.
Number particle released: 10,000
Emission rate: 3333.333333 mass units/hour
Emission duration: 3 hours
Total mass emitted (10,000 = 3333.333333 * 3) is equivalent to the total number of particles released (10,000).
1 particle = 1 mass unit
ICHEM=5 (Deposit particles rather than reducing their mass)
Total run time (simulation duration): 25 hours
After the simulation, I looked at the MESSAGE file. It showed how many particles and how much mass were in the atmosphere.
Hour 1: 3335 (particles) vs. 3000.531 (mass units)
Hour 2: 6336 (particles) vs. 5826.308 (mass units)
Hour 3: 9162 (particles) vs. 8442.341 (mass units)
Hour 25: 4781 (particles) vs. 4709.621 (mass units)
Question 1: The total number of particles did not match the total mass units exactly, but they were similar. Does this actually how HYSPLIT works during simulations?
Question 2: After Hour 3, there should be 10,000 rather than 9162 in the MESSAGE file. It seems that the particles started depositing on the ground during the first three hours, although they were still in emission duration. Would I be correct?
Question 3: In this simulation example, there is for instance a concentration value (3.67 mass/m2) for a grid cell (25 m2) on the ground level.
the total mass of particles = the average concentration * the size of a grid cell = 91.75 mass units
Since I made 1 particle = 1 mass unit, would I be able to interpret that there were approximately 92 particles in this grid cell?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
 Attachments

 MESSAGE_dry_ichem.txt
 (20.31 KiB) Downloaded 100 times
Re: Time Step and Concentration Calculation
I’m going to answer your questions slightly out of order.
Question 2: The deposition calculation begins at the start of the simulation. During a three hour simulation with continuous emissions, some computational particles will be created at each time step. Once the particles are in the simulation, they may be deposited (or lose mass to deposition depending on your options). HYSPLIT does not wait until all the particles are emitted to begin deposition.
Question 1: The message file is writing out the number of particles before the particles with 0 mass have been officially removed from the simulation. 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. However, the MESSAGE file is written before that happens. So that is why your MESSAGE file is close but not exact for number of particles matching the amount of mass that you would expect. Some of those 3335 particles in hour 1 actually have 0 mass.
Question 3: I’m not quite sure what you are trying to do here so I will try to be as clear as possible.
A. The deposition layer is giving you mass/area not concentration which is mass/volume.
B. There were actually many more particles “In” (above?) the grid cell since not all the particles would have deposited.
C. Also if you are looking at a time average you have to remember that it will be an average number of particles over that time. So if you had 30 particles for time step of 20 minutes, 20 particles for time step of 30 minutes and 10 particles for time step of 10 minutes, you would end up with hour average of
30 * (20/60) + 20 * (30/60) + 10 (10/60) = 26.6 (NOT 30 + 20 + 10 = 60)
Question 2: The deposition calculation begins at the start of the simulation. During a three hour simulation with continuous emissions, some computational particles will be created at each time step. Once the particles are in the simulation, they may be deposited (or lose mass to deposition depending on your options). HYSPLIT does not wait until all the particles are emitted to begin deposition.
Question 1: The message file is writing out the number of particles before the particles with 0 mass have been officially removed from the simulation. 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. However, the MESSAGE file is written before that happens. So that is why your MESSAGE file is close but not exact for number of particles matching the amount of mass that you would expect. Some of those 3335 particles in hour 1 actually have 0 mass.
Question 3: I’m not quite sure what you are trying to do here so I will try to be as clear as possible.
A. The deposition layer is giving you mass/area not concentration which is mass/volume.
B. There were actually many more particles “In” (above?) the grid cell since not all the particles would have deposited.
C. Also if you are looking at a time average you have to remember that it will be an average number of particles over that time. So if you had 30 particles for time step of 20 minutes, 20 particles for time step of 30 minutes and 10 particles for time step of 10 minutes, you would end up with hour average of
30 * (20/60) + 20 * (30/60) + 10 (10/60) = 26.6 (NOT 30 + 20 + 10 = 60)
Re: Time Step and Concentration Calculation
Dear alicec,
Thank you very much for your reply.
I am very sorry that I did not make Question 3 clear.
For Question 3, I would only like to calculate the number of particles via concentration/density (mass/area) on the deposition layer.
As you suggested before, I would be able to work backward to get the number of particles from the concentration as long as the mass on the particles is not changing. To achieve that, I set ICHEM=5 in HYSPLIT, and made each particle equivalent to 1 mass unit.
If I get a concentration value (mass/area) for a grid cell on the ground level, would I be able to to work backward to get the number of particles from the concentration? Or are there any other ways to do that?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Thank you very much for your reply.
I am very sorry that I did not make Question 3 clear.
For Question 3, I would only like to calculate the number of particles via concentration/density (mass/area) on the deposition layer.
As you suggested before, I would be able to work backward to get the number of particles from the concentration as long as the mass on the particles is not changing. To achieve that, I set ICHEM=5 in HYSPLIT, and made each particle equivalent to 1 mass unit.
If I get a concentration value (mass/area) for a grid cell on the ground level, would I be able to to work backward to get the number of particles from the concentration? Or are there any other ways to do that?
Your help would be much appreciated. Thanks.
Kind Regards,
Ming
Re: Time Step and Concentration Calculation
I don't know what you mean by the number of particles on the deposition layer.
Maybe if you explain what you think this number is going to represent physically in the real world or what information
you hope to get from this number I can help you better.
Maybe if you explain what you think this number is going to represent physically in the real world or what information
you hope to get from this number I can help you better.