Need help with a MATLAB Graph

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bryson1018
bryson1018 el 1 de Mayo de 2016
Editada: bryson1018 el 20 de Jul. de 2016
Hi everyone,
I'm currently stuck on a MATLAB graph problem. Both the problem and the code are seen below. I've done #1 easily, but I am stuck on the second problem.. I am unsure as to how I would go about doing it. Thanks!
%%circular spiral 1
t = linspace (0,8*pi,1000);
x = (2*t).*cos(t);
y = (2*t).*sin(t);
plot(x,y)
%%square spiral 1
t = linspace (0,8*pi,17);
x = (2*t).*cos(t);
y = (2*t).*sin(t);
plot(x,y)
%%circular spiral
2
a = 0-3.75*360;
r = (14.3/360)*a;
Now what???
  1 comentario
Stephen23
Stephen23 el 7 de Mayo de 2016
@Avery Buehler: please do not edit your questions and remove the test like this. We are not your own personal MATLAB service, we are volunteers who give our time to help everyone who reads this website. When you delete your question our answers stop making any sense. When we decide to help you, we have chosen to do so knowing that our answers can be useful to all browsers of this forum: when you delete your text you make our work meaningless, and you show that you do not value our goals or time, because you show that you do not care if our answers are useful to a anyone else. You are treating us as your own disposable asset.

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Walter Roberson
Walter Roberson el 1 de Mayo de 2016
The part where it says "theta = 0-3.75*360" is giving a range of theta, 0 to 3.75*360 degrees.
But it is not clear what your question is. The problem description says that both specifications are for the same design, so it appears that there should be only two spirals produced.
  1 comentario
Walter Roberson
Walter Roberson el 1 de Mayo de 2016
Draw one segment at a time. Reduce the radius after each segment.

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Más respuestas (1)

John BG
John BG el 2 de Mayo de 2016
Editada: John BG el 2 de Mayo de 2016
You haven't scaled the radius yet.
First, reproduce exactly the diagram of the spiral so you know that at the end of the angle range you have 100mm.
t = linspace (0,8*pi-pi/2,1000);
x = t.*cos(t);
y = t.*sin(t);
figure(1);plot(x,y)
grid on; axis equal
now, focus on the last lap, the biggest radius:
right at the end the angle is 8*pi-pi/2, here y is minimum, and pi rad earlier y is max.
Your current y range is:
range(y)
=
44.006631955593235
Now scale the radius with this factor:
t = linspace (0,8*pi-pi/2,1000);
x = 100/44.006631955593235*t.*cos(t);
y = 100/44.006631955593235*t.*sin(t);
figure(1);plot(x,y)
grid on; axis equal
now your spiral outer range matches the sketch requirement of outer diameter of 100mm
range(y)
=
9.999999999999999e+01
You can check that the the condition 14mm between top largest turn and the one before of 14mm is also met. I checked with the marker, but you may want to check
y2-y1=14
where
y2=y(th=8*pi-3*pi/2)
y1=y(th=7*pi-3*pi/2)
This was the answer to point 1
Do you want give it a go to point 2 now before I do?
If you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John

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