Need a tangent line in my plot like the example provided but I cannot seem to do it

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ryan on 30 May 2025

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https://in.mathworks.com/matlabcentral/answers/2177431-need-a-tangent-line-in-my-plot-like-the-example-provided-but-i-cannot-seem-to-do-it

Edited: Star Strider on 1 Jun 2025

I am trying to make a tangent or line of best fit t this data but only from the second half of the plot on should be focused on. Below is an example from the professor

Need smething like this...

which translates to something like this...

But for some reason I keep getting this...

By using this code...

%% Calculation Problem 2
%Given Data Problems 1-4
dd = load('NEWPROB6.txt') %Data loading for airfoil
Error using load
Unable to find file or directory 'NEWPROB6.txt'.
alpha = dd(:,1); %Angle of attacks (Degreed)
CL = dd(:,2); %Lift coefficient
CD = dd(:,3); %Drag coefficient
Rh = 0.75; %Turbine hub radius (m)
Rt = 15; %Turbine blade tip radius (m)
Rht = linspace(Rh,Rt,10); %r/R radius hub to tip (m)
Rr = Rht/Rt;
B = 3; %Number of blades on the turbine
TSR = 8; %Tips-speed ratio
%Calculation Problem 1
CLD = CL./CD %Lift-to-Drag Ratio
TCL = 1.0280 %Lift coefficient associated with maximum lift-to-drag ratio
Talpha = 6 %Target angle of attack (degrees)
%Calculation Problem 2
Phi = atand(2./(3*TSR*(Rr))) %Local inflow angle (degrees)
CrR = Rt*(8*pi*sind(Phi))./(3*B*TCL*TSR); %chord distribution
Beta = Phi-Talpha
subplot(2,1,1)
plot(Rht,CrR, 'b.-', 'LineWidth',2, 'MarkerSize', 20)
halfpoints = round(numel(CrR)/2)
xlabel('r/R')
ylabel('Chord Distribution (m)')
grid on
hold on
% Assume x and y are your data vectors
xline(Rht(halfpoints), 'Color', 'r')
% Get halfway index
RhtHalf = Rht(halfpoints : end)
CrRHalf = CrR(halfpoints : end)
coefficients = polyfit(RhtHalf, CrRHalf, 1)
RhtFitted = linspace(5, 10, 1)
CrRFitted = polyval(coefficients, RhtFitted);
plot(RhtFitted, CrRFitted, 'c--', 'LineWidth', 2)
grid on
legend('Original Data', 'Linear Fit (2nd Half)');
hold off

Does anyone know how I can fix this?

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Walter Roberson on 31 May 2025

It is not clear what the difference is between this and https://www.mathworks.com/matlabcentral/answers/2177356-how-can-i-place-a-linear-regression-line-line-of-best-fit-through-both-plots-in-this-code-but-onl ?

ryan on 31 May 2025

Edited: Walter Roberson on 1 Jun 2025

Open in MATLAB Online

The txt file NewProb6

0 0.3983 0.0066 0.0017 -0.0927 0.6330 0.6222
5 0.4536 0.0066 0.0018 -0.0926 0.6172 0.6320
0 0.5091 0.0068 0.0018 -0.0925 0.6013 0.6389
5 0.5641 0.0069 0.0019 -0.0924 0.5857 0.6453
0 0.6189 0.0070 0.0020 -0.0923 0.5699 0.6517
5 0.6733 0.0072 0.0021 -0.0921 0.5540 0.6577
0 0.7270 0.0073 0.0023 -0.0917 0.5358 0.6644
5 0.7792 0.0075 0.0024 -0.0911 0.5135 0.6714
0 0.8310 0.0078 0.0026 -0.0905 0.4921 0.6782
5 0.8822 0.0080 0.0028 -0.0897 0.4682 0.6864
0 0.9328 0.0083 0.0031 -0.0889 0.4464 0.6948
5 0.9812 0.0086 0.0034 -0.0877 0.4199 0.7041
0 1.0280 0.0090 0.0037 -0.0861 0.3896 0.7145
5 1.0728 0.0094 0.0041 -0.0843 0.3581 0.7262
0 1.1138 0.0100 0.0045 -0.0817 0.3214 0.7394
5 1.1459 0.0107 0.0051 -0.0776 0.2766 0.7556
0 1.1724 0.0117 0.0059 -0.0726 0.2228 0.7750
5 1.1957 0.0128 0.0068 -0.0673 0.1745 0.7994
0 1.2193 0.0139 0.0079 -0.0622 0.1382 0.8326
5 1.2403 0.0149 0.0089 -0.0568 0.1102 0.9219
0 1.2659 0.0161 0.0102 -0.0528 0.0907 0.9825
5 1.2906 0.0175 0.0114 -0.0489 0.0728 1.0000
0 1.3137 0.0191 0.0130 -0.0452 0.0602 1.0000
5 1.3367 0.0208 0.0147 -0.0419 0.0518 1.0000
0 1.3589 0.0226 0.0166 -0.0387 0.0458 1.0000
5 1.3796 0.0247 0.0187 -0.0358 0.0412 1.0000
0 1.3951 0.0273 0.0213 -0.0328 0.0368 1.0000
5 1.4159 0.0297 0.0238 -0.0307 0.0345 1.0000
0 1.4268 0.0330 0.0271 -0.0282 0.0316 1.0000
5 1.4433 0.0360 0.0303 -0.0266 0.0300 1.0000
0 1.4559 0.0396 0.0340 -0.0251 0.0282 1.0000
5 1.4590 0.0443 0.0388 -0.0239 0.0264 1.0000
0 1.4703 0.0485 0.0431 -0.0233 0.0253 1.0000
5 1.4782 0.0532 0.0480 -0.0231 0.0241 1.0000
0 1.4794 0.0590 0.0539 -0.0233 0.0230 1.0000
5 1.4707 0.0663 0.0614 -0.0242 0.0220 1.0000
0 1.4743 0.0723 0.0676 -0.0254 0.0214 1.0000
5 1.4732 0.0792 0.0746 -0.0272 0.0208 1.0000
0 1.4701 0.0867 0.0822 -0.0294 0.0201 1.0000
5 1.4626 0.0950 0.0907 -0.0324 0.0196 1.0000
0 1.4501 0.1043 0.1002 -0.0360 0.0190 1.0000

