Area of trapezoid formed by slicing a cylinder

[magnetpedro]

What is the equation of the area of the trapezoid when a cylinder of radius R is cut by a plane inclined at an angle α? (Orange area of the figure)

How can I relate it with A_base=π*R² or with A_inside=2*R*L?

Any hints please?

Thank you very much in advance.

 

[berkeman]

What is the equation of the area of the trapezoid when a cylinder of radius R is cut by a plane inclined at an angle α? (Orange area of the figure)

How can I relate it with A_base=π*R² or with A_inside=2*R*L?

Any hints please?

Thank you very much in advance.
View attachment 84229

That looks pretty straightforward unless I'm missing something. Have you taken trig yet?

 

[magnetpedro]

That looks pretty straightforward unless I'm missing something. Have you taken trig yet?

Yes, I think I have but, unless I'm not grasping the whole problem, it's not quite simple.

I know that one of the sides of the trapezoid is 2R, but I don't know anything else.
I first thought that the length of the trapezoid would be L*cosα, but I think taht L*cosα it's only the projection of the length of the trapezoid in the xy plane (the plane of the screen of the computer).

 

[Mark44]

Draw a line from the center of the left end of the cylinder to the center of the chord on the right end of the cylinder. Using the angle $\alpha$ you should be able to use trig to find the vertical distance (h in the figure below) between the central axis of the cylinder to the center of the chord.

Once you know h, you can find the coordinates of the chord on the circular end of the cylinder.

 

[Ibix]

I don't think the orange region is a trapezoid. If you imagine extending the cylinder in both directions, the orange slice extends into an ellipse (get a sausage, cut it on the diagonal and look at the face you expose). Reducing the cylinder back to its original length simply trims the ends off the ellipse. So the ends of the shape are straight lines, but the sides are curves. Integration may be needed.

 

[SteamKing]

What is the equation of the area of the trapezoid when a cylinder of radius R is cut by a plane inclined at an angle α? (Orange area of the figure)

How can I relate it with A_base=π*R² or with A_inside=2*R*L?

Any hints please?

Thank you very much in advance.
View attachment 84229

It's not clear what you are calling A_base.

In any event, if you constructed a true size view looking directly at the plane cutting thru the horizontal cylinder, you would see that the long sides of the "trapezoid" were not straight lines; in fact, these sides should be curved, possibly parts of a parabola.

One end of the "trapezoid" has length 2R only because of its vertical position on the end of the cylinder. At the opposite end, the length of the "trapezoid" is less than 2R.

If you make a series of cuts along the length of the cylinder, you can calculate how wide the "trapezoid" is at any arbitrary point. If you have the width at the ends and in the middle of the cylinder, you can use Simpson's Rule to calculate the area pretty accurately.

 

[berkeman]

Yes, I think I have but, unless I'm not grasping the whole problem, it's not quite simple.

I know that one of the sides of the trapezoid is 2R, but I don't know anything else.
I first thought that the length of the trapezoid would be L*cosα, but I think taht L*cosα it's only the projection of the length of the trapezoid in the xy plane (the plane of the screen of the computer).

Can you give us some context to this question? Is it for a real-world application, or just a curiosity? Or is it for schoolwork? Thanks.

 

[RUber]

I would integrate with respect to the height of the "trapezoid," i.e. $A = 2\int_0^H w(x) dx$ where H is the total height of the shape and w is the half-width of the "trapezoid".

If you can get it into the form of
$A = m\int_0^H \sqrt{c^2 - x^2} dx$
then the integral evaluates to:
$\frac m 2 \left( H \sqrt {c^2 -H^2}+c^2 \sin^{-1} ( H/c ) \right)$
I had to look that one up, so maybe it'll save you some time.

 

[Mark44]

My interpretation of this problem, which could be wrong, is that the diameter at the left end of the cylinder and the chord at the right end, define a trapezoid. In my interpretation, it's immaterial whether the nonparallel sides of the trapezoid are inside the cylinder or outside it.

The problem is somewhat ambiguous. The OP originally had "Area of ellipse..." in the title, but in post 1, asked about the area of the trapezoid. I changed the thread title to be consistent with the question in post 1.

