How do I plot the angle

25 ビュー (過去 30 日間)
Umut Ayyildiz
Umut Ayyildiz 2019 年 1 月 29 日
回答済み: Brian Hart 2019 年 1 月 30 日
A boat moves at 20kmh^−1 along a straight path described by
y =11x/15+43/3
starting at x = −10; y = 7. Plot the angle (in degrees) of the line of sight from an observer at the coordinate origin to the boat as a function of time for 3 hours.
This is what I have so far
x=-10:20:50
y=((11*x)/15)+(43/3)
plot(x,y)
hold on
a=atand(y/x)
legend('boat')

採用された回答

Brian Hart
Brian Hart 2019 年 1 月 30 日
Hi Umut,
This is a neat problem! Some nice physics with rectangular-to-polar conversions and trig asymptotes. My code to solve the problem is below. There's a fair amount of background knowledge needed to understand the math.
%define constants
vel = 20 / 60; %km/min
slope = 11/15;
lineAngle = atan(slope); %radians
yInt = 43/3;
xStart = -10;
yStart = 7;
tMax = 3*60; %max time in minutes
tInc = 5; %compute for five minute increments
%Make a vector of time values for which to compute x, y position:
tVec = 0:tInc:tMax;
%Compute the radial distance from the starting point at each time interval:
distArr = tVec * vel;
% Compute the x, y position at each time interval:
xArr = xStart + cos(lineAngle) * distArr;
yArr = yStart + sin(lineAngle) * distArr;
figure;plot(xArr,yArr,'x');title('Boat position at each time interval')
origToPtAngle = rad2deg(atan(yArr ./ xArr));
%Because the trajectory crosses x=0, we get a phase error. Crude
%correction:
origToPtAngle(find(origToPtAngle<0))=origToPtAngle(find(origToPtAngle<0)) + 180;
figure;plot(tVec,origToPtAngle,'.-');title('Line-of-sight from origin to boat position vs time in minutes (x-axis = 0°)')
With this I get:
untitled.bmp
untitled1.bmp
There's probably a more elegant way to handle the rectangular-to-polar stuff.

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeGeneral Applications についてさらに検索

タグ

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by