I am trying to plot data based on the specimen id, I have two types of speciments. 0.12C and 0.07C Steel, as seen in the photo here is the table.

I would like to plot the data with different markers for each Plate ID, but i can’t seem to figure out how to plot them. I have tried calling out the rows and varible when plotting but it won’t work.
Here is how i load the data:
Table4 = readtable(‘Grain Size.xlsx’,’sheet’,’KIC Master’,’VariableNamingRule’,’preserve’);
Here is how I am trying to plot:
plot(ax2,Table4(1:7,"Ferrite Grain Size-1/2 (in.)"),Table4(1:7,"Provisional To °F"),’Color’,’black’,’LineStyle’,’none’,’Marker’,’x’,’MarkerEdgeColor’,’black’,’MarkerFaceColor’,’black’);
plot(ax2,Table4(8:11,"Ferrite Grain Size-1/2 (in.)"),Table4(8:11,"Provisional To °F"),’Color’,’black’,’LineStyle’,’none’,’Marker’,’S’,’MarkerEdgeColor’,’black’,’MarkerFaceColor’,’black’);
I get this error:
Error using plot
Invalid subscript for Y. A table variable subscript must be a numeric array containing real positive integers, a logical array, a character vector, a string array, a cell
array of character vectors, or a pattern scalar used to match variable names.Error using plot
Invalid subscript for Y. A table variable subscript must be a numeric array containing real positive integers, a logical array, a character vector, a string array, a cell
array of character vectors, or a pattern scalar used to match variable names.

What am I missing here? The only way i have gotten this kind of thing to work in the past is by creating seaperate tables for each specimen type then plotting each table individually. I would like to at least keep them in the same table and be able to call our which rows I want to plot.
Thank You!I am trying to plot data based on the specimen id, I have two types of speciments. 0.12C and 0.07C Steel, as seen in the photo here is the table.

I would like to plot the data with different markers for each Plate ID, but i can’t seem to figure out how to plot them. I have tried calling out the rows and varible when plotting but it won’t work.
Here is how i load the data:
Table4 = readtable(‘Grain Size.xlsx’,’sheet’,’KIC Master’,’VariableNamingRule’,’preserve’);
Here is how I am trying to plot:
plot(ax2,Table4(1:7,"Ferrite Grain Size-1/2 (in.)"),Table4(1:7,"Provisional To °F"),’Color’,’black’,’LineStyle’,’none’,’Marker’,’x’,’MarkerEdgeColor’,’black’,’MarkerFaceColor’,’black’);
plot(ax2,Table4(8:11,"Ferrite Grain Size-1/2 (in.)"),Table4(8:11,"Provisional To °F"),’Color’,’black’,’LineStyle’,’none’,’Marker’,’S’,’MarkerEdgeColor’,’black’,’MarkerFaceColor’,’black’);
I get this error:
Error using plot
Invalid subscript for Y. A table variable subscript must be a numeric array containing real positive integers, a logical array, a character vector, a string array, a cell
array of character vectors, or a pattern scalar used to match variable names.Error using plot
Invalid subscript for Y. A table variable subscript must be a numeric array containing real positive integers, a logical array, a character vector, a string array, a cell
array of character vectors, or a pattern scalar used to match variable names.

What am I missing here? The only way i have gotten this kind of thing to work in the past is by creating seaperate tables for each specimen type then plotting each table individually. I would like to at least keep them in the same table and be able to call our which rows I want to plot.
Thank You! I am trying to plot data based on the specimen id, I have two types of speciments. 0.12C and 0.07C Steel, as seen in the photo here is the table.

I would like to plot the data with different markers for each Plate ID, but i can’t seem to figure out how to plot them. I have tried calling out the rows and varible when plotting but it won’t work.
Here is how i load the data:
Table4 = readtable(‘Grain Size.xlsx’,’sheet’,’KIC Master’,’VariableNamingRule’,’preserve’);
Here is how I am trying to plot:
plot(ax2,Table4(1:7,"Ferrite Grain Size-1/2 (in.)"),Table4(1:7,"Provisional To °F"),’Color’,’black’,’LineStyle’,’none’,’Marker’,’x’,’MarkerEdgeColor’,’black’,’MarkerFaceColor’,’black’);
plot(ax2,Table4(8:11,"Ferrite Grain Size-1/2 (in.)"),Table4(8:11,"Provisional To °F"),’Color’,’black’,’LineStyle’,’none’,’Marker’,’S’,’MarkerEdgeColor’,’black’,’MarkerFaceColor’,’black’);
I get this error:
Error using plot
Invalid subscript for Y. A table variable subscript must be a numeric array containing real positive integers, a logical array, a character vector, a string array, a cell
array of character vectors, or a pattern scalar used to match variable names.Error using plot
Invalid subscript for Y. A table variable subscript must be a numeric array containing real positive integers, a logical array, a character vector, a string array, a cell
array of character vectors, or a pattern scalar used to match variable names.

What am I missing here? The only way i have gotten this kind of thing to work in the past is by creating seaperate tables for each specimen type then plotting each table individually. I would like to at least keep them in the same table and be able to call our which rows I want to plot.
Thank You! plotting MATLAB Answers — New Questions

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Hi everyone
I am a linux user and i have an arch installation running with X11 desktop management on my pc. I always struggled to run matlab on my laptop (interface wont open and stuff like that) since arch is not officially supported and i deferred to using the online version. After a bit i fixed the problems and started using the desktop version without issue up untill the last month when the glibc packag updated to the 2.4.1 while matlab needs the 2.4.0 and everythng broke again. Tired of all these problems i decided to opt for the dockerized version, i downloded the official ubuntu 24.04 docker from the docker hub and as said in the website i ran the following commands:
$ xhost +
$ docker run -it –rm -e DISPLAY=$DISPLAY -v /tmp/.X11-unix:/tmp/.X11-unix:ro –shm-size=512M mathworks/matlab:r2025a
and the terminal is hanging and no interface is opening. I thought it might be a licencing problem but my licence is given from my university and i dont know how to retreive the licence file. Can someone help me fix the problem (and identify it if its different from my suppositions)?
Also extra quick question, after solving this problem how do i download add-ons and libs? Same as in the desktop version?
Thanks in advance :)Hi everyone
I am a linux user and i have an arch installation running with X11 desktop management on my pc. I always struggled to run matlab on my laptop (interface wont open and stuff like that) since arch is not officially supported and i deferred to using the online version. After a bit i fixed the problems and started using the desktop version without issue up untill the last month when the glibc packag updated to the 2.4.1 while matlab needs the 2.4.0 and everythng broke again. Tired of all these problems i decided to opt for the dockerized version, i downloded the official ubuntu 24.04 docker from the docker hub and as said in the website i ran the following commands:
$ xhost +
$ docker run -it –rm -e DISPLAY=$DISPLAY -v /tmp/.X11-unix:/tmp/.X11-unix:ro –shm-size=512M mathworks/matlab:r2025a
and the terminal is hanging and no interface is opening. I thought it might be a licencing problem but my licence is given from my university and i dont know how to retreive the licence file. Can someone help me fix the problem (and identify it if its different from my suppositions)?
Also extra quick question, after solving this problem how do i download add-ons and libs? Same as in the desktop version?
Thanks in advance 🙂 Hi everyone
I am a linux user and i have an arch installation running with X11 desktop management on my pc. I always struggled to run matlab on my laptop (interface wont open and stuff like that) since arch is not officially supported and i deferred to using the online version. After a bit i fixed the problems and started using the desktop version without issue up untill the last month when the glibc packag updated to the 2.4.1 while matlab needs the 2.4.0 and everythng broke again. Tired of all these problems i decided to opt for the dockerized version, i downloded the official ubuntu 24.04 docker from the docker hub and as said in the website i ran the following commands:
$ xhost +
$ docker run -it –rm -e DISPLAY=$DISPLAY -v /tmp/.X11-unix:/tmp/.X11-unix:ro –shm-size=512M mathworks/matlab:r2025a
and the terminal is hanging and no interface is opening. I thought it might be a licencing problem but my licence is given from my university and i dont know how to retreive the licence file. Can someone help me fix the problem (and identify it if its different from my suppositions)?
Also extra quick question, after solving this problem how do i download add-ons and libs? Same as in the desktop version?
Thanks in advance 🙂 docker, matlab, linux MATLAB Answers — New Questions