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Answer 1

Star Strider on 31 May 2025

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https://in.mathworks.com/matlabcentral/answers/2177431-need-a-tangent-line-in-my-plot-like-the-example-provided-but-i-cannot-seem-to-do-it#answer_1565703

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Calculate the numerical derivative, find its value at the desired point on the curve, and calculate the tangent line parameters.

Try this --

x = linspace(0, 15);

f = @(x) 10./(x+0.1);

y = f(x);

dfdx = gradient(y,x);

figure

plot(x, y, DisplayName='f(x)')

hold on

plot(x, dfdx, DisplayName = 'df(x)/dx')

hold off

grid

legend(Location='best')

ylim([-4 4])

title('Data & Derivative')

xpt = 7.5;

ypt = f(xpt);

slope = interp1(x, dfdx, xpt)

slope = -0.1733

intercept = ypt - slope*xpt

intercept = 2.6152

figure

plot(x, y, DisplayName='f(x)')

hold on

plot(xpt, ypt,'sr', MarkerFaceColor='r', DisplayName='Tangent Point')

plot(x, slope*x+intercept, '--r', DisplayName='Tangent Line')

hold off

grid

legend(Location='best')

ylim([0 4])

title('Data & Tangent Line')

.

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Image Analyst on 31 May 2025

@ryan, tangent line is the slope at a particular point, while a fit is a line through a range of points. Make it clear what you want: the tangent or the fitted line.

