Having problems understanding the command y( size(t) ) in a plot.

2011 12 15
2 个回答
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The whole command is given as:
plot(x,y,’k’,xmax,y(size(t)),’o’,xmax/2,ymax,’o’)
x, y, xmax, t, and ymax are all assigned variables. However I have no idea what the size command does. Does anyone know what this does?

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回答（2 个）

Walter Roberson
Walter Roberson 2011-12-15

1 个投票

It is a fancy (and confusing) way to draw a vertical line at xmax that shows the range of y values. It would be clearer to use y([1 end])
bym
bym 2011-12-15

0 个投票

it just plots xmax vs y(size(t)) as a second trace on the plot, but with the same markings as [xmax/2,ymax]...Have to see more code to figure why...

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Alex
Alex 2011-12-15
Here is the program, the command is all the way in the last line. How would the value of y(size(t)) be calculated in this case.
The projectile problem with zero air resistance
% in a gravitational field with constant g.
%
% Written by D. T. Valentine ........ September 2006
% Revised by D. T. Valentine ........ November 2008
% An eight-step structure plan applied in MATLAB:
%
% 1. Definition of the input variables.
%
g = 9.81; % Gravity in m/s/s.
vo = input(’What is the launch speed in m/s? ’);
tho = input(’What is the launch angle in degrees? ’);
tho = pi*tho/180; % Conversion of degrees to radians.
%
% 2. Calculate the range and duration of the flight.
%
txmax = (2*vo/g) * sin(tho);
xmax = txmax * vo * cos(tho);
%
% 3. Calculate the sequence of time steps to compute trajectory.
%
dt = txmax/100;
t = 0:dt:txmax;
% 4. Compute the trajectory.
%
x = (vo * cos(tho)) .* t;
y = (vo * sin(tho)) .* t –(g/2) .* t.ˆ2;
%
% 5. Compute the speed and angular direction of the projectile.
% Note that vx = dx/dt, vy = dy/dt.
%
vx = vo * cos(tho);
vy = vo * sin(tho) –g .* t;
v = sqrt(vx.*vx + vy.*vy);
th = (180/pi) .* atan2(vy,vx);
%
% 6. Compute the time, horizontal distance at maximum altitude.
%
tymax = (vo/g) * sin(tho);
xymax = xmax/2;
ymax = (vo/2) * tymax * sin(tho);
%
% 7. Display ouput.
%
disp([’ Range in m = ’,num2str(xmax), ...
’ Duration in s = ’, num2str(txmax)])
disp(’ ’)
disp([’ Maximum altitude in m = ’,num2str(ymax), ...
’ Arrival in s = ’, num2str(tymax)])
plot(x,y,’k’,xmax,y(size(t)),’o’,xmax/2,ymax,’o’)

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Alex
Alex
2011-12-15

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