area of specific shape based on ellipse

[niksirat]

Is it possible to calculate the area marked with yellow color?
An ellipse with radius a and b and inner closed curve is parallel with ellipse with length x0

 

[Deleted member 4993]

Is it possible to calculate the area marked with yellow color?
An ellipse with radius a and b and inner closed curve is parallel with ellipse with length x0
View attachment 34213

What is that area of the outer ellipse?

What are the lengths of major and minor axes of the inner ellipse?

Continue...,..

 

[niksirat]

What is that area of the outer ellipse?

What are the lengths of major and minor axes of the inner ellipse?

Continue...,..

When I specified a and b, the area of the outer ellipse is $\pi a b$.

Inner shape is not an ellipse , just parallel with outer ellipse with distance x0.

Secondly, if I can solve this problem, I don't write this question here, definitely.

 

[Deleted member 4993]

When I specified a and b, the area of the outer ellipse is $\pi a b$.

Inner shape is not an ellipse , just parallel with outer ellipse with distance x0.

Secondly, if I can solve this problem, I don't write this question here, definitely.

The inner shape is an ellipse - you should be able to prove it from the conditions given. I stand corrected - inner shape is NOT an ellipse.

 

[Dr.Peterson]

When I specified a and b, the area of the outer ellipse is $\pi a b$.

Inner shape is not an ellipse , just parallel with outer ellipse with distance x0.

Secondly, if I can solve this problem, I don't write this question here, definitely.

You're right that the inner curve is not an ellipse:

Ellipse Parallel Curves -- from Wolfram MathWorld

The parallel curves for (outward) offset k of an ellipse with semi-axis lengths a and b are given by x_p = (a+(bk)/(sqrt(a^2sin^2t+b^2cos^2t)))cost (1) y_p = (b+(ak)/(sqrt(a^2sin^2t+b^2cos^2t)))sint. (2) The plots above show ellipse parallel curves for ellipses with a/b=0.5 and 0.9. Each...

mathworld.wolfram.com

If there were a nice formula for area, it would be found there. You will probably need to approximate in some way.

Where does the problem come from?

 

[niksirat]

Where does the problem come from?

In the company where I work, a wooden product is produced in high circulation, and shape of the product from above is like the one I drew.

The yellow part is the part of the product that is dyed. To calculate the price of the finished paint, I must have its area.

 

[jonah2.0]

Beer induced recall and suggestion follows.

Is it possible to calculate the area marked with yellow color?
An ellipse with radius a and b and inner closed curve is parallel with ellipse with length x0
View attachment 34213

When I specified a and b, the area of the outer ellipse is $\pi a b$.

Inner shape is not an ellipse , just parallel with outer ellipse with distance x0.

Secondly, if I can solve this problem, I don't write this question here, definitely.

In the company where I work, a wooden product is produced in high circulation, and shape of the product from above is like the one I drew.

The yellow part is the part of the product that is dyed. To calculate the price of the finished paint, I must have its area.

You could approximate very closely with a Planimeter.
Search for it on YouTube.

 

[blamocur]

In the company where I work, a wooden product is produced in high circulation, and shape of the product from above is like the one I drew.

The yellow part is the part of the product that is dyed. To calculate the price of the finished paint, I must have its area.

What is a typical value of $\frac{x_0}{b}$ ?

 

[Cubist]

When I specified a and b, the area of the outer ellipse is $\pi a b$.

Inner shape is not an ellipse , just parallel with outer ellipse with distance x0.

Secondly, if I can solve this problem, I don't write this question here, definitely.

Are you absolutely sure the inner shape is a parallel curve to the ellipse? Have you seen what happens in the link in post#5 when "x0" becomes large - the parallel curve starts to intersect/ cross itself.

Could it be that the curve you're actually after is the inner edge of the coloured area you'd get by drawing the ellipse with a really thick (circular nib) pen, like this...

...note that this inner curve would never self-intersect.

 

[blamocur]

Are you absolutely sure the inner shape is a parallel curve to the ellipse? Have you seen what happens in the link in post#5 when "x0" becomes large - the parallel curve starts to intersect/ cross itself.

