Multiple projectile plots

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Stuart
Stuart 2012 年 3 月 23 日
Hi, I'm trying to plot multiple projectiles on a graph. The projectiles at launched between 60 and 80 degrees with a one degree increment.
The launch velocity is specified by the Lanunch angle, to ensure it reaches the desired coordinates (6.7,3.05).
So the (x,y)graph I wanted would show 21 different trajectories.
However when I put my code in, Matblab only plots one line and I'm not really sure what it represents.
My code is:
%Angles of Projectile theta=[60:1:80];
%Launch velocity from given theta u=((((9.81/2)*1.25.^2)*(1+(tand(theta)).^2)./((1.25*tand(theta))-0.55))').^(0.5);
%Horizontal positions x=[0:0.335:6.7];
%Vertical Positions y=(x*tand(theta)')-((9.81/2)*(x.^2))./(((u.^2).*(cosd(theta))'.^2)')+2.5;
%Graph of projectiles plot(x,y)
Can anyone point out my error or what I should do? Thanks.

回答 (1 件)

Thomas
Thomas 2012 年 3 月 23 日
My understanding is that you change the theta and the x values..
declare theta and x
for change in theta
for change in x
compute u % remember to get u as vector u(i) or u(j)
compute y % remember to get y as vector y(i) or y(j)
end % to compute at this theta and all values of x giving you
%one set of values of y
plot(x,y) % to plot the first curve
hold on % to show more curves on same graph
end % to go to next value of theta and repeat
  2 件のコメント
Stuart
Stuart 2012 年 3 月 23 日
Thanks Thomas,
I'm not really sure how I should add this into my code.
Should I enter u(i) and y(i)?
Thomas
Thomas 2012 年 3 月 23 日
Think about how you get a vector for y?
You can svae the output in a vector or matrix as shown,
Eg:
for i=1:10
y(i)=i+rand; % us y(i) so that it is written as a vector
end
or you could use:
z=[];
for i=1:10
z=[z i+rand];
end
both should give you similar results..

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