 

[magnetpedro]

Can you give us some context to this question? Is it for a real-world application, or just a curiosity? Or is it for schoolwork? Thanks.

Yes, of course. It's a model for a real-world application.

Imagine a magnetic flux (or any type of flux, for that matter) that passes through that cylinder.
If the cylinder's axis is parallel to the x-axis and parallel to the magnetic flux vector, the area crossed by the flux is the A_base=π*R2, and the length of the path of the magnetic flux is L. On the other hand, if the magnetic flux vector is orthogonal to the cylinder, the area crossed by the flux is A_inside=2*R*L and the length of the path of the magnetic flux is 2*R, it's diameter.

Now, when the cylinder is inclined with an angle α to the x axis, the magnetic flux through the cylinder must be decomposed in two parts:

-one is the projection of the cylinder in the x axis, where the area crossed by the flux is the area of an ellipse (A_base/cos(α) and the length of the path of the magnetic flux is L*cos(α);
-the other is the projection of the cylinder in the y-axis (orthogonal) where the area crossed by the flux is in my opinion the area of the trapezoid that I want to find (and that I need/I think to have a sin(α) in the expression) and the length of the path of the magnetic flux is L*sin(α);

Thanks for your help.

 

[magnetpedro]

My interpretation of this problem, which could be wrong, is that the diameter at the left end of the cylinder and the chord at the right end, define a trapezoid. In my interpretation, it's immaterial whether the nonparallel sides of the trapezoid are inside the cylinder or outside it.

The problem is somewhat ambiguous. The OP originally had "Area of ellipse..." in the title, but in post 1, asked about the area of the trapezoid. I changed the thread title to be consistent with the question in post 1.

It was my mystake, the new titles fits much better. Thanks!

 

[magnetpedro]

It's not clear what you are calling A_base.

In any event, if you constructed a true size view looking directly at the plane cutting thru the horizontal cylinder, you would see that the long sides of the "trapezoid" were not straight lines; in fact, these sides should be curved, possibly parts of a parabola.

One end of the "trapezoid" has length 2R only because of its vertical position on the end of the cylinder. At the opposite end, the length of the "trapezoid" is less than 2R.

If you make a series of cuts along the length of the cylinder, you can calculate how wide the "trapezoid" is at any arbitrary point. If you have the width at the ends and in the middle of the cylinder, you can use Simpson's Rule to calculate the area pretty accurately.

A_base is the area of the base of the cylinder, hence π*R2.

Your chain of thought is similar to mine, but I've lost myself along your description. Could you be so kind to give me some start on the expressions?
As I said, I think that the area of the trapezoid is somehow related (or I wish it'd be for a matter of easyness of the treatment of the expressions) with the sin(α).

 

[magnetpedro]

I would integrate with respect to the height of the "trapezoid," i.e. $A = 2\int_0^H w(x) dx$ where H is the total height of the shape and w is the half-width of the "trapezoid".

If you can get it into the form of
$A = m\int_0^H \sqrt{c^2 - x^2} dx$
then the integral evaluates to:
$\frac m 2 \left( H \sqrt {c^2 -H^2}+c^2 \sin^{-1} ( H/c ) \right)$
I had to look that one up, so maybe it'll save you some time.

Thanks for your help.
I'm sorry but I don't understand. When you consider w the half-width of the trapezoid, are you referring to the length of the base of the trapezoid? In this case which base? The minor or the major?

 

[certainly]

one is the projection of the cylinder in the x axis, where the area crossed by the flux is the area of an ellipse (Abase/cos(α) and the length of the path of the magnetic flux is L*cos(α);

It's actually the projection of the inclined cylinder on the y-z plane. (the side view)

the other is the projection of the cylinder in the y-axis (orthogonal) where the area crossed by the flux is in my opinion the area of the trapezoid that I want to find (and that I need/I think to have a sin(α) in the expression) and the length of the path of the magnetic flux is L*sin(α);

The projection is in the x-y plane (the top view). What you're doing is cutting a straight cylinder with an inclined plane and taking the projection of the cut-section. As I understand it, what you want is the projection of a cylinder that is inclined. The two are very different things! Think about viewing a cylinder that is inclined with the x-axis from the top versus what you were doing. The top view of a tilted cylinder is simply a rectangle with sides $2R$ and $Lcos(\alpha)$.