Could it be that the curve you're actually after is the inner edge of the coloured area you'd get by drawing the ellipse with a really thick (circular nib) pen, like this...
View attachment 34214
...note that this inner curve would never self-intersect.

I believe the inner curve (the edge of your thick pen) would self-intersect, but in your picture it is covered by ink. In the picture below the blue and the green curves would be drawn by the edges of the diameter which is orthogonal to the ellipse:

 

[Dr.Peterson]

In the company where I work, a wooden product is produced in high circulation, and shape of the product from above is like the one I drew.

The yellow part is the part of the product that is dyed. To calculate the price of the finished paint, I must have its area.

You don't need a mathematically precise area. If the thickness $x_0$ is relatively small, you could just pretend the inner curve is an ellipse, which would lead to an easy calculation that is likely to be reasonably accurate.

That's the likely reason for this question:

What is a typical value of $\frac{x_0}{b}$ ?

 

[blamocur]

This is a pretty interesting numerical problem. I believe I can see the approach, but
would appreciate some sanity checking from this forum.

If we use the standard parametrization of an ellipse then the area in question can be
expressed by this integral:

$$S_h = \int_0^{2\pi} h\left(1-\frac{h\kappa(t)}{2} \right) \dot s (t) dt$$where:

• $h$ is the width of the strip (same as $x_0$ in the original post),
• $\dot s(t)$ is the derivative of the arc's length
• $\kappa(t)$ is the curvature of the ellipse.

Also, we assume that $h\kappa(t) < 1$ for all $t$, i.e. no self-intersection for the curve.
For the reference, with the curvature formula taken from Wikipedia:
$$x = a\sin t$$$$y = b\cos t$$$$\dot x = a\cos t$$$$\dot y = -b\sin t$$$$\dot s = \sqrt{\dot x^2 + \dot y^2}$$$$\kappa = \frac{1}{a^2b^2}\left( \frac{x^2}{a^4} + \frac{y^2}{b^4} \right)^{-\frac{3}{2}}$$

 

[Cubist]

I believe the inner curve (the edge of your thick pen) would self-intersect, but in your picture it is covered by ink. In the picture below the blue and the green curves would be drawn by the edges of the diameter which is orthogonal to the ellipse:

Yes, I think you're correct.

If the OP wanted to disregard the intersection areas (as I suspect) then the whole parallel curve wouldn't be required. One approach would be to consider half the ellipse (and then double the result afterwards). The area could be broken down into A,B,C as shown in the following diagram...


(Obviously A and C would be the same).

 

[niksirat]

Are you absolutely sure the inner shape is a parallel curve to the ellipse? Have you seen what happens in the link in post#5 when "x0" becomes large - the parallel curve starts to intersect/ cross itself.

x0 is between 1 and 2 cm. Precisely, when the painter paints the body around the product, some color is also used around and above the product, which becomes a significant number in the high number of products, because polyester and thinner are also used.

 

[jonah2.0]

Beer induced recall and ramblings follow.

...
The yellow part is the part of the product that is dyed. To calculate the price of the finished paint, I must have its area.

Price problems are usually the province of cost accountants. Splitting hair over this area dilemma becomes meaningless for them average minded practitioners as they would probably just do that, that is, compute the amount of paint consumed on a production run and do some minor average reckoning.

 

[blamocur]

x0 is between 1 and 2 cm.

What about 'b' ?

 

[Cubist]

If the outer ellipse has a=3 and b=2 then here's a graph of the "yellow" area calculated two ways (with the thickness x0 along the bottom)...