​

Ideally this would be using the same format as the save command (.mat file), but instead of writing to disk directly, it would write it to data in memory.
The reason is I need to access the saved data directly to implement a custom hashing method on the variable, and have no need to write it to disk in order to achieve this, and would rather not go through the administratively-heavy option of creating a ram disk through ramfs.
I’d actually be OK with a version-change of MATLAB invalidating the object in my particular case (the idea being that any software updates may influence results, however unlikely).
Assuming full .mat support is not an option, I still need fairly general support, which is why the (intermediate) output of save seems like a good option: e.g., nested structs, cells, arrays. I could probably do without Java objects in this particular case.Ideally this would be using the same format as the save command (.mat file), but instead of writing to disk directly, it would write it to data in memory.
The reason is I need to access the saved data directly to implement a custom hashing method on the variable, and have no need to write it to disk in order to achieve this, and would rather not go through the administratively-heavy option of creating a ram disk through ramfs.
I’d actually be OK with a version-change of MATLAB invalidating the object in my particular case (the idea being that any software updates may influence results, however unlikely).
Assuming full .mat support is not an option, I still need fairly general support, which is why the (intermediate) output of save seems like a good option: e.g., nested structs, cells, arrays. I could probably do without Java objects in this particular case. Ideally this would be using the same format as the save command (.mat file), but instead of writing to disk directly, it would write it to data in memory.
The reason is I need to access the saved data directly to implement a custom hashing method on the variable, and have no need to write it to disk in order to achieve this, and would rather not go through the administratively-heavy option of creating a ram disk through ramfs.
I’d actually be OK with a version-change of MATLAB invalidating the object in my particular case (the idea being that any software updates may influence results, however unlikely).
Assuming full .mat support is not an option, I still need fairly general support, which is why the (intermediate) output of save seems like a good option: e.g., nested structs, cells, arrays. I could probably do without Java objects in this particular case. save, serialization, pickling, hashing, undocumented, serialize MATLAB Answers — New Questions

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MATLAB encountered a crash. How do I locate the crash log files on my computer. MATLAB encountered a crash. How do I locate the crash log files on my computer.  MATLAB encountered a crash. How do I locate the crash log files on my computer.   MATLAB Answers — New Questions

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Hey guys,
i’m trying to add controls to the attached model. I’d like to control the flow-rate of the 2 flow-rate-sources in order to change the heat/cold transfered to the MA-Network. When doing so manually there’s not the change i expected, so i figured, that i also need to change the nominal rate of heat-transfer of the system-level-heat-exchanger accordingly to the flow-rate. Also when the flow-rate is too low, i get an error. I assume it’s because the model thinks of the rate of heat-transfer as constant, so by chaning the flow-rate deltaT becomes incredibly high.
I tried to do so with a Matlab script, that influenced the parameters over time. But the graphs showed me this doesn’t work like inteded. And i’ve found this, saying basicly it’s not possible to add the controls i had in mind, am i wrong?
https://de.mathworks.com/help/simscape/ug/how-simscape-run-time-parameters-differs-from-simulink.html

Can somebody help me please? Maybe i’m just not seeing the obvious solution here 😀

Cheers,Hey guys,
i’m trying to add controls to the attached model. I’d like to control the flow-rate of the 2 flow-rate-sources in order to change the heat/cold transfered to the MA-Network. When doing so manually there’s not the change i expected, so i figured, that i also need to change the nominal rate of heat-transfer of the system-level-heat-exchanger accordingly to the flow-rate. Also when the flow-rate is too low, i get an error. I assume it’s because the model thinks of the rate of heat-transfer as constant, so by chaning the flow-rate deltaT becomes incredibly high.
I tried to do so with a Matlab script, that influenced the parameters over time. But the graphs showed me this doesn’t work like inteded. And i’ve found this, saying basicly it’s not possible to add the controls i had in mind, am i wrong?
https://de.mathworks.com/help/simscape/ug/how-simscape-run-time-parameters-differs-from-simulink.html

Can somebody help me please? Maybe i’m just not seeing the obvious solution here 😀

Cheers, Hey guys,
i’m trying to add controls to the attached model. I’d like to control the flow-rate of the 2 flow-rate-sources in order to change the heat/cold transfered to the MA-Network. When doing so manually there’s not the change i expected, so i figured, that i also need to change the nominal rate of heat-transfer of the system-level-heat-exchanger accordingly to the flow-rate. Also when the flow-rate is too low, i get an error. I assume it’s because the model thinks of the rate of heat-transfer as constant, so by chaning the flow-rate deltaT becomes incredibly high.
I tried to do so with a Matlab script, that influenced the parameters over time. But the graphs showed me this doesn’t work like inteded. And i’ve found this, saying basicly it’s not possible to add the controls i had in mind, am i wrong?
https://de.mathworks.com/help/simscape/ug/how-simscape-run-time-parameters-differs-from-simulink.html

Can somebody help me please? Maybe i’m just not seeing the obvious solution here 😀

Cheers, simscape, simulink, heatexchanger, fluids MATLAB Answers — New Questions

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Dear community,
Is it possible to create an electrical model using Simscape blocks (blue ones) and solve it in Discrete Phasor domain, i.e. RMS quantities (as it can be done using Specialized Power Systems)?
If not, is there any other alternative?
Thank you very much in advance.
Best regards,
Víctor Sánchez SuárezDear community,
Is it possible to create an electrical model using Simscape blocks (blue ones) and solve it in Discrete Phasor domain, i.e. RMS quantities (as it can be done using Specialized Power Systems)?
If not, is there any other alternative?
Thank you very much in advance.
Best regards,
Víctor Sánchez Suárez Dear community,
Is it possible to create an electrical model using Simscape blocks (blue ones) and solve it in Discrete Phasor domain, i.e. RMS quantities (as it can be done using Specialized Power Systems)?
If not, is there any other alternative?
Thank you very much in advance.
Best regards,
Víctor Sánchez Suárez simscape, discrete phasor, rms, specialized power systems MATLAB Answers — New Questions

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Hi all, we’ve had to transfer some Matlab code to a new computer for analysis of .abf files and moved from version R2018b to R2025a. However abfload is now giving us the error message below. Can anybody suggest a work around? The error message is below.

Error using :
Colon operands must be real scalars.
Error in
abfload (line 304)
h.protocolName=char(BigString(tmpIx1:tmpIx2(1)+3))’;
^Hi all, we’ve had to transfer some Matlab code to a new computer for analysis of .abf files and moved from version R2018b to R2025a. However abfload is now giving us the error message below. Can anybody suggest a work around? The error message is below.

