MATLAB plotting problems...

I have been trying to plot this graph on matlab or figure out how to go about doing it but everything I seem to do so far does not work. I was hoping I could get some help.

%%Define Variables
g = 9.8;
c = 14;
v=35;
%%Calculations
figure; hold on
for m = 60:70
    y(m) = (14*35)/(m*9.8);
end

figure(1); plot(m,y(m));
xlim([60 70]);

[Stephen23]

See Jan Simon's answer for the simplest solution.

Answer 1

[John BG]

Hi David

this is John BG <mailto:[e-mail] [e-mail]>

your code doesn't plot because m is scalar while y is a vector.

command plot requires either a vector, in this case would be y, or 2 vectors.

If you pass a scalar and a vector plot basically does nothing.

try either

plot([60:70],y)

or simply

plot(y)

if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:[e-mail] [e-mail]>

[David Oshidero]

Thank you, I'll try work on changing this then

[Jan]

 If you pass a scalar and a vector plot basically does nothing

No. Try this:

plot(1:10, 5, 'o')

You see, it does draw something. But if you do not specify the marker, the single points are not visible. With plot(60:70, y) a polygon is drawn through the points, which is visible also, even if there are no markers.

[Walter Roberson]

Note that because y(m) is being assigned to and m starts at 60, y(1) through y(59) are going to exist and be 0, which is going to throw off the plotting. plot([60:70], y) would fail because y would not be the same length as [60:70]

Also note that [60:70] is a more expensive way of writing 60:70 or (60:70). My timing tests suggest that using [60:70] is on the order of 40% slower than using 60:70 (though for an array that small, there is a lot of variation in the timings.)

[John BG]

the 'does nothing' is a way to speak.

The point is that neither passing a single value against a function represented with the many values of a vector, nor the multicolour bubble plot all-to-one mapping are what was being asked.

The y(m) of the question is an m-to-m mapping common to the majority functions.

John BG <mailto:[e-mail] [e-mail]>

[Walter Roberson]

"The y(m) of the question is an m-to-m mapping common to the majority functions."

In the user's code, it is a (max(m)-min(m)+1) to max(m) mapping -- that is, 70-60+1 = 11 different m values, and 70 vector locations created, the first min(m)-1 = 59 of which would be 0.

[Jan]

@John BG: Your "does nothing" is accompanied by "your code doesn't plot because m is scalar while y is a vector" and "command plot requires either a vector [...] or 2 vectors". Both statements are wrong and therefore confusing - independent from the fact, if this is or is not the point of the original question.

[John BG]

the reason why this question was posted is that the plot was invoked with a scalar in the reference vector while passing the vector y in the function field of the plot.

My answer is not confusing, it addresses the attempt to use a scalar as reference to multiple values in y.

The code posted in the question didn't plot, but with the correct reference vector input in plot, now it plots ok, what's so confusing?

Answer 2

[Walter Roberson]

After the for loop m will be the scalar 70, not the range.

[David Oshidero]

how would I go about changing it so m is the range?

[Walter Roberson]

The following general pattern is useful:

%%Define Variables
g = 9.8;
c = 14;
v=35;
%%Calculations
figure; hold on
m_vals = 60:70;
num_m = length(m_vals);
y = zeros(1, num_m);
for m_idx = 1 : num_m
  m = m_vals(m_idx);
  y(m_idx) = (14*35)/(m*9.8);
end

plot(m_vals, y)

[Walter Roberson]

In the specialized case, the above can be abbreviated as

%%Define Variables
g = 9.8;
c = 14;
v=35;
%%Calculations
figure; hold on
y = zeros(1, 11);
for m = 60:70
  y(m-60+1) = (14*35)/(m*9.8);
end

plot(60:70, y)

[David Oshidero]

Okay, thanks I'll retry taking all this into account

[David Oshidero]

I wanted to ask what does the y = zeros do in the eq?

[Jan]

@David: It is a pre-allocation. The output array is created at once, because this is much cheaper than letting the array grow iteratively. Example:

x = [];
for k = 1:1e7
  x(k) = k + rand;
end

Now in each iteration Matlab has to create a new array in the memory, which is 1 element larger than the old one, copy the former contents and insert the new value. Finally Matlab did not reserve memory for 1e6 elements, but for sum(1:1e7). For a double this needs 8 byte per element, such that 400 TB must be allocated and copied. This wastes a lot of time. To avoid this, the array is allocated before the loop:

x = zeros(1, 1e7);
for k = 1:1e7
  x(k) = k + rand;
end

While the first version takes 1.75 sec, the second needs 0.45 sec only.

Well, this is the theory. Fortunately Matlab tries to reduce the drawbacks as good as possible. Even a fast processor could allocate and copy 400 TB in 1.75 seconds.

In your case the runtime does not matter. But the strategy in the forum is to teach good programming styles in general.

Answer 3

[Jan]

A simplified version of your code:

g = 9.8;
c = 14;
v = 35;
t = 60:70;
y = (14*35) ./ (m * 9.8);   % [EDITED, fixed typo, thanks Walter!]
figure;
plot(m, y);
xlim([60 70]);

This is called "vectorizing": Matlab can perform the calculations with the vector m directly. This can be faster and nicer than the loop version, and it is less prone to bugs of indexing.

[Walter Roberson]

The lines

t = 60:70;
y = (14*35) / (m * 9.8);

should be

m = 60:70;
y = (14*35) ./ (m * 9.8);

[Jan]

Thank you, Walter.