Star Strider on 1 Jun 2025

Edited: Star Strider on 1 Jun 2025

Open in MATLAB Online

Using the provided data , try this --

dd = [0.0 0.3983 0.0066 0.0017 -0.0927 0.6330 0.6222

0.5 0.4536 0.0066 0.0018 -0.0926 0.6172 0.6320

1.0 0.5091 0.0068 0.0018 -0.0925 0.6013 0.6389

1.5 0.5641 0.0069 0.0019 -0.0924 0.5857 0.6453

2.0 0.6189 0.0070 0.0020 -0.0923 0.5699 0.6517

2.5 0.6733 0.0072 0.0021 -0.0921 0.5540 0.6577

3.0 0.7270 0.0073 0.0023 -0.0917 0.5358 0.6644

3.5 0.7792 0.0075 0.0024 -0.0911 0.5135 0.6714

4.0 0.8310 0.0078 0.0026 -0.0905 0.4921 0.6782

4.5 0.8822 0.0080 0.0028 -0.0897 0.4682 0.6864

5.0 0.9328 0.0083 0.0031 -0.0889 0.4464 0.6948

5.5 0.9812 0.0086 0.0034 -0.0877 0.4199 0.7041

6.0 1.0280 0.0090 0.0037 -0.0861 0.3896 0.7145

6.5 1.0728 0.0094 0.0041 -0.0843 0.3581 0.7262

7.0 1.1138 0.0100 0.0045 -0.0817 0.3214 0.7394

7.5 1.1459 0.0107 0.0051 -0.0776 0.2766 0.7556

8.0 1.1724 0.0117 0.0059 -0.0726 0.2228 0.7750

8.5 1.1957 0.0128 0.0068 -0.0673 0.1745 0.7994

9.0 1.2193 0.0139 0.0079 -0.0622 0.1382 0.8326

9.5 1.2403 0.0149 0.0089 -0.0568 0.1102 0.9219

10.0 1.2659 0.0161 0.0102 -0.0528 0.0907 0.9825

10.5 1.2906 0.0175 0.0114 -0.0489 0.0728 1.0000

11.0 1.3137 0.0191 0.0130 -0.0452 0.0602 1.0000

11.5 1.3367 0.0208 0.0147 -0.0419 0.0518 1.0000

12.0 1.3589 0.0226 0.0166 -0.0387 0.0458 1.0000

12.5 1.3796 0.0247 0.0187 -0.0358 0.0412 1.0000

13.0 1.3951 0.0273 0.0213 -0.0328 0.0368 1.0000

13.5 1.4159 0.0297 0.0238 -0.0307 0.0345 1.0000

14.0 1.4268 0.0330 0.0271 -0.0282 0.0316 1.0000

14.5 1.4433 0.0360 0.0303 -0.0266 0.0300 1.0000

15.0 1.4559 0.0396 0.0340 -0.0251 0.0282 1.0000

15.5 1.4590 0.0443 0.0388 -0.0239 0.0264 1.0000

16.0 1.4703 0.0485 0.0431 -0.0233 0.0253 1.0000

16.5 1.4782 0.0532 0.0480 -0.0231 0.0241 1.0000

17.0 1.4794 0.0590 0.0539 -0.0233 0.0230 1.0000

17.5 1.4707 0.0663 0.0614 -0.0242 0.0220 1.0000

18.0 1.4743 0.0723 0.0676 -0.0254 0.0214 1.0000

18.5 1.4732 0.0792 0.0746 -0.0272 0.0208 1.0000

19.0 1.4701 0.0867 0.0822 -0.0294 0.0201 1.0000

19.5 1.4626 0.0950 0.0907 -0.0324 0.0196 1.0000

20.0 1.4501 0.1043 0.1002 -0.0360 0.0190 1.0000];

alpha = dd(:,1); %Angle of attacks (Degreed)

CL = dd(:,2); %Lift coefficient

CD = dd(:,3); %Drag coefficient

Rh = 0.75; %Turbine hub radius (m)

Rt = 15; %Turbine blade tip radius (m)

Rht = linspace(Rh,Rt,10); %r/R radius hub to tip (m)

Rr = Rht/Rt;

B = 3; %Number of blades on the turbine

TSR = 8; %Tips-speed ratio

%Calculation Problem 1

CLD = CL./CD %Lift-to-Drag Ratio

CLD = 41×1

60.3485 68.7273 74.8676 81.7536 88.4143 93.5139 99.5890 103.8933 106.5385 110.2750 112.3855 114.0930 114.2222 114.1277 111.3800

<mw-icon class=""></mw-icon>

TCL = 1.0280 %Lift coefficient associated with maximum lift-to-drag ratio

TCL = 1.0280

Talpha = 6 %Target angle of attack (degrees)

Talpha = 6

%Calculation Problem 2

Phi = atand(2./(3*TSR*(Rr))) %Local inflow angle (degrees)

Phi = 1×10

59.0362 28.1786 17.7004 12.8043 10.0080 8.2072 6.9530 6.0300 5.3228 4.7636

<mw-icon class=""></mw-icon>

CrR = Rt*(8*pi*sind(Phi))./(3*B*TCL*TSR); %chord distribution

Beta = Phi-Talpha

Beta = 1×10

53.0362 22.1786 11.7004 6.8043 4.0080 2.2072 0.9530 0.0300 -0.6772 -1.2364

<mw-icon class=""></mw-icon>

dydx = gradient(CrR, Rht);

xpt = Rht(round(numel(CrR)/2));

ypt = interp1(Rht, CrR, xpt);

slope = interp1(Rht, dydx, xpt)

slope = -0.1269

intercept = ypt - slope*xpt

intercept = 1.7837

figure

subplot(2,1,1)

plot(Rht,CrR, 'b.-', 'LineWidth',2, 'MarkerSize', 20)

halfpoints = round(numel(CrR)/2)

halfpoints = 5

xlabel('r/R')

ylabel('Chord Distribution (m)')

grid on

hold on

% Assume x and y are your data vectors

xline(Rht(halfpoints), 'Color', 'r')

% Get halfway index

RhtHalf = Rht(halfpoints : end)

RhtHalf = 1×6

7.0833 8.6667 10.2500 11.8333 13.4167 15.0000

●

CrRHalf = CrR(halfpoints : end)

CrRHalf = 1×6

0.8852 0.7271 0.6166 0.5351 0.4725 0.4230

coefficients = polyfit(RhtHalf, CrRHalf, 1)

coefficients = 1×2

-0.0570 1.2388

RhtFitted = linspace(5, 10, 1)

RhtFitted = 10

CrRFitted = polyval(coefficients, RhtFitted);

plot(Rht, slope*Rht+intercept, '--m', DisplayName='Tangent Line', LineWidth=1.5)

plot(RhtFitted, CrRFitted, 'c--', 'LineWidth', 2)

grid on

legend('Original Data', 'Linear Fit (2nd Half)','Tangent Line');

hold off

.

EDIT -- Minor tweak.

.

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Answer 2

Image Analyst on 1 Jun 2025

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RhtFitted = linspace(5, 10, 1)
RhtFitted = 10

gives only a single value, so of course no line will be plotted. You need at least two points to plot a line.

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Need a tangent line in my plot like the example provided but I cannot seem to do it

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Need a tangent line in my plot like the example provided but I cannot seem to do it

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