Error using :
Colon operands must be real scalars.
Error in
abfload (line 304)
h.protocolName=char(BigString(tmpIx1:tmpIx2(1)+3))’;
^ Hi all, we’ve had to transfer some Matlab code to a new computer for analysis of .abf files and moved from version R2018b to R2025a. However abfload is now giving us the error message below. Can anybody suggest a work around? The error message is below.

Error using :
Colon operands must be real scalars.
Error in
abfload (line 304)
h.protocolName=char(BigString(tmpIx1:tmpIx2(1)+3))’;
^ abfload, colon operands MATLAB Answers — New Questions

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Hi,
I am given a contineuous water level data measured and measured cross section to calculate discharge by Manning equation. All these data were checked for gaps, and found contineuous. When running the following script the discharge having gaps, creats a jump which looks strange. Tried for several weeks to find it, but didn’t. Hope for your help.
%this script reads Cross and crocss-sections data from tributaries:
%Tzipori, Yiftachel, Tzvi, Oz, Mizra, Adashim, Taanach and Keini
%It 1. reads divers data 2. change them to be water level 3. calculate
%discharge.
%Discharge calculation requires:
%1. find the water level in the current time step 2. caculate discharge using manning
tic
clear all
close all

%%%%%%%%%%%% Part A: find water level of all sites%%%%%%%%%%%%%%%%%%%
% Get a list of all txt files in the current folder, or subfolders of it.

%Read discharge
lines1 = readlines("trib_names.txt");
trib_name=(categorical(lines1));

for i=1:length(trib_name)
trib1{i}=trib_name(i)
end

trib=string(trib1)+’level’; %for the case of water level;
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischdivers’;
lvl=NaN(418907,16);%columns: longest diver data, rows/2= # of tributaries. each st. ha two columns: for date and data of stations
AllDivresfiles = natsortfiles(dir(‘*.txt’));

for k =1:size(AllDivresfiles,1)
% k=5%shut ‘end’ at the end of loop. Find thie in ‘kishon_catchment_discharge_calc_for1trib’
thisfilename1 = AllDivresfiles(k).name; %just the name
thisdata1 = load(thisfilename1); %load just this file
Lng_lvl(k)=length(thisdata1);
date1=x2mdate(thisdata1(:,1));
%plot(date1,thisdata1(:,2));ylabel(‘Level (cm)’)
num_days=length(unique(thisdata1(:,1)));
interval=720;%720 is the daily period of the time interval of data: 2 min. length(thisdata1(:,1))/(num_days*24);%Model has 1 hours intervals, so this is the interval obtained here. Obtain intervals according to 10 min intrval
baseP=min(thisdata1(:,2));%background pressure
curr_lvl1=(thisdata1(:,2)-baseP)/100;%units change to meter%for smoothing turn curr_lvl to curr_lvl1
curr_lvl=smoothdata(curr_lvl1,’movmean’,interval);
%curr_lvl=smoothdata(curr_lvl,’gaussian’,interval);
lvl(1:size(thisdata1,1),(2*k-1):2*k)=[date1 curr_lvl];%Removing diver’s bias (correct only forephemeral tributaries) and converting to m
lngt(k)=length(curr_lvl);
end

in0=find((lvl)==0);
lvl(in0)=NaN;
figure(‘Name’,’Adashim’);adash=lvl(:,1:2);plot(adash(:,1),adash(:,2));datetick;title=’Adashim lvl’;
figure(‘Name’,’Keini’);keini=lvl(:,3:4);plot(keini(:,1),keini(:,2));title=’Keini lvl’;datetick;
figure(‘Name’,’KishonMaale’);KishonMaale=lvl(:,5:6);plot(KishonMaale(:,1),KishonMaale(:,2));title=’KishonMaale lvl’;datetick
figure(‘Name’,’Mizra’);mizra=lvl(:,7:8);plot(mizra(:,1),mizra(:,2));title=’Mizra lvl’;datetick;
figure(‘Name’,’Oz’);oz=lvl(:,9:10);plot(oz(:,1),oz(:,2));title=’Oz lvl’;datetick;
figure(‘Name’,’Taanach’);taanach=lvl(:,11:12);plot(taanach(:,1),taanach(:,2));title=’Taanach lvl’;datetick;
figure(‘Name’,’Tzvi’);title=’Tzvi lvl’;tzvi=lvl(:,13:14);plot(tzvi(:,1),tzvi(:,2));datetick;
figure(‘Name’,’Yiftachel’);title=’Yiftachel lvl’;yiftachel=lvl(:,15:16);plot(yiftachel(:,1),yiftachel(:,2));title=’Yiftachel lvl’;datetick;

%%%%%%%%%%%% Part B: find cross section%%%%%%%%%%%%%%%%%%%
%Read cross sections. In most tributaries the divers are in the downstream
%All files with ‘1’ are upstrream and all with ‘2’ are downstreaצ. For example: OzUp=Oz1, OzDown=Oz2

trib=length(AllDivresfiles);
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischcrossectionsDownstream’;
crs=NaN(max(lngt),5*trib);%columns: longest cross section data, rows-# of stations*5
AllCrossfiles = natsortfiles(dir(‘*.txt’));
min_for_slope1=NaN(trib,3);

%Collect all downstream crossection data together
for l = 1:length(AllCrossfiles)
%l=1 %shut ‘end’ at the end of loop
thisfilename2 = AllCrossfiles(l).name; %just the name
thisdata2 = load(thisfilename2); %load current file
crs(1:length(thisdata2),5*l-4:l*5)=thisdata2;
[MIN,I0]=min(thisdata2(:,3));%sea level elevation to have absolute elevation
min_for_slope1(l,:)=thisdata2(I0,1:3);%The slope is obtained by the slope bwtween two min of adjacent croo sections.
Lng(l)=length(thisdata2);
% figure(‘Name’,[char(trib_name(l)) ‘_crs_dn’])
% plot(crs(:,5*l-4),crs(:,5*l))
end

% adash1_crs=crs(:,4:5);keini1_crs=crs(:,9:10);mizra1_crs=crs(:,14:15);oz1_crs=crs(:,19:20);taanach1_crs=crs(:,24:25);tzvi1_crs=crs(:,29:30);yiftachel1_crs=crs(:,34:35);
% figure(11);plot(adash1_crs(:,1),adash1_crs(:,2));figure(22); plot(keini1_crs(:,1),keini1_crs(:,2));figure(33);plot(mizra1_crs(:,1),mizra1_crs(:,2));figure(44);plot(oz1_crs(:,1),oz1_crs(:,2));figure(55);plot(taanach1_crs(:,1),taanach1_crs(:,2));figure(66);plot(tzvi1_crs(:,1),tzvi1_crs(:,2));figure(77);plot(yiftachel1_crs(:,1),yiftachel1_crs(:,2));

%%%%% Part B1: Reading upstream for slope%%%%%%%%
min_for_slope2=NaN(trib,3);%coordinates (columns 1 and 2) and elevation (column3). of minimum in the cross sectioncolumns: longest diver data, rows/2= # of tributaries. each st. ha two columns: for date and data of stations
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischcrossections’;

%min_for_slope=NaN(trib*2,2);%columns: longest cross section data, rows-# of stations
Allupstrmfiles = orderfields(dir(‘*.txt’));
for j = 1 : trib
upstrmName = Allupstrmfiles(j).name; %just the name; j or l
thisupstrm = load(upstrmName); %load just this file
[MIN1,I1]=min(thisupstrm(:,3));%Sea level to get absolute elevation
min_for_slope2(j,:)=(thisupstrm(I1,1:3));
% figure(‘Name’,[char(trib_name(j)) ‘_crs_up’])
% plot(thisupstrm(:,4),thisupstrm(:,3))
end