 

[RUber]

Thanks for your help.
I'm sorry but I don't understand. When you consider w the half-width of the trapezoid, are you referring to the length of the base of the trapezoid? In this case which base? The minor or the major?

In my breakdown of it, I was calling w the half-width of the short side of the trapezoid.

 

[magnetpedro]

In my breakdown of it, I was calling w the half-width of the short side of the trapezoid.
View attachment 84341

I see and agree.
In that case my only last question is what is c.

 

[certainly]

Would both of you care to explain, exactly how is the section of the cylinder relevant when you're actually tilting the cylinder itself ?

 

[magnetpedro]

Would both of you care to explain, exactly how is the section of the cylinder relevant when you're actually tilting the cylinder itself ?

Imagine slicing the tilted cylinder from one base to another. That's how the flux "see/crosses" the cylinder when it is tilted.

 

[stedwards]

?? As lbix pointed out. The plane figure is not a trapezoid but a truncated ellipse. ??

---Unless it's the trapezoid contained within the truncated ellipse that is of interest.

 

[RUber]

I see and agree.
In that case my only last question is what is c.

c is a constant for the formula, just like m. When I worked it out, it ended up being some combination of initial parameters R and sin(alpha).

 

[magnetpedro]

?? As lbix pointed out. The plane figure is not a trapezoid but a truncated ellipse. ??

---Unless it's the trapezoid contained within the truncated ellipse that is of interest.

In my opinion it is not. It's a trapezoid because a flux would cross all the length of the cylinder.

 

[certainly]

Here is my attempt at explaining why the cut-section won't be a trapezium:-

When the cutting plane is horizontal the section is obviously a circle.
When the plane is vertical the section is again obviously a rectangle.

Now imagine we start with a plane that is completely horizontal (lowermost cut) and slowly tilt it until it becomes completely vertical. At first you will get ellipses (upper 2 cuts)

Now if we keep tilting the cutting plane, at one point it will cut one of the bases of the cylinder as well and one curvy end of the ellipse is as ibix says "trimmed off", even more tilting, and the cutting plane will cut both of the bases, as in the original picture, and both the ends will be trimmed off, however the other two sides still remain curved, and they will continue to remain curved until the cutting plane is completely perpendicular. The only difference is, is that as you keep tilting the cutting plane their curve is becoming less and less apparent.
P.S. I still can't see what all this has to do with your original problem.

 

[magnetpedro]

Here is my attempt at explaining why the cut-section won't be a trapezium:-

When the cutting plane is horizontal the section is obviously a circle.
When the plane is vertical the section is again obviously a rectangle.

Now imagine we start with a plane that is completely horizontal (lowermost cut) and slowly tilt it until it becomes completely vertical. At first you will get ellipses (upper 2 cuts)

Now if we keep tilting the cutting plane, at one point it will cut one of the bases of the cylinder as well and one curvy end of the ellipse is as ibix says "trimmed off", even more tilting, and the cutting plane will cut both of the bases, as in the original picture, and both the ends will be trimmed off, however the other two sides still remain curved, and they will continue to remain curved until the cutting plane is completely perpendicular. The only difference is is that as you keep tilting the cutting plane their curve is becoming less and less apparent.
P.S. I still can't see what all this has to do with your original problem.

I understand your point, and it has kept me thinking for a while. Non the less, imagine this as shown in the picture,

These arrows represent the flux crossing the cylinder, so it has to cross all the cylinder.

 

[certainly]

These arrows represent the flux crossing the cylinder, so it has to cross all the cylinder.

Which is why I said in my first post that you should be thinking about the area of the top view of the whole cylinder and not the area of the cut-section.
Still, I don't see how this bears any relation to your argument that the area of the cut-section is a trapezium.

 

[magnetpedro]

Which is why I said in my first post that you should be thinking about the area of the top view of the whole cylinder and not the area of the cut-section.
Still, I don't see how this bears any relation to your argument that the area of the cut-section is a trapezium.

