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Re: No bug in peakflow

Silvia Franceschi-2
Dear Matteo,
third column of the peakflow output file is not the precipitation as you probably would like to have, that was a check in the volumes that I did when I changed peakflow and needed something to check the results.
In fact we should delete this column, but always forgot, so your seems to be more a suggestion than a bug... we could use the third column of the peakflow file to store the precipitation in the next versions of the program.

For the second question... Jeff is in fact the effective precipitation and for now we have the most conservative approach that is Jeff==J, for me Jeff is the precipitation that contributes to the discharge, only in this case Jeff=J.
We are thinking to add some method for rainfall abstraction like SCS and so on, so at that time the J that will be used as input in peakflow will be Jeff != J.

Ciao

Silvia


Dear JGrass developers,
I have a question about the function h.peakflow.
I have applied h.peakflow to a basin and I have considered a_idf=33.6, n_idf=0.49 and timestep=100 [s]. From the simulation one obtains that tpmax=88811sec
As far as know, h.peakflow considers an event of a constant precipitation that lasts for tpmax. Using the rain curves, one should obtain a total rain of:

33.6*(88811/3600)^0.49=161.62 [mm]

I have opened the output file and I have summed all the values in the precipitation column (third column) obtaining:

sum(output[,3])
[1] 143.5201

which is less than 161.62 [mm].

I have looked to the code and, as far as I understand, the third column is calculated in the file QStatistic.java:

Q[j][3] = h;

The variable h is calculated in the file StatisticJeff.java at the line 87:

double h = a_idf * Math.pow(tpmax / 3600.0, n_idf)/1000;

where tpmax is the precipitation time, and a_idf and n_idf are the parameters of the rain curves. 

I think this is not correct. In fact, this h represents the total height of precipitation fallen in tpmax and is expressed in meters [m]. It results in:
> 33.6*(88811/3600)^0.49/1000
[1] 0.1616217

In fact the third column of the output file is a series of  0.16162 until tpmax. According to the code, this would mean that a rain of 0.16 meters is fallen on the basin for a duration of tpmax which counts for 143.5 meters of rain!!!

For this reason, I think the correct line should be:

Q[j][3] = J*1000*t; // [mm]

which represents rain fallen on the basin during the timestep given in input. J is correctly calculated as:

double J = a_idf * Math.pow(tpmax / 3600.0, n_idf - 1) / (1000.0 * 3600.0); // [m/s]


In this case one would obtain in the output file a series of 0.1819839 [mm] which, multiplied by (88811/100) gives exactly 161.62 [mm] of rain.

Another thing I would like to ask refers to the name Jeff: to me Jeff sounds like "effective rainfall intensity". However in the code the rain is given all in input, i.e. Jeff is equal to J.
Is maybe the name misleading?

Thanks in advance,

Matteo Dall'Amico



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Re: No bug in peakflow

Matteo Dall'Amico-2
Dear Silvia, 
thank you for the answer.
I think that a column with the input precipitation [mm] during the time_step would be highly appreciated.
The user could have a direct view of the rain hyetograph in input to the model. At this stage it is constant, but in the future it could change.
I also suggest to put a header in the output file, explaining the column and the measurement unit.
Regards,

Matteo

*****************************************
Ing. Matteo Dall'Amico
MOUNTAIN-EERING S.r.l. - G.m.b.H.
Spin Off dell'Università di Trento - Startup del TIS innovation park di Bolzano
web: www.mountain-eering.com
Sede legale: Siemensstr. 19 via Siemens, Bozen 39100 Bolzano - Italy
Ufficio Tecnico: Via Giusti 10, 38100 Trento - Italy
Mobile: +39 328 4235367

On Feb 18, 2010, at 3:22, Silvia Franceschi wrote:

Dear Matteo,
third column of the peakflow output file is not the precipitation as you probably would like to have, that was a check in the volumes that I did when I changed peakflow and needed something to check the results.
In fact we should delete this column, but always forgot, so your seems to be more a suggestion than a bug... we could use the third column of the peakflow file to store the precipitation in the next versions of the program.

For the second question... Jeff is in fact the effective precipitation and for now we have the most conservative approach that is Jeff==J, for me Jeff is the precipitation that contributes to the discharge, only in this case Jeff=J.
We are thinking to add some method for rainfall abstraction like SCS and so on, so at that time the J that will be used as input in peakflow will be Jeff != J.