%%%%%%%%%%%%%%%%NEW 29052025%%%%%%%%%%%%%%%%%%%%%%%%%%%
all_lvl=NaN(max(lngt),trib);
discharge_all=NaN(418907,trib*2);%longest data
discharge=NaN(418907,2);
g=0;
f=0;

for t=1:trib
x=crs(1:Lng(t),5*t-4);
y=crs(1:Lng(t),5*t-3);
crsc=crs(1:Lng(t),5*t);
x0=x(1);y0=y(1);
distance1=((x-x0).^2+(y-y0).^2).^(1/2);%crs(1:Lng(t),5*t-1);
%Interpolation to create smoother discharge calculation
DF=abs(min(crsc(1:length(crsc)-1)-crsc(2:length(crsc))))/300;%Order of magnitude of distance difference and elveation are similar
basic_distance=distance1(1):DF:distance1(end);
distance11=1:(length(crsc));
distance=interp1(distance11,distance1,basic_distance);%to do interpolation to distance , since t is not a stright line
smth_crs=interp1(distance1,crsc,distance);
% smth_x=interp1(distance1,x,distance);
% smth_y=interp1(distance1,y,distance);

for m=1:Lng_lvl(t)%m=233648

if isnan(lvl(m,2*t))
discharge(m,1:2)=[lvl(m,2*t-1) NaN];
f=f+1;
continue
end

[M lvl_ind1]=min(abs(lvl(m,2*t)-smth_crs));%Find current water level in lvl and the current cross section

if smth_crs(lvl_ind1)==min(smth_crs)
discharge(m,1:2)=[lvl(m,2*t-1) 0];
g=g+1;
continue
end

lvl_ind2=find(smth_crs<=smth_crs(lvl_ind1));
distance2=distance(lvl_ind2);smth_crs2=smth_crs(lvl_ind2);
P1=((distance2(2:end)-distance2(1:end-1)).^2+(smth_crs2(2:end)-smth_crs2(1:end-1)).^2).^(1/2);
% lvl_ind2=find(smth_crs<=smth_crs(lvl_ind1));
% lvl_ind22=crsc<=min(crsc);
% crsc_lvl=crsc(lvl_ind22);
% x_lvl=x(lvl_ind22);%Due to wetted perimeter.
% y_lvl=y(lvl_ind22);%Due to wetted perimeter.
% % smth_x_lvl=smth_x(lvl_ind2);%Due to wetted perimeter.
% % smth_y_lvl=smth_y(lvl_ind2);%Due to wetted perimeter.
% smth_crs_lvl=smth_crs(lvl_ind2);%Due to wetted perimeter.

%Wetted perimeter:
% P1=((x_lvl(2:end)-x_lvl(1:end-1)).^2+(y_lvl(2:end)-y_lvl(1:end-1)).^2+(crsc_lvl(2:end)-crsc_lvl(1:end-1)).^2).^(1/2);
P(m)=sum(P1);

if isnan(P(m))
discharge(m,1:2)=[lvl(m,2*t-1) 0];
q=q+1;
continue
end

%Slope
dx = min_for_slope2(t,1) – min_for_slope1(t,1);
dy = min_for_slope2(t,2) – min_for_slope1(t,2);
dz(t) = min_for_slope2(t,3) – min_for_slope1(t,3);
distance3D = sqrt(dx^2 + dy^2 + dz(t)^2);
S(t) = abs(dz(t)) / distance3D;
% S(t)=min_for_slope2(t,3)-min_for_slope1(t,3)/(min_for_slope2(t,1)-min_for_slope1(t,1))^2+(min_for_slope2(t,2)-min_for_slope1(t,2))^2+…
% (min_for_slope2(t,3)-min_for_slope1(t,3))^2;%Slope. Constant for crossection

%Area
A(m) = trapz(distance(lvl_ind2), smth_crs(lvl_ind1) – smth_crs(lvl_ind2)); % assumes smth_crs is bed elevation
%A(m)=trapz(distance-smth_crs);
R(m)=A(m)/P(m);%Hydraulic radius;

%Manning
%Effect of water level on n of Manning:

if smth_crs(lvl_ind1) < 0.3
N = 0.1;
elseif and(smth_crs(lvl_ind1) >= 0.3, smth_crs(lvl_ind1) < 0.7)
N = 0.06;
else
N = 0.03;
end

discharge1=(A(m)*(1/N)*R(m)^(2/3))*S(t)^(1/2);
discharge(m,1:2)=[lvl(m,2*t-1) discharge1];

end%end ‘for m=…’
% figure(‘Name’,[char(trib_name(t)) ‘_smth_crs’])
% plot(smth_crs)

discharge_all(1:size(discharge,1),2*t-1:2*t)=discharge;
% [discharge2,ia,~] = unique(discharge1(:,2));
% discharge=[lvl(ia,2*t-1) discharge2];
% discharge_all(1:size(discharge,1),2*t-1:2*t)=discharge;

% plot(date1,discharge_all);
n=t;
discharge_all(1:length(discharge),2*n-1:2*n)=discharge;
% figure(n)
% plot(discharge_all(1:Lng_lvl(n),2*n-1),discharge_all(1:Lng_lvl(n),2*n))
% title=trib_name(n);
% ylabel(‘Q (m^3/s)’)
% datetick

figure(‘name’,char(trib_name(t)))
plot(lvl(:,2*t-1),discharge_all(:,2*t),’.’)
%title(names(t));
ylabel(‘Q (m^3/s)’)
datetick
end

%end %end for t=, m=…
discharge_all_smth=discharge_all;
filename = ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischdischarge_all_smth.mat’;
save( filename, ‘discharge_all_smth’ );
tocHi,
I am given a contineuous water level data measured and measured cross section to calculate discharge by Manning equation. All these data were checked for gaps, and found contineuous. When running the following script the discharge having gaps, creats a jump which looks strange. Tried for several weeks to find it, but didn’t. Hope for your help.
%this script reads Cross and crocss-sections data from tributaries:
%Tzipori, Yiftachel, Tzvi, Oz, Mizra, Adashim, Taanach and Keini
%It 1. reads divers data 2. change them to be water level 3. calculate
%discharge.
%Discharge calculation requires:
%1. find the water level in the current time step 2. caculate discharge using manning
tic
clear all
close all

%%%%%%%%%%%% Part A: find water level of all sites%%%%%%%%%%%%%%%%%%%
% Get a list of all txt files in the current folder, or subfolders of it.

