Hello all,
I have done a model for controlling roll, pitch, yaw and Z of a quadcopter, I did it using the 6DoF Euler Angles Simulink´s Block, and I have solved the controll both, with using fuzzy controllers and also with a PID.
Now I am trying to use neural networks for control it, and I have tryed the control blocks (Model-Reference and predictive controller), but I don´t know how to use them for this scenario on a propper way, or how I should redo the plant. Because until now, with the other controllers I pick each of the outputs and I do deacoupled control for each one and join them again.
I will be very thanked for any help, thank you.
Regards.Hello all,
I have done a model for controlling roll, pitch, yaw and Z of a quadcopter, I did it using the 6DoF Euler Angles Simulink´s Block, and I have solved the controll both, with using fuzzy controllers and also with a PID.
Now I am trying to use neural networks for control it, and I have tryed the control blocks (Model-Reference and predictive controller), but I don´t know how to use them for this scenario on a propper way, or how I should redo the plant. Because until now, with the other controllers I pick each of the outputs and I do deacoupled control for each one and join them again.
I will be very thanked for any help, thank you.
Regards. Hello all,
I have done a model for controlling roll, pitch, yaw and Z of a quadcopter, I did it using the 6DoF Euler Angles Simulink´s Block, and I have solved the controll both, with using fuzzy controllers and also with a PID.
Now I am trying to use neural networks for control it, and I have tryed the control blocks (Model-Reference and predictive controller), but I don´t know how to use them for this scenario on a propper way, or how I should redo the plant. Because until now, with the other controllers I pick each of the outputs and I do deacoupled control for each one and join them again.
I will be very thanked for any help, thank you.
Regards. neural networks control, 6dof euler angles controller MATLAB Answers — New Questions

​

Look this example:
selected = findobj(allLines,’Type’,’line’,’LineWidth’,0.5);
[cc]=selected;
selected 13×1
xx selected 13×1
this is ok
now i try to use selected property:
selected.Color
it give me 13×1 color!
now i try to record information in array but it’s not possible

[cc]=selected.Color
now i see cc have only the first element!
can i do it?Look this example:
selected = findobj(allLines,’Type’,’line’,’LineWidth’,0.5);
[cc]=selected;
selected 13×1
xx selected 13×1
this is ok
now i try to use selected property:
selected.Color
it give me 13×1 color!
now i try to record information in array but it’s not possible

[cc]=selected.Color
now i see cc have only the first element!
can i do it? Look this example:
selected = findobj(allLines,’Type’,’line’,’LineWidth’,0.5);
[cc]=selected;
selected 13×1
xx selected 13×1
this is ok
now i try to use selected property:
selected.Color
it give me 13×1 color!
now i try to record information in array but it’s not possible

[cc]=selected.Color
now i see cc have only the first element!
can i do it? how to build array using matlab.graphics.chart. MATLAB Answers — New Questions

​

According to the doc page assumptions
To clear all objects in the MATLAB workspace and close the Symbolic Math Toolbox™ engine associated with the MATLAB workspace resetting all its assumptions, use this command
clear all
Comment: I think the doc page would be clearer if that clause replaced "resetting" with "clearing".
But
syms x
assume(x < 1)
assumptions
clear all
assumptions
I’m quite certain that in previous versions (like 2022a?) the command "clear all" did clear the assumptions from the engine.
Is this a relatively-newly-introduced bug or am I misremembering/misunderstanding the situation?
Answers is running 2024b, I see the same behavior on 2024a.According to the doc page assumptions
To clear all objects in the MATLAB workspace and close the Symbolic Math Toolbox™ engine associated with the MATLAB workspace resetting all its assumptions, use this command
clear all
Comment: I think the doc page would be clearer if that clause replaced "resetting" with "clearing".
But
syms x
assume(x < 1)
assumptions
clear all
assumptions
I’m quite certain that in previous versions (like 2022a?) the command "clear all" did clear the assumptions from the engine.
Is this a relatively-newly-introduced bug or am I misremembering/misunderstanding the situation?
Answers is running 2024b, I see the same behavior on 2024a. According to the doc page assumptions
To clear all objects in the MATLAB workspace and close the Symbolic Math Toolbox™ engine associated with the MATLAB workspace resetting all its assumptions, use this command
clear all
Comment: I think the doc page would be clearer if that clause replaced "resetting" with "clearing".
But
syms x
assume(x < 1)
assumptions
clear all
assumptions
I’m quite certain that in previous versions (like 2022a?) the command "clear all" did clear the assumptions from the engine.
Is this a relatively-newly-introduced bug or am I misremembering/misunderstanding the situation?
Answers is running 2024b, I see the same behavior on 2024a. asusmptions, clear MATLAB Answers — New Questions