Perhaps you were right, but in any case, how can I relate this area that is crossed by the flux with the angle a?
Thanks a lot for your help once again.

 

[certainly]

see post #14.

 

[RUber]

Above is a quick plot of the trapezoid cut from a cylinder using R = 2, L =2, alpha = 30 degrees. The outer lines represent the actual width of the cylinder along the path, while the inner lines represent the straight trapezoid.
Once you work out the required trig to find the proper values of your variables, this either reduces to a simple expression (trapezoid), or the integral I posted above (truncated ellipse).
You can draw right triangles everywhere to find the right values based on R, L, and alpha.

 

[magnetpedro]

View attachment 84347
Above is a quick plot of the trapezoid cut from a cylinder using R = 2, L =2, alpha = 30 degrees. The outer lines represent the actual width of the cylinder along the path, while the inner lines represent the straight trapezoid.
Once you work out the required trig to find the proper values of your variables, this either reduces to a simple expression (trapezoid), or the integral I posted above (truncated ellipse).
You can draw right triangles everywhere to find the right values based on R, L, and alpha.

This is awesome. Have you done it in Matlab?
Would you be so kind to lend me the code you've used?

 

[RUber]

Here is the code I used in Matlab. It is not elegant, but does the trick.

r = 2;
alp = pi/6;
L =2;
H = 2/cos(alp); %length of trapezoid
h = 2*tan(alp); %height from middle at opposing side
w = sqrt(r^2 - h^2); %half-width at opposing side

x = linspace(-r,r);
y = zeros(size(x));

for i = 1:length(x)
if x(i) < -w %left edge
y(i) = H/(r-w)*(x(i)+2);
elseif and(x(i)>= -w, x(i) <= w) %top
y(i) = H;
else
y(i) = -H/(r-w)*(x(i)-2); %right edge
end
end
end

y2 = linspace(0,H);
wplus = sqrt(r^2-y2.^2*sin(alp)^2);
wmin = - wplus;

plot(x,y) %trapezoid
hold on
plot([-r,r],[0,0]) %bottom line
plot(wplus,y2) %right side cylinder
plot(wmin, y2) %left side cylinder
hold off

 

[magnetpedro]

Here is the code I used in Matlab. It is not elegant, but does the trick.

r = 2;
alp = pi/6;
L =2;
H = 2/cos(alp); %length of trapezoid
h = 2*tan(alp); %height from middle at opposing side
w = sqrt(r^2 - h^2); %half-width at opposing side

x = linspace(-r,r);
y = zeros(size(x));

for i = 1:length(x)
if x(i) < -w %left edge
y(i) = H/(r-w)*(x(i)+2);
elseif and(x(i)>= -w, x(i) <= w) %top
y(i) = H;
else
y(i) = -H/(r-w)*(x(i)-2); %right edge
end
end
end

y2 = linspace(0,H);
wplus = sqrt(r^2-y2.^2*sin(alp)^2);
wmin = - wplus;

plot(x,y) %trapezoid
hold on
plot([-r,r],[0,0]) %bottom line
plot(wplus,y2) %right side cylinder
plot(wmin, y2) %left side cylinder
hold off

Thank you!

 

[stedwards]

In my opinion it is not. It's a trapezoid because a flux would cross all the length of the cylinder.

Now your talking about the projection of a cylinder into a plane, I believe. The projection would have two parallel sides corresponding to the sides of the cylinder with elliptical end caps. (Whether a projection or a slice at some non-zero angle, you don't get a trapezoid.)

Does the flux pass through the entire cylinder?

 

[magnetpedro]

Now your talking about the projection of a cylinder into a plane, I believe. The projection would have two parallel sides corresponding to the sides of the cylinder with elliptical end caps. (Whether a projection or a slice at some non-zero angle, you don't get a trapezoid.)

Does the flux pass through the entire cylinder?

Yes, it does!

 

[stedwards]

Then, I guess you are asking about a projection, which is a long rectangle with the short ends replaced by half an ellipse. If you have a small light source across the room, it's the shadow cast by a soup can near the wall.