%Read discharge
lines1 = readlines("trib_names.txt");
trib_name=(categorical(lines1));

for i=1:length(trib_name)
trib1{i}=trib_name(i)
end

trib=string(trib1)+’level’; %for the case of water level;
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischdivers’;
lvl=NaN(418907,16);%columns: longest diver data, rows/2= # of tributaries. each st. ha two columns: for date and data of stations
AllDivresfiles = natsortfiles(dir(‘*.txt’));

for k =1:size(AllDivresfiles,1)
% k=5%shut ‘end’ at the end of loop. Find thie in ‘kishon_catchment_discharge_calc_for1trib’
thisfilename1 = AllDivresfiles(k).name; %just the name
thisdata1 = load(thisfilename1); %load just this file
Lng_lvl(k)=length(thisdata1);
date1=x2mdate(thisdata1(:,1));
%plot(date1,thisdata1(:,2));ylabel(‘Level (cm)’)
num_days=length(unique(thisdata1(:,1)));
interval=720;%720 is the daily period of the time interval of data: 2 min. length(thisdata1(:,1))/(num_days*24);%Model has 1 hours intervals, so this is the interval obtained here. Obtain intervals according to 10 min intrval
baseP=min(thisdata1(:,2));%background pressure
curr_lvl1=(thisdata1(:,2)-baseP)/100;%units change to meter%for smoothing turn curr_lvl to curr_lvl1
curr_lvl=smoothdata(curr_lvl1,’movmean’,interval);
%curr_lvl=smoothdata(curr_lvl,’gaussian’,interval);
lvl(1:size(thisdata1,1),(2*k-1):2*k)=[date1 curr_lvl];%Removing diver’s bias (correct only forephemeral tributaries) and converting to m
lngt(k)=length(curr_lvl);
end

in0=find((lvl)==0);
lvl(in0)=NaN;
figure(‘Name’,’Adashim’);adash=lvl(:,1:2);plot(adash(:,1),adash(:,2));datetick;title=’Adashim lvl’;
figure(‘Name’,’Keini’);keini=lvl(:,3:4);plot(keini(:,1),keini(:,2));title=’Keini lvl’;datetick;
figure(‘Name’,’KishonMaale’);KishonMaale=lvl(:,5:6);plot(KishonMaale(:,1),KishonMaale(:,2));title=’KishonMaale lvl’;datetick
figure(‘Name’,’Mizra’);mizra=lvl(:,7:8);plot(mizra(:,1),mizra(:,2));title=’Mizra lvl’;datetick;
figure(‘Name’,’Oz’);oz=lvl(:,9:10);plot(oz(:,1),oz(:,2));title=’Oz lvl’;datetick;
figure(‘Name’,’Taanach’);taanach=lvl(:,11:12);plot(taanach(:,1),taanach(:,2));title=’Taanach lvl’;datetick;
figure(‘Name’,’Tzvi’);title=’Tzvi lvl’;tzvi=lvl(:,13:14);plot(tzvi(:,1),tzvi(:,2));datetick;
figure(‘Name’,’Yiftachel’);title=’Yiftachel lvl’;yiftachel=lvl(:,15:16);plot(yiftachel(:,1),yiftachel(:,2));title=’Yiftachel lvl’;datetick;

%%%%%%%%%%%% Part B: find cross section%%%%%%%%%%%%%%%%%%%
%Read cross sections. In most tributaries the divers are in the downstream
%All files with ‘1’ are upstrream and all with ‘2’ are downstreaצ. For example: OzUp=Oz1, OzDown=Oz2

trib=length(AllDivresfiles);
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischcrossectionsDownstream’;
crs=NaN(max(lngt),5*trib);%columns: longest cross section data, rows-# of stations*5
AllCrossfiles = natsortfiles(dir(‘*.txt’));
min_for_slope1=NaN(trib,3);

%Collect all downstream crossection data together
for l = 1:length(AllCrossfiles)
%l=1 %shut ‘end’ at the end of loop
thisfilename2 = AllCrossfiles(l).name; %just the name
thisdata2 = load(thisfilename2); %load current file
crs(1:length(thisdata2),5*l-4:l*5)=thisdata2;
[MIN,I0]=min(thisdata2(:,3));%sea level elevation to have absolute elevation
min_for_slope1(l,:)=thisdata2(I0,1:3);%The slope is obtained by the slope bwtween two min of adjacent croo sections.
Lng(l)=length(thisdata2);
% figure(‘Name’,[char(trib_name(l)) ‘_crs_dn’])
% plot(crs(:,5*l-4),crs(:,5*l))
end

% adash1_crs=crs(:,4:5);keini1_crs=crs(:,9:10);mizra1_crs=crs(:,14:15);oz1_crs=crs(:,19:20);taanach1_crs=crs(:,24:25);tzvi1_crs=crs(:,29:30);yiftachel1_crs=crs(:,34:35);
% figure(11);plot(adash1_crs(:,1),adash1_crs(:,2));figure(22); plot(keini1_crs(:,1),keini1_crs(:,2));figure(33);plot(mizra1_crs(:,1),mizra1_crs(:,2));figure(44);plot(oz1_crs(:,1),oz1_crs(:,2));figure(55);plot(taanach1_crs(:,1),taanach1_crs(:,2));figure(66);plot(tzvi1_crs(:,1),tzvi1_crs(:,2));figure(77);plot(yiftachel1_crs(:,1),yiftachel1_crs(:,2));

%%%%% Part B1: Reading upstream for slope%%%%%%%%
min_for_slope2=NaN(trib,3);%coordinates (columns 1 and 2) and elevation (column3). of minimum in the cross sectioncolumns: longest diver data, rows/2= # of tributaries. each st. ha two columns: for date and data of stations
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischcrossections’;

%min_for_slope=NaN(trib*2,2);%columns: longest cross section data, rows-# of stations
Allupstrmfiles = orderfields(dir(‘*.txt’));
for j = 1 : trib
upstrmName = Allupstrmfiles(j).name; %just the name; j or l
thisupstrm = load(upstrmName); %load just this file
[MIN1,I1]=min(thisupstrm(:,3));%Sea level to get absolute elevation
min_for_slope2(j,:)=(thisupstrm(I1,1:3));
% figure(‘Name’,[char(trib_name(j)) ‘_crs_up’])
% plot(thisupstrm(:,4),thisupstrm(:,3))
end

%%%%%%%%%%%%%%%%NEW 29052025%%%%%%%%%%%%%%%%%%%%%%%%%%%
all_lvl=NaN(max(lngt),trib);
discharge_all=NaN(418907,trib*2);%longest data
discharge=NaN(418907,2);
g=0;
f=0;

for t=1:trib
x=crs(1:Lng(t),5*t-4);
y=crs(1:Lng(t),5*t-3);
crsc=crs(1:Lng(t),5*t);
x0=x(1);y0=y(1);
distance1=((x-x0).^2+(y-y0).^2).^(1/2);%crs(1:Lng(t),5*t-1);
%Interpolation to create smoother discharge calculation
DF=abs(min(crsc(1:length(crsc)-1)-crsc(2:length(crsc))))/300;%Order of magnitude of distance difference and elveation are similar
basic_distance=distance1(1):DF:distance1(end);
distance11=1:(length(crsc));
distance=interp1(distance11,distance1,basic_distance);%to do interpolation to distance , since t is not a stright line
smth_crs=interp1(distance1,crsc,distance);
% smth_x=interp1(distance1,x,distance);
% smth_y=interp1(distance1,y,distance);

for m=1:Lng_lvl(t)%m=233648

if isnan(lvl(m,2*t))
discharge(m,1:2)=[lvl(m,2*t-1) NaN];
f=f+1;
continue
end

[M lvl_ind1]=min(abs(lvl(m,2*t)-smth_crs));%Find current water level in lvl and the current cross section

if smth_crs(lvl_ind1)==min(smth_crs)
discharge(m,1:2)=[lvl(m,2*t-1) 0];
g=g+1;
continue
end

lvl_ind2=find(smth_crs<=smth_crs(lvl_ind1));
distance2=distance(lvl_ind2);smth_crs2=smth_crs(lvl_ind2);
P1=((distance2(2:end)-distance2(1:end-1)).^2+(smth_crs2(2:end)-smth_crs2(1:end-1)).^2).^(1/2);
% lvl_ind2=find(smth_crs<=smth_crs(lvl_ind1));
% lvl_ind22=crsc<=min(crsc);
% crsc_lvl=crsc(lvl_ind22);
% x_lvl=x(lvl_ind22);%Due to wetted perimeter.
% y_lvl=y(lvl_ind22);%Due to wetted perimeter.
% % smth_x_lvl=smth_x(lvl_ind2);%Due to wetted perimeter.
% % smth_y_lvl=smth_y(lvl_ind2);%Due to wetted perimeter.
% smth_crs_lvl=smth_crs(lvl_ind2);%Due to wetted perimeter.