​

I have an application that plots time against pressure values. Time is of the format datetime. I plot using these lines of code.
plot(app.UIAxes_9,pressureTime,Pressure);
app.UIAxes_9.XGrid ="on";
app.UIAxes_9.XMinorGrid ="on";
app.UIAxes_9.YGrid ="on";
app.UIAxes_9.YMinorGrid ="on";
app.UIAxes_9.LineWidth = 2;
app.UIAxes_9.XLabel.String = ‘Time&Date’;
app.UIAxes_9.XLabel.FontSize = 12;
app.UIAxes_9.XLabel.FontWeight ="bold";
app.UIAxes_9.YLabel.String = ‘Pressure[bar]’;
app.UIAxes_9.YLabel.FontSize = 12;
app.UIAxes_9.YLabel.FontWeight ="bold";
app.UIAxes_9.Title.String = ‘Pressure Cycle’;
app.UIAxes_9.Title.FontSize = 14;
app.UIAxes_9.Title.FontWeight ="bold";
[maxValue, maxIndex] = max(Pressure);
maxY = Pressure(maxIndex);
maxX = pressureTime(maxIndex);
text1 = ‘Max Pressure[bar]: ‘;
text2 = num2str(maxValue);
finalText = strcat(text1,text2);
text(‘Parent’,app.UIAxes_9,’String’,finalText,’HorizontalAlignment’, ‘left’,’VerticalAlignment’,’bottom’,’Position’,[maxX maxY]);
The problem is the last line, where I tried to put a text on the UIAxes. That is
text(‘Parent’,app.UIAxes_9,’String’,finalText,’HorizontalAlignment’, ‘left’,’VerticalAlignment’,’bottom’,’Position’,[maxX maxY]);
where I tried to set the position of the text at maxX and maxY. But [maxX maxY] is giving me an error :
Error using datetime/horzcat (line 1468)
All inputs must be datetimes or date/time character vectors or date/time strings.
So what I understood is you cant combine a double and datetime to form an array. So how can I do this for a app.UIaxes. For a normal plot it is easy, but here I need for app.UIaxes. How to do this. Thank you.I have an application that plots time against pressure values. Time is of the format datetime. I plot using these lines of code.
plot(app.UIAxes_9,pressureTime,Pressure);
app.UIAxes_9.XGrid ="on";
app.UIAxes_9.XMinorGrid ="on";
app.UIAxes_9.YGrid ="on";
app.UIAxes_9.YMinorGrid ="on";
app.UIAxes_9.LineWidth = 2;
app.UIAxes_9.XLabel.String = ‘Time&Date’;
app.UIAxes_9.XLabel.FontSize = 12;
app.UIAxes_9.XLabel.FontWeight ="bold";
app.UIAxes_9.YLabel.String = ‘Pressure[bar]’;
app.UIAxes_9.YLabel.FontSize = 12;
app.UIAxes_9.YLabel.FontWeight ="bold";
app.UIAxes_9.Title.String = ‘Pressure Cycle’;
app.UIAxes_9.Title.FontSize = 14;
app.UIAxes_9.Title.FontWeight ="bold";
[maxValue, maxIndex] = max(Pressure);
maxY = Pressure(maxIndex);
maxX = pressureTime(maxIndex);
text1 = ‘Max Pressure[bar]: ‘;
text2 = num2str(maxValue);
finalText = strcat(text1,text2);
text(‘Parent’,app.UIAxes_9,’String’,finalText,’HorizontalAlignment’, ‘left’,’VerticalAlignment’,’bottom’,’Position’,[maxX maxY]);
The problem is the last line, where I tried to put a text on the UIAxes. That is
text(‘Parent’,app.UIAxes_9,’String’,finalText,’HorizontalAlignment’, ‘left’,’VerticalAlignment’,’bottom’,’Position’,[maxX maxY]);
where I tried to set the position of the text at maxX and maxY. But [maxX maxY] is giving me an error :
Error using datetime/horzcat (line 1468)
All inputs must be datetimes or date/time character vectors or date/time strings.
So what I understood is you cant combine a double and datetime to form an array. So how can I do this for a app.UIaxes. For a normal plot it is easy, but here I need for app.UIaxes. How to do this. Thank you. I have an application that plots time against pressure values. Time is of the format datetime. I plot using these lines of code.
plot(app.UIAxes_9,pressureTime,Pressure);
app.UIAxes_9.XGrid ="on";
app.UIAxes_9.XMinorGrid ="on";
app.UIAxes_9.YGrid ="on";
app.UIAxes_9.YMinorGrid ="on";
app.UIAxes_9.LineWidth = 2;
app.UIAxes_9.XLabel.String = ‘Time&Date’;
app.UIAxes_9.XLabel.FontSize = 12;
app.UIAxes_9.XLabel.FontWeight ="bold";
app.UIAxes_9.YLabel.String = ‘Pressure[bar]’;
app.UIAxes_9.YLabel.FontSize = 12;
app.UIAxes_9.YLabel.FontWeight ="bold";
app.UIAxes_9.Title.String = ‘Pressure Cycle’;
app.UIAxes_9.Title.FontSize = 14;
app.UIAxes_9.Title.FontWeight ="bold";
[maxValue, maxIndex] = max(Pressure);
maxY = Pressure(maxIndex);
maxX = pressureTime(maxIndex);
text1 = ‘Max Pressure[bar]: ‘;
text2 = num2str(maxValue);
finalText = strcat(text1,text2);
text(‘Parent’,app.UIAxes_9,’String’,finalText,’HorizontalAlignment’, ‘left’,’VerticalAlignment’,’bottom’,’Position’,[maxX maxY]);
The problem is the last line, where I tried to put a text on the UIAxes. That is
text(‘Parent’,app.UIAxes_9,’String’,finalText,’HorizontalAlignment’, ‘left’,’VerticalAlignment’,’bottom’,’Position’,[maxX maxY]);
where I tried to set the position of the text at maxX and maxY. But [maxX maxY] is giving me an error :
Error using datetime/horzcat (line 1468)
All inputs must be datetimes or date/time character vectors or date/time strings.
So what I understood is you cant combine a double and datetime to form an array. So how can I do this for a app.UIaxes. For a normal plot it is easy, but here I need for app.UIaxes. How to do this. Thank you. matlab, appdesigner MATLAB Answers — New Questions

