return iteration number from selected for loop value?

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Andrew Poissant
Andrew Poissant 2017 年 8 月 21 日
編集済み: Jan 2017 年 8 月 21 日
I have a for loop that calculates distance that a projectile travels given different acceleration values. What I want is to return the iteration number, i, for a specified output. So for example, lets say I want to know which index value corresponds to dx=201.8842904397075. This answer would be i=3. If I input a value of dx or dy, I want to return the corresponding iteration, i, value.
vi = 23;
theta = 35;
phi = 10;
xnew = 44;
ynew = 33;
dz = 0 - 150;
v_z = -vi*sind(phi);
a(1) = 0;
ax(1) = a(1)*cosd(theta)*cosd(phi);
ay(1) = a(1)*sind(theta)*cosd(phi);
az(1) = a(1)*sind(phi) - 9.8;
v_zi = -vi*sind(phi);
dt(1) = (-sqrt(2*az(1)*dz + v_zi^2) + v_zi)/az(1);
vf(1) = vi + a(1)*dt(1);
v_xf(1) = vf(1)*cosd(theta)*cosd(phi);
v_yf(1) = vf(1)*sind(theta)*cosd(phi);
v_zf(1) = v_zi + az(1)*dt(1);
dx(1) = v_xf(1)*dt(1) + 0.5*ax(1)*dt(1)^2;
dy(1) = v_yf(1)*dt(1) + 0.5*ay(1)*dt(1)^2;
for i = 2:10
a(i) = a(i-1) + 1;
ax(i) = a(i)*cosd(theta)*cosd(phi);
ay(i) = a(i)*sind(theta)*cosd(phi);
az(i) = a(i)*sind(phi) - 9.8;
dt(i) = (-sqrt(2*az(i)*dz + v_zi^2) + v_zi)/az(i);
vf(i) = vi + a(i)*dt(i);
v_zf(i) = v_zi + az(i)*dt(i);
v_xf(i) = vf(i)*cosd(theta)*cosd(phi);
v_yf(i) = vf(i)*sind(theta)*cosd(phi);
dx(i) = v_xf(i)*dt(i) + 0.5*ax(i)*dt(i)^2
dy(i) = v_yf(i)*dt(i) + 0.5*ay(i)*dt(i)^2;
end
  1 件のコメント
Jan
Jan 2017 年 8 月 21 日
編集済み: Jan 2017 年 8 月 21 日
It is useful to post the relevant part of the code only. As far as I can see, this would be enough to explain the problem:
for i = 1:20
dx(i) = rand + i;
end
Now find the position of a specific element.

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採用された回答

Jan
Jan 2017 年 8 月 21 日
編集済み: Jan 2017 年 8 月 21 日
Find the index with the minimal distance of dx and a given value:
adx = 201.8842904397075;
[~, index] = min(abs(dx - adx));
If you know the value exactly without any rounding errors (and this is rarely the case):
index = find(dx == adx)
  1 件のコメント
Andrew Poissant
Andrew Poissant 2017 年 8 月 21 日
ah, very clever! Thank you, works like a charm!

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