%Wetted perimeter:
% P1=((x_lvl(2:end)-x_lvl(1:end-1)).^2+(y_lvl(2:end)-y_lvl(1:end-1)).^2+(crsc_lvl(2:end)-crsc_lvl(1:end-1)).^2).^(1/2);
P(m)=sum(P1);

if isnan(P(m))
discharge(m,1:2)=[lvl(m,2*t-1) 0];
q=q+1;
continue
end

%Slope
dx = min_for_slope2(t,1) – min_for_slope1(t,1);
dy = min_for_slope2(t,2) – min_for_slope1(t,2);
dz(t) = min_for_slope2(t,3) – min_for_slope1(t,3);
distance3D = sqrt(dx^2 + dy^2 + dz(t)^2);
S(t) = abs(dz(t)) / distance3D;
% S(t)=min_for_slope2(t,3)-min_for_slope1(t,3)/(min_for_slope2(t,1)-min_for_slope1(t,1))^2+(min_for_slope2(t,2)-min_for_slope1(t,2))^2+…
% (min_for_slope2(t,3)-min_for_slope1(t,3))^2;%Slope. Constant for crossection

%Area
A(m) = trapz(distance(lvl_ind2), smth_crs(lvl_ind1) – smth_crs(lvl_ind2)); % assumes smth_crs is bed elevation
%A(m)=trapz(distance-smth_crs);
R(m)=A(m)/P(m);%Hydraulic radius;

%Manning
%Effect of water level on n of Manning:

if smth_crs(lvl_ind1) < 0.3
N = 0.1;
elseif and(smth_crs(lvl_ind1) >= 0.3, smth_crs(lvl_ind1) < 0.7)
N = 0.06;
else
N = 0.03;
end

discharge1=(A(m)*(1/N)*R(m)^(2/3))*S(t)^(1/2);
discharge(m,1:2)=[lvl(m,2*t-1) discharge1];

end%end ‘for m=…’
% figure(‘Name’,[char(trib_name(t)) ‘_smth_crs’])
% plot(smth_crs)

discharge_all(1:size(discharge,1),2*t-1:2*t)=discharge;
% [discharge2,ia,~] = unique(discharge1(:,2));
% discharge=[lvl(ia,2*t-1) discharge2];
% discharge_all(1:size(discharge,1),2*t-1:2*t)=discharge;

% plot(date1,discharge_all);
n=t;
discharge_all(1:length(discharge),2*n-1:2*n)=discharge;
% figure(n)
% plot(discharge_all(1:Lng_lvl(n),2*n-1),discharge_all(1:Lng_lvl(n),2*n))
% title=trib_name(n);
% ylabel(‘Q (m^3/s)’)
% datetick

figure(‘name’,char(trib_name(t)))
plot(lvl(:,2*t-1),discharge_all(:,2*t),’.’)
%title(names(t));
ylabel(‘Q (m^3/s)’)
datetick
end

%end %end for t=, m=…
discharge_all_smth=discharge_all;
filename = ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischdischarge_all_smth.mat’;
save( filename, ‘discharge_all_smth’ );
toc Hi,
I am given a contineuous water level data measured and measured cross section to calculate discharge by Manning equation. All these data were checked for gaps, and found contineuous. When running the following script the discharge having gaps, creats a jump which looks strange. Tried for several weeks to find it, but didn’t. Hope for your help.
%this script reads Cross and crocss-sections data from tributaries:
%Tzipori, Yiftachel, Tzvi, Oz, Mizra, Adashim, Taanach and Keini
%It 1. reads divers data 2. change them to be water level 3. calculate
%discharge.
%Discharge calculation requires:
%1. find the water level in the current time step 2. caculate discharge using manning
tic
clear all
close all

%%%%%%%%%%%% Part A: find water level of all sites%%%%%%%%%%%%%%%%%%%
% Get a list of all txt files in the current folder, or subfolders of it.

%Read discharge
lines1 = readlines("trib_names.txt");
trib_name=(categorical(lines1));

for i=1:length(trib_name)
trib1{i}=trib_name(i)
end

trib=string(trib1)+’level’; %for the case of water level;
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischdivers’;
lvl=NaN(418907,16);%columns: longest diver data, rows/2= # of tributaries. each st. ha two columns: for date and data of stations
AllDivresfiles = natsortfiles(dir(‘*.txt’));

for k =1:size(AllDivresfiles,1)
% k=5%shut ‘end’ at the end of loop. Find thie in ‘kishon_catchment_discharge_calc_for1trib’
thisfilename1 = AllDivresfiles(k).name; %just the name
thisdata1 = load(thisfilename1); %load just this file
Lng_lvl(k)=length(thisdata1);
date1=x2mdate(thisdata1(:,1));
%plot(date1,thisdata1(:,2));ylabel(‘Level (cm)’)
num_days=length(unique(thisdata1(:,1)));
interval=720;%720 is the daily period of the time interval of data: 2 min. length(thisdata1(:,1))/(num_days*24);%Model has 1 hours intervals, so this is the interval obtained here. Obtain intervals according to 10 min intrval
baseP=min(thisdata1(:,2));%background pressure
curr_lvl1=(thisdata1(:,2)-baseP)/100;%units change to meter%for smoothing turn curr_lvl to curr_lvl1
curr_lvl=smoothdata(curr_lvl1,’movmean’,interval);
%curr_lvl=smoothdata(curr_lvl,’gaussian’,interval);
lvl(1:size(thisdata1,1),(2*k-1):2*k)=[date1 curr_lvl];%Removing diver’s bias (correct only forephemeral tributaries) and converting to m
lngt(k)=length(curr_lvl);
end

in0=find((lvl)==0);
lvl(in0)=NaN;
figure(‘Name’,’Adashim’);adash=lvl(:,1:2);plot(adash(:,1),adash(:,2));datetick;title=’Adashim lvl’;
figure(‘Name’,’Keini’);keini=lvl(:,3:4);plot(keini(:,1),keini(:,2));title=’Keini lvl’;datetick;
figure(‘Name’,’KishonMaale’);KishonMaale=lvl(:,5:6);plot(KishonMaale(:,1),KishonMaale(:,2));title=’KishonMaale lvl’;datetick
figure(‘Name’,’Mizra’);mizra=lvl(:,7:8);plot(mizra(:,1),mizra(:,2));title=’Mizra lvl’;datetick;
figure(‘Name’,’Oz’);oz=lvl(:,9:10);plot(oz(:,1),oz(:,2));title=’Oz lvl’;datetick;
figure(‘Name’,’Taanach’);taanach=lvl(:,11:12);plot(taanach(:,1),taanach(:,2));title=’Taanach lvl’;datetick;
figure(‘Name’,’Tzvi’);title=’Tzvi lvl’;tzvi=lvl(:,13:14);plot(tzvi(:,1),tzvi(:,2));datetick;
figure(‘Name’,’Yiftachel’);title=’Yiftachel lvl’;yiftachel=lvl(:,15:16);plot(yiftachel(:,1),yiftachel(:,2));title=’Yiftachel lvl’;datetick;