​

When I send my first request to the MATLAB Production Server in MATLAB R2019a, it takes ~30 seconds to process it. However, the subsequent requests take less than a second to process. Why is this behavior encountered?When I send my first request to the MATLAB Production Server in MATLAB R2019a, it takes ~30 seconds to process it. However, the subsequent requests take less than a second to process. Why is this behavior encountered? When I send my first request to the MATLAB Production Server in MATLAB R2019a, it takes ~30 seconds to process it. However, the subsequent requests take less than a second to process. Why is this behavior encountered? mps, process, workers, startup MATLAB Answers — New Questions

​

Dear all,

I am using the SoC Blockset add-on to create a design for ZCU111 evaluation board.

1) The RF Data converter block accepts only int16/uint16 at adcTxChy inport. How does it determine the amplitude of the received signal when deployed on hardware? Or, to put question differently, which amplitudes of the received signal are converted to which uint16 instances? Am I correct, that in the simulation it does not matter what the amplitude of the signal is – it receives and passes raw bits?
2) The output adcTxChyData of RFDC can be of types uint16 | uint32 | uint64 | fixed point. I am curious about in which cases it is a fixed point datatype? Does the amplitude of the received signal equal real world value of this fixed point data type when deployed on hardware?

Thank you!Dear all,

I am using the SoC Blockset add-on to create a design for ZCU111 evaluation board.

1) The RF Data converter block accepts only int16/uint16 at adcTxChy inport. How does it determine the amplitude of the received signal when deployed on hardware? Or, to put question differently, which amplitudes of the received signal are converted to which uint16 instances? Am I correct, that in the simulation it does not matter what the amplitude of the signal is – it receives and passes raw bits?
2) The output adcTxChyData of RFDC can be of types uint16 | uint32 | uint64 | fixed point. I am curious about in which cases it is a fixed point datatype? Does the amplitude of the received signal equal real world value of this fixed point data type when deployed on hardware?

Thank you! Dear all,

I am using the SoC Blockset add-on to create a design for ZCU111 evaluation board.

1) The RF Data converter block accepts only int16/uint16 at adcTxChy inport. How does it determine the amplitude of the received signal when deployed on hardware? Or, to put question differently, which amplitudes of the received signal are converted to which uint16 instances? Am I correct, that in the simulation it does not matter what the amplitude of the signal is – it receives and passes raw bits?
2) The output adcTxChyData of RFDC can be of types uint16 | uint32 | uint64 | fixed point. I am curious about in which cases it is a fixed point datatype? Does the amplitude of the received signal equal real world value of this fixed point data type when deployed on hardware?