Ciao

Silvia


Dear JGrass developers,
I have a question about the function h.peakflow.
I have applied h.peakflow to a basin and I have considered a_idf=33.6, n_idf=0.49 and timestep=100 [s]. From the simulation one obtains that tpmax=88811sec
As far as know, h.peakflow considers an event of a constant precipitation that lasts for tpmax. Using the rain curves, one should obtain a total rain of:

33.6*(88811/3600)^0.49=161.62 [mm]

I have opened the output file and I have summed all the values in the precipitation column (third column) obtaining:

sum(output[,3])
[1] 143.5201

which is less than 161.62 [mm].

I have looked to the code and, as far as I understand, the third column is calculated in the file QStatistic.java:

Q[j][3] = h;

The variable h is calculated in the file StatisticJeff.java at the line 87:

double h = a_idf * Math.pow(tpmax / 3600.0, n_idf)/1000;

where tpmax is the precipitation time, and a_idf and n_idf are the parameters of the rain curves. 

I think this is not correct. In fact, this h represents the total height of precipitation fallen in tpmax and is expressed in meters [m]. It results in:
> 33.6*(88811/3600)^0.49/1000
[1] 0.1616217

In fact the third column of the output file is a series of  0.16162 until tpmax. According to the code, this would mean that a rain of 0.16 meters is fallen on the basin for a duration of tpmax which counts for 143.5 meters of rain!!!

For this reason, I think the correct line should be:

Q[j][3] = J*1000*t; // [mm]

which represents rain fallen on the basin during the timestep given in input. J is correctly calculated as:

double J = a_idf * Math.pow(tpmax / 3600.0, n_idf - 1) / (1000.0 * 3600.0); // [m/s]


In this case one would obtain in the output file a series of 0.1819839 [mm] which, multiplied by (88811/100) gives exactly 161.62 [mm] of rain.

Another thing I would like to ask refers to the name Jeff: to me Jeff sounds like "effective rainfall intensity". However in the code the rain is given all in input, i.e. Jeff is equal to J.
Is maybe the name misleading?

Thanks in advance,

Matteo Dall'Amico



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JGrass-users mailing list
[hidden email]
https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users


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[hidden email]
https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users


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Re: No bug in peakflow

Andrea Antonello
Hi Matteo,
we are always glad to accept patches for enhancements.
So if you would like to see changes in the module, I suggest to open a
report in the JGrass tracker and perhaps since you are able to browse
and understand the code, attach the patch for the change.

Ciao
Andrea



On Thu, Feb 18, 2010 at 6:07 PM, Matteo Dall'Amico
<[hidden email]> wrote:

> Dear Silvia,
> thank you for the answer.
> I think that a column with the input precipitation [mm] during the time_step
> would be highly appreciated.
> The user could have a direct view of the rain hyetograph in input to the
> model. At this stage it is constant, but in the future it could change.
> I also suggest to put a header in the output file, explaining the column and
> the measurement unit.
> Regards,
> Matteo
> *****************************************
> Ing. Matteo Dall'Amico
> MOUNTAIN-EERING S.r.l. - G.m.b.H.
> Spin Off dell'Università di Trento - Startup del TIS innovation park di
> Bolzano
> web: www.mountain-eering.com
> Sede legale: Siemensstr. 19 via Siemens, Bozen 39100 Bolzano - Italy
> Ufficio Tecnico: Via Giusti 10, 38100 Trento - Italy
> Mobile: +39 328 4235367
> e-mail: [hidden email]
> On Feb 18, 2010, at 3:22, Silvia Franceschi wrote:
>
> Dear Matteo,
> third column of the peakflow output file is not the precipitation as you
> probably would like to have, that was a check in the volumes that I did when
> I changed peakflow and needed something to check the results.
> In fact we should delete this column, but always forgot, so your seems to be
> more a suggestion than a bug... we could use the third column of the
> peakflow file to store the precipitation in the next versions of the
> program.
>
> For the second question... Jeff is in fact the effective precipitation and
> for now we have the most conservative approach that is Jeff==J, for me Jeff
> is the precipitation that contributes to the discharge, only in this case
> Jeff=J.
> We are thinking to add some method for rainfall abstraction like SCS and so
> on, so at that time the J that will be used as input in peakflow will be
> Jeff != J.
>
> Ciao
>
> Silvia
>
>
>> Dear JGrass developers,
>> I have a question about the function h.peakflow.
>> I have applied h.peakflow to a basin and I have considered a_idf=33.6,
>> n_idf=0.49 and timestep=100 [s]. From the simulation one obtains that
>> tpmax=88811sec
>> As far as know, h.peakflow considers an event of a constant precipitation
>> that lasts for tpmax. Using the rain curves, one should obtain a total rain
>> of:
>> 33.6*(88811/3600)^0.49=161.62 [mm]
>> I have opened the output file and I have summed all the values in the
>> precipitation column (third column) obtaining:
>> sum(output[,3])
>> [1] 143.5201
>> which is less than 161.62 [mm].
>> I have looked to the code and, as far as I understand, the third column is
>> calculated in the file QStatistic.java:
>> Q[j][3] = h;
>> The variable h is calculated in the file StatisticJeff.java at the line
>> 87:
>> double h = a_idf * Math.pow(tpmax / 3600.0, n_idf)/1000;
>> where tpmax is the precipitation time, and a_idf and n_idf are the
>> parameters of the rain curves.
>> I think this is not correct. In fact, this h represents the total height
>> of precipitation fallen in tpmax and is expressed in meters [m]. It results
>> in:
>> > 33.6*(88811/3600)^0.49/1000
>> [1] 0.1616217
>> In fact the third column of the output file is a series of  0.16162 until
>> tpmax. According to the code, this would mean that a rain of 0.16 meters is
>> fallen on the basin for a duration of tpmax which counts for 143.5 meters of
>> rain!!!
>> For this reason, I think the correct line should be:
>> Q[j][3] = J*1000*t; // [mm]
>> which represents rain fallen on the basin during the timestep given in
>> input. J is correctly calculated as:
>> double J = a_idf * Math.pow(tpmax / 3600.0, n_idf - 1) / (1000.0 *
>> 3600.0); // [m/s]
>>
>> In this case one would obtain in the output file a series of 0.1819839
>> [mm] which, multiplied by (88811/100) gives exactly 161.62 [mm] of rain.
>> Another thing I would like to ask refers to the name Jeff: to me Jeff
>> sounds like "effective rainfall intensity". However in the code the rain is
>> given all in input, i.e. Jeff is equal to J.
>> Is maybe the name misleading?
>> Thanks in advance,
>> Matteo Dall'Amico
>>
>>
>> _______________________________________________
>> JGrass-users mailing list
>> [hidden email]
>> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>>
>
> _______________________________________________
> JGrass-users mailing list
> [hidden email]
> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>
>
> _______________________________________________
> JGrass-users mailing list
> [hidden email]
> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>
>
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Re: No bug in peakflow

Matteo Dall'Amico-2
Hey,
thanks Andrea. I don't know if I am "good" enough to attach the patch  
for the change but I will try to do it :)
I think that I will find the information on the procedures in the  
wiki. Otherwise, please let me know.
Ciao!

Matteo

On Feb 20, 2010, at 12:50, andrea antonello wrote:

> Hi Matteo,
> we are always glad to accept patches for enhancements.
> So if you would like to see changes in the module, I suggest to open a
> report in the JGrass tracker and perhaps since you are able to browse
> and understand the code, attach the patch for the change.
>
> Ciao
> Andrea
>
>
>
> On Thu, Feb 18, 2010 at 6:07 PM, Matteo Dall'Amico
> <[hidden email]> wrote:
>> Dear Silvia,
>> thank you for the answer.
>> I think that a column with the input precipitation [mm] during the  
>> time_step
>> would be highly appreciated.
>> The user could have a direct view of the rain hyetograph in input  
>> to the
>> model. At this stage it is constant, but in the future it could  
>> change.
>> I also suggest to put a header in the output file, explaining the  
>> column and
>> the measurement unit.
>> Regards,
>> Matteo
>> *****************************************
>> Ing. Matteo Dall'Amico
>> MOUNTAIN-EERING S.r.l. - G.m.b.H.
>> Spin Off dell'Università di Trento - Startup del TIS innovation  
>> park di
>> Bolzano
>> web: www.mountain-eering.com
>> Sede legale: Siemensstr. 19 via Siemens, Bozen 39100 Bolzano - Italy
>> Ufficio Tecnico: Via Giusti 10, 38100 Trento - Italy
>> Mobile: +39 328 4235367
>> e-mail: [hidden email]
>> On Feb 18, 2010, at 3:22, Silvia Franceschi wrote:
>>
>> Dear Matteo,
>> third column of the peakflow output file is not the precipitation  
>> as you
>> probably would like to have, that was a check in the volumes that I  
>> did when
>> I changed peakflow and needed something to check the results.
>> In fact we should delete this column, but always forgot, so your  
>> seems to be
>> more a suggestion than a bug... we could use the third column of the
>> peakflow file to store the precipitation in the next versions of the
>> program.
>>
>> For the second question... Jeff is in fact the effective  
>> precipitation and
>> for now we have the most conservative approach that is Jeff==J, for  
>> me Jeff
>> is the precipitation that contributes to the discharge, only in  
>> this case
>> Jeff=J.
>> We are thinking to add some method for rainfall abstraction like  
>> SCS and so
>> on, so at that time the J that will be used as input in peakflow  
>> will be
>> Jeff != J.
>>
>> Ciao
>>
>> Silvia
>>
>>
>>> Dear JGrass developers,
>>> I have a question about the function h.peakflow.
>>> I have applied h.peakflow to a basin and I have considered  
>>> a_idf=33.6,
>>> n_idf=0.49 and timestep=100 [s]. From the simulation one obtains  
>>> that
>>> tpmax=88811sec
>>> As far as know, h.peakflow considers an event of a constant  
>>> precipitation
>>> that lasts for tpmax. Using the rain curves, one should obtain a  
>>> total rain
>>> of:
>>> 33.6*(88811/3600)^0.49=161.62 [mm]
>>> I have opened the output file and I have summed all the values in  
>>> the
>>> precipitation column (third column) obtaining:
>>> sum(output[,3])
>>> [1] 143.5201
>>> which is less than 161.62 [mm].
>>> I have looked to the code and, as far as I understand, the third  
>>> column is
>>> calculated in the file QStatistic.java:
>>> Q[j][3] = h;
>>> The variable h is calculated in the file StatisticJeff.java at the  
>>> line
>>> 87:
>>> double h = a_idf * Math.pow(tpmax / 3600.0, n_idf)/1000;
>>> where tpmax is the precipitation time, and a_idf and n_idf are the
>>> parameters of the rain curves.
>>> I think this is not correct. In fact, this h represents the total  
>>> height
>>> of precipitation fallen in tpmax and is expressed in meters [m].  
>>> It results
>>> in:
>>>> 33.6*(88811/3600)^0.49/1000
>>> [1] 0.1616217
>>> In fact the third column of the output file is a series of  
>>> 0.16162 until
>>> tpmax. According to the code, this would mean that a rain of 0.16  
>>> meters is
>>> fallen on the basin for a duration of tpmax which counts for 143.5  
>>> meters of
>>> rain!!!
>>> For this reason, I think the correct line should be:
>>> Q[j][3] = J*1000*t; // [mm]
>>> which represents rain fallen on the basin during the timestep  
>>> given in
>>> input. J is correctly calculated as:
>>> double J = a_idf * Math.pow(tpmax / 3600.0, n_idf - 1) / (1000.0 *
>>> 3600.0); // [m/s]
>>>
>>> In this case one would obtain in the output file a series of  
>>> 0.1819839
>>> [mm] which, multiplied by (88811/100) gives exactly 161.62 [mm] of  
>>> rain.
>>> Another thing I would like to ask refers to the name Jeff: to me  
>>> Jeff
>>> sounds like "effective rainfall intensity". However in the code  
>>> the rain is
>>> given all in input, i.e. Jeff is equal to J.
>>> Is maybe the name misleading?
>>> Thanks in advance,
>>> Matteo Dall'Amico
>>>
>>>
>>> _______________________________________________
>>> JGrass-users mailing list
>>> [hidden email]
>>> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>>>
>>
>> _______________________________________________
>> JGrass-users mailing list
>> [hidden email]
>> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>>
>>
>> _______________________________________________
>> JGrass-users mailing list
>> [hidden email]
>> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>>
>>
> _______________________________________________
> JGrass-users mailing list
> [hidden email]
> https://dev.fsc.bz.it/cgi-bin/mailman/listinfo/jgrass-users
>

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