%%%%%%%%%%%% Part B: find cross section%%%%%%%%%%%%%%%%%%%
%Read cross sections. In most tributaries the divers are in the downstream
%All files with ‘1’ are upstrream and all with ‘2’ are downstreaצ. For example: OzUp=Oz1, OzDown=Oz2

trib=length(AllDivresfiles);
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischcrossectionsDownstream’;
crs=NaN(max(lngt),5*trib);%columns: longest cross section data, rows-# of stations*5
AllCrossfiles = natsortfiles(dir(‘*.txt’));
min_for_slope1=NaN(trib,3);

%Collect all downstream crossection data together
for l = 1:length(AllCrossfiles)
%l=1 %shut ‘end’ at the end of loop
thisfilename2 = AllCrossfiles(l).name; %just the name
thisdata2 = load(thisfilename2); %load current file
crs(1:length(thisdata2),5*l-4:l*5)=thisdata2;
[MIN,I0]=min(thisdata2(:,3));%sea level elevation to have absolute elevation
min_for_slope1(l,:)=thisdata2(I0,1:3);%The slope is obtained by the slope bwtween two min of adjacent croo sections.
Lng(l)=length(thisdata2);
% figure(‘Name’,[char(trib_name(l)) ‘_crs_dn’])
% plot(crs(:,5*l-4),crs(:,5*l))
end

% adash1_crs=crs(:,4:5);keini1_crs=crs(:,9:10);mizra1_crs=crs(:,14:15);oz1_crs=crs(:,19:20);taanach1_crs=crs(:,24:25);tzvi1_crs=crs(:,29:30);yiftachel1_crs=crs(:,34:35);
% figure(11);plot(adash1_crs(:,1),adash1_crs(:,2));figure(22); plot(keini1_crs(:,1),keini1_crs(:,2));figure(33);plot(mizra1_crs(:,1),mizra1_crs(:,2));figure(44);plot(oz1_crs(:,1),oz1_crs(:,2));figure(55);plot(taanach1_crs(:,1),taanach1_crs(:,2));figure(66);plot(tzvi1_crs(:,1),tzvi1_crs(:,2));figure(77);plot(yiftachel1_crs(:,1),yiftachel1_crs(:,2));

%%%%% Part B1: Reading upstream for slope%%%%%%%%
min_for_slope2=NaN(trib,3);%coordinates (columns 1 and 2) and elevation (column3). of minimum in the cross sectioncolumns: longest diver data, rows/2= # of tributaries. each st. ha two columns: for date and data of stations
cd ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischcrossections’;

%min_for_slope=NaN(trib*2,2);%columns: longest cross section data, rows-# of stations
Allupstrmfiles = orderfields(dir(‘*.txt’));
for j = 1 : trib
upstrmName = Allupstrmfiles(j).name; %just the name; j or l
thisupstrm = load(upstrmName); %load just this file
[MIN1,I1]=min(thisupstrm(:,3));%Sea level to get absolute elevation
min_for_slope2(j,:)=(thisupstrm(I1,1:3));
% figure(‘Name’,[char(trib_name(j)) ‘_crs_up’])
% plot(thisupstrm(:,4),thisupstrm(:,3))
end

%%%%%%%%%%%%%%%%NEW 29052025%%%%%%%%%%%%%%%%%%%%%%%%%%%
all_lvl=NaN(max(lngt),trib);
discharge_all=NaN(418907,trib*2);%longest data
discharge=NaN(418907,2);
g=0;
f=0;

for t=1:trib
x=crs(1:Lng(t),5*t-4);
y=crs(1:Lng(t),5*t-3);
crsc=crs(1:Lng(t),5*t);
x0=x(1);y0=y(1);
distance1=((x-x0).^2+(y-y0).^2).^(1/2);%crs(1:Lng(t),5*t-1);
%Interpolation to create smoother discharge calculation
DF=abs(min(crsc(1:length(crsc)-1)-crsc(2:length(crsc))))/300;%Order of magnitude of distance difference and elveation are similar
basic_distance=distance1(1):DF:distance1(end);
distance11=1:(length(crsc));
distance=interp1(distance11,distance1,basic_distance);%to do interpolation to distance , since t is not a stright line
smth_crs=interp1(distance1,crsc,distance);
% smth_x=interp1(distance1,x,distance);
% smth_y=interp1(distance1,y,distance);

for m=1:Lng_lvl(t)%m=233648

if isnan(lvl(m,2*t))
discharge(m,1:2)=[lvl(m,2*t-1) NaN];
f=f+1;
continue
end

[M lvl_ind1]=min(abs(lvl(m,2*t)-smth_crs));%Find current water level in lvl and the current cross section

if smth_crs(lvl_ind1)==min(smth_crs)
discharge(m,1:2)=[lvl(m,2*t-1) 0];
g=g+1;
continue
end

lvl_ind2=find(smth_crs<=smth_crs(lvl_ind1));
distance2=distance(lvl_ind2);smth_crs2=smth_crs(lvl_ind2);
P1=((distance2(2:end)-distance2(1:end-1)).^2+(smth_crs2(2:end)-smth_crs2(1:end-1)).^2).^(1/2);
% lvl_ind2=find(smth_crs<=smth_crs(lvl_ind1));
% lvl_ind22=crsc<=min(crsc);
% crsc_lvl=crsc(lvl_ind22);
% x_lvl=x(lvl_ind22);%Due to wetted perimeter.
% y_lvl=y(lvl_ind22);%Due to wetted perimeter.
% % smth_x_lvl=smth_x(lvl_ind2);%Due to wetted perimeter.
% % smth_y_lvl=smth_y(lvl_ind2);%Due to wetted perimeter.
% smth_crs_lvl=smth_crs(lvl_ind2);%Due to wetted perimeter.

%Wetted perimeter:
% P1=((x_lvl(2:end)-x_lvl(1:end-1)).^2+(y_lvl(2:end)-y_lvl(1:end-1)).^2+(crsc_lvl(2:end)-crsc_lvl(1:end-1)).^2).^(1/2);
P(m)=sum(P1);

if isnan(P(m))
discharge(m,1:2)=[lvl(m,2*t-1) 0];
q=q+1;
continue
end

%Slope
dx = min_for_slope2(t,1) – min_for_slope1(t,1);
dy = min_for_slope2(t,2) – min_for_slope1(t,2);
dz(t) = min_for_slope2(t,3) – min_for_slope1(t,3);
distance3D = sqrt(dx^2 + dy^2 + dz(t)^2);
S(t) = abs(dz(t)) / distance3D;
% S(t)=min_for_slope2(t,3)-min_for_slope1(t,3)/(min_for_slope2(t,1)-min_for_slope1(t,1))^2+(min_for_slope2(t,2)-min_for_slope1(t,2))^2+…
% (min_for_slope2(t,3)-min_for_slope1(t,3))^2;%Slope. Constant for crossection

%Area
A(m) = trapz(distance(lvl_ind2), smth_crs(lvl_ind1) – smth_crs(lvl_ind2)); % assumes smth_crs is bed elevation
%A(m)=trapz(distance-smth_crs);
R(m)=A(m)/P(m);%Hydraulic radius;

%Manning
%Effect of water level on n of Manning:

if smth_crs(lvl_ind1) < 0.3
N = 0.1;
elseif and(smth_crs(lvl_ind1) >= 0.3, smth_crs(lvl_ind1) < 0.7)
N = 0.06;
else
N = 0.03;
end

discharge1=(A(m)*(1/N)*R(m)^(2/3))*S(t)^(1/2);
discharge(m,1:2)=[lvl(m,2*t-1) discharge1];

end%end ‘for m=…’
% figure(‘Name’,[char(trib_name(t)) ‘_smth_crs’])
% plot(smth_crs)

discharge_all(1:size(discharge,1),2*t-1:2*t)=discharge;
% [discharge2,ia,~] = unique(discharge1(:,2));
% discharge=[lvl(ia,2*t-1) discharge2];
% discharge_all(1:size(discharge,1),2*t-1:2*t)=discharge;