Thank you! fixed-point, soc blockset, simulink, datatypes, rf data converter, rfdc, adc MATLAB Answers — New Questions

​

close all
clear all
clc
tspan = [0 100];
y0 = 0.7; % initial value of state variable x1
r0 = 0.7; % initial value of state variable x2
w0 = 0.8; % initial value of state variable x3
x0 = [y0; r0; w0];
[t, x] = ode45(@odefcn, tspan, x0);

%% Plot x2 (y-axis) vs. x1 (x-axis)
subplot(3,1,1);
plot(t,x(:,1));
xlabel(‘Time’)
ylabel(‘Output’);
subplot(3,1,2);
plot(t,x(:,2));
xlabel(‘Time’)
ylabel(‘Policy Rate’);
subplot(3,1,3);
plot(t,x(:,3));
xlabel(‘Time’)
ylabel(‘Wage Share’);
y0 = [0.7; 0.7; 0.8];

%% System of three differential equations
function dx = odefcn(t, x)
% definitions
y = x(1);
r = x(2);
w = x(3);
% parameters
alpha = 1.0;
beta = 1.0;
gamma = 0.1;
delta = 0.3;
mu = 1.0;
lambda = 0.3;
theta = 0.1;
omega = 0.2;
% ODEs
dx(1,1) = alpha * y – beta * r * y;
dx(2,1) = – omega * r + gamma * w * r + delta * y * r;
dx(3,1) = – theta * w + lambda * y * w – mu * w * w;
endclose all
clear all
clc
tspan = [0 100];
y0 = 0.7; % initial value of state variable x1
r0 = 0.7; % initial value of state variable x2
w0 = 0.8; % initial value of state variable x3
x0 = [y0; r0; w0];
[t, x] = ode45(@odefcn, tspan, x0);

%% Plot x2 (y-axis) vs. x1 (x-axis)
subplot(3,1,1);
plot(t,x(:,1));
xlabel(‘Time’)
ylabel(‘Output’);
subplot(3,1,2);
plot(t,x(:,2));
xlabel(‘Time’)
ylabel(‘Policy Rate’);
subplot(3,1,3);
plot(t,x(:,3));
xlabel(‘Time’)
ylabel(‘Wage Share’);
y0 = [0.7; 0.7; 0.8];

%% System of three differential equations
function dx = odefcn(t, x)
% definitions
y = x(1);
r = x(2);
w = x(3);
% parameters
alpha = 1.0;
beta = 1.0;
gamma = 0.1;
delta = 0.3;
mu = 1.0;
lambda = 0.3;
theta = 0.1;
omega = 0.2;
% ODEs
dx(1,1) = alpha * y – beta * r * y;
dx(2,1) = – omega * r + gamma * w * r + delta * y * r;
dx(3,1) = – theta * w + lambda * y * w – mu * w * w;
end close all
clear all
clc
tspan = [0 100];
y0 = 0.7; % initial value of state variable x1
r0 = 0.7; % initial value of state variable x2
w0 = 0.8; % initial value of state variable x3
x0 = [y0; r0; w0];
[t, x] = ode45(@odefcn, tspan, x0);

%% Plot x2 (y-axis) vs. x1 (x-axis)
subplot(3,1,1);
plot(t,x(:,1));
xlabel(‘Time’)
ylabel(‘Output’);
subplot(3,1,2);
plot(t,x(:,2));
xlabel(‘Time’)
ylabel(‘Policy Rate’);
subplot(3,1,3);
plot(t,x(:,3));
xlabel(‘Time’)
ylabel(‘Wage Share’);
y0 = [0.7; 0.7; 0.8];

%% System of three differential equations
function dx = odefcn(t, x)
% definitions
y = x(1);
r = x(2);
w = x(3);
% parameters
alpha = 1.0;
beta = 1.0;
gamma = 0.1;
delta = 0.3;
mu = 1.0;
lambda = 0.3;
theta = 0.1;
omega = 0.2;
% ODEs
dx(1,1) = alpha * y – beta * r * y;
dx(2,1) = – omega * r + gamma * w * r + delta * y * r;
dx(3,1) = – theta * w + lambda * y * w – mu * w * w;
end graphics, subplot, plotting, plot, differential equations MATLAB Answers — New Questions

​

I want entity_batch_creator block to take no. of entities from its previous block without explicitly defining it in the dialog box.

That is if for a specific time period no of entities generated are 20 then the entity batch creator takes the batch of 20 entities automatically.I want entity_batch_creator block to take no. of entities from its previous block without explicitly defining it in the dialog box.

That is if for a specific time period no of entities generated are 20 then the entity batch creator takes the batch of 20 entities automatically. I want entity_batch_creator block to take no. of entities from its previous block without explicitly defining it in the dialog box.

That is if for a specific time period no of entities generated are 20 then the entity batch creator takes the batch of 20 entities automatically. entity_batch_creator, aggregation MATLAB Answers — New Questions

​

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

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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

​