% plot(date1,discharge_all);
n=t;
discharge_all(1:length(discharge),2*n-1:2*n)=discharge;
% figure(n)
% plot(discharge_all(1:Lng_lvl(n),2*n-1),discharge_all(1:Lng_lvl(n),2*n))
% title=trib_name(n);
% ylabel(‘Q (m^3/s)’)
% datetick

figure(‘name’,char(trib_name(t)))
plot(lvl(:,2*t-1),discharge_all(:,2*t),’.’)
%title(names(t));
ylabel(‘Q (m^3/s)’)
datetick
end

%end %end for t=, m=…
discharge_all_smth=discharge_all;
filename = ‘C:UsersuserDocumentsShulamitPhDMatlab_PhDHydrologyDischdischarge_all_smth.mat’;
save( filename, ‘discharge_all_smth’ );
toc hydrology manning water level gap in discharge MATLAB Answers — New Questions

​

I’m making a set of tables from a 4x4x4 array, like so:
T{jj} = table(data(jj,:,1)’,data(jj,:,2)’,data(jj,:,3)’,data(jj,:,4)’,’VariableNames’,colName,’RowNames’,rowName);
The annoying thing is, I have to change this line every time there’s a different number of columns. For example, for 3 columns, the command is:
T{jj} = table(data(jj,:,1)’,data(jj,:,2)’,data(jj,:,3)’,’VariableNames’,colName,’RowNames’,rowName);
I know I could do a switch on the number of columns and have a case for every possible number of columns I might ever have, but is some way around that? The obvious solution of giving it the data as an array doesn’t work if you want named rows and columns:

T{jj} = table(data(jj,:,:),’VariableNames’,colName,’RowNames’,rowName);
Error using table (line 369)
The VariableNames property must contain one name for each variable in the table.

I get the exact same error back if I use slice():
T{jj} = table(slice(data(jj,:,:)),’VariableNames’,colName,’RowNames’,rowName);

But it works fine without row / column names
T{jj} = table(data(jj,:,:));
Am I missing something simple, or is this maybe a design oversight?I’m making a set of tables from a 4x4x4 array, like so:
T{jj} = table(data(jj,:,1)’,data(jj,:,2)’,data(jj,:,3)’,data(jj,:,4)’,’VariableNames’,colName,’RowNames’,rowName);
The annoying thing is, I have to change this line every time there’s a different number of columns. For example, for 3 columns, the command is:
T{jj} = table(data(jj,:,1)’,data(jj,:,2)’,data(jj,:,3)’,’VariableNames’,colName,’RowNames’,rowName);
I know I could do a switch on the number of columns and have a case for every possible number of columns I might ever have, but is some way around that? The obvious solution of giving it the data as an array doesn’t work if you want named rows and columns:

T{jj} = table(data(jj,:,:),’VariableNames’,colName,’RowNames’,rowName);
Error using table (line 369)
The VariableNames property must contain one name for each variable in the table.

I get the exact same error back if I use slice():
T{jj} = table(slice(data(jj,:,:)),’VariableNames’,colName,’RowNames’,rowName);

But it works fine without row / column names
T{jj} = table(data(jj,:,:));
Am I missing something simple, or is this maybe a design oversight? I’m making a set of tables from a 4x4x4 array, like so:
T{jj} = table(data(jj,:,1)’,data(jj,:,2)’,data(jj,:,3)’,data(jj,:,4)’,’VariableNames’,colName,’RowNames’,rowName);
The annoying thing is, I have to change this line every time there’s a different number of columns. For example, for 3 columns, the command is:
T{jj} = table(data(jj,:,1)’,data(jj,:,2)’,data(jj,:,3)’,’VariableNames’,colName,’RowNames’,rowName);
I know I could do a switch on the number of columns and have a case for every possible number of columns I might ever have, but is some way around that? The obvious solution of giving it the data as an array doesn’t work if you want named rows and columns:

T{jj} = table(data(jj,:,:),’VariableNames’,colName,’RowNames’,rowName);
Error using table (line 369)
The VariableNames property must contain one name for each variable in the table.

I get the exact same error back if I use slice():
T{jj} = table(slice(data(jj,:,:)),’VariableNames’,colName,’RowNames’,rowName);

But it works fine without row / column names
T{jj} = table(data(jj,:,:));
Am I missing something simple, or is this maybe a design oversight? table MATLAB Answers — New Questions

​

Hello,
I am performing a Processor-in-the-Loop (PIL) simulation on a control subsystem of a fuel-cell model using Simulink/Simscape and an STM32F429 target with Embedded Coder.
I used the option "Create SIL/PIL block" for the subsystem, followed by C/C++ Code > Build This Subsystem, and then added the generated block to my model for testing(fuel-cell model).
When I run the simulation, I get the following warning:
"Top model does not contain a referenced model. ‘Simulation’ and ‘SIL/PIL Simulation’ will behave in the same way. Change ‘system under test’ or add a model block to the top model."
At the same time, the simulation output curves for normal simulation and PIL are identical (confondues), which is expected, but this warning makes me unsure.
My questions:
Does this warning indicate that the PIL simulation is not actually running on the target hardware?
Since I’m testing only a control subsystem using a PIL block, and not the entire model, is it safe to ignore this warning?
Do I need to replace the subsystem with a referenced model (Model block) to properly use PIL at the top level?
Thank you in advance for your help!Hello,
I am performing a Processor-in-the-Loop (PIL) simulation on a control subsystem of a fuel-cell model using Simulink/Simscape and an STM32F429 target with Embedded Coder.
I used the option "Create SIL/PIL block" for the subsystem, followed by C/C++ Code > Build This Subsystem, and then added the generated block to my model for testing(fuel-cell model).
When I run the simulation, I get the following warning:
"Top model does not contain a referenced model. ‘Simulation’ and ‘SIL/PIL Simulation’ will behave in the same way. Change ‘system under test’ or add a model block to the top model."
At the same time, the simulation output curves for normal simulation and PIL are identical (confondues), which is expected, but this warning makes me unsure.
My questions:
Does this warning indicate that the PIL simulation is not actually running on the target hardware?
Since I’m testing only a control subsystem using a PIL block, and not the entire model, is it safe to ignore this warning?
Do I need to replace the subsystem with a referenced model (Model block) to properly use PIL at the top level?
Thank you in advance for your help! Hello,
I am performing a Processor-in-the-Loop (PIL) simulation on a control subsystem of a fuel-cell model using Simulink/Simscape and an STM32F429 target with Embedded Coder.
I used the option "Create SIL/PIL block" for the subsystem, followed by C/C++ Code > Build This Subsystem, and then added the generated block to my model for testing(fuel-cell model).
When I run the simulation, I get the following warning:
"Top model does not contain a referenced model. ‘Simulation’ and ‘SIL/PIL Simulation’ will behave in the same way. Change ‘system under test’ or add a model block to the top model."
At the same time, the simulation output curves for normal simulation and PIL are identical (confondues), which is expected, but this warning makes me unsure.
My questions:
Does this warning indicate that the PIL simulation is not actually running on the target hardware?
Since I’m testing only a control subsystem using a PIL block, and not the entire model, is it safe to ignore this warning?
Do I need to replace the subsystem with a referenced model (Model block) to properly use PIL at the top level?
Thank you in advance for your help! pil MATLAB Answers — New Questions

​