The red curve is calculated by assuming the inner curve is an ellipse.
The blue curve is calculated via numerical integration (inner parallel curve shape)

Here's the python code that calculates the area. NOTE: I'm not at all sure that it's correct. It is supplied with no warranty of any kind

Spoiler: code

import sys
import math

# configure here...
a=3
b=2 # Must be less than a
k=1.7 # thickness
n=1000 # increase this for more accuracy, up to a sensible amount

############

def binarySearch(func, target, bot, top, acc):
    bv = func(bot) - target >= 0
    tv = func(top) - target >= 0
    if tv == bv: sys.exit(-1)
    while top - bot > acc:
        mid = (top + bot)*0.5
        mv = func(mid)-target >= 0
        if mv == bv:
            bot = mid
        else:
            top = mid
    return (top + bot)*0.5

############

def get_xy(t):
    s = k/math.sqrt((a*math.sin(t))**2 + (b*math.cos(t))**2)
    return [ (a - b*s)*math.cos(t), (b - a*s)*math.sin(t) ]

############

min_t = 0
max_t = math.pi*0.5

# Adjust min_t if there's an intersection
def gtz(t): return (a*math.sin(t))**2 + (b*math.cos(t))**2 -  (a*k/b)**2
if gtz(0) < 0:
    small = 1e-15
    min_t = binarySearch(gtz, 0, 0,math.pi*0.5, small)
    # print("New min_t", min_t)

area = math.pi*a*b # area of the ellipse
print(area - math.pi*(a-k)*(b-k), "<- result based on two ellipses")
if k < b:
    inner_area = 0
    # Calculate an approximate inner area using Trapezoidal rule
    # https://en.wikipedia.org/wiki/Trapezoidal_rule#Non-uniform_grid
    [last_x, last_y] = get_xy(max_t)
    for i in range(1, n+1):
        t = max_t - (max_t - min_t)*i/n
        [x,y] = get_xy(t)
        # print(x, y, inner_area)
        inner_area += (y + last_y) * 0.5 * (x - last_x)
        last_x = x
        last_y = y
    inner_area *= 4 # the above just integrates over one quadrant
    # print(area, inner_area)
    area -= inner_area

print(area, "<- result from numerical integration")

 

[jonah2.0]

Beer soaked genuine gratitude follows.

... It is supplied with no warranty of any kind.

That's beautiful.
I used to have a similar disclaimer.

 

[blamocur]

Here's the python code that calculates the area. NOTE: I'm not at all sure that it's correct. It is supplied with no warranty of any kind

I added some plots to your code, and got these numbers:

k = 1.96​

18.718865667149423 <- result based on two ellipses​

18.801842694398868 <- result from numerical integration​

​

But when I plot both the inner ellipse and the actual curve I got the pictures below (the whole and zoomed inner curves), which I don't know how relate to the numbers. BTW, if you want to avoid self-intersection you want to use $k \leq \frac{b^2}{a}$ instead of $k < b$.

Disclaimer: I did some (I believe) insignificant changes to the code, like switching from 'math' to 'numpy', but the pictures look reasonable to me, and the numbers match the ones produced by your script.

 

[Cubist]

But when I plot both the inner ellipse and the actual curve I got the pictures below (the whole and zoomed inner curves), which I don't know how relate to the numbers.

Thanks for looking at my code! I intentionally programmed it to only consider the part of the parallel curve shown in green below, the red part is deliberately excluded because I considered it as being hidden by the "thick pen" discussed in post#9. Only the area within the green lines below would remain unpainted. The red lines would be painted over (I think)

BTW, if you want to avoid self-intersection you want to use $k \leq \frac{b^2}{a}$ instead of $k < b$.

I just wanted to ensure that a ≥ b ≥ k. I did these separately in...
- a comment
- and the line of code that you're talking about
In this update I remove the conditional, and just add an extra comment against "k" since the conditional wouldn't have worked properly anyway)

Spoiler: update

# No warranty of any kind
import sys
import math

# configure here...
a=3
b=2 # Must be less than a
k=1.96 # thickness, must be less than b
n=1000 # increase this for more accuracy, up to a sensible amount

############

def binarySearch(func, target, bot, top, acc):
    bv = func(bot) - target >= 0
    tv = func(top) - target >= 0
    if tv == bv: sys.exit(-1)
    while top - bot > acc:
        mid = (top + bot)*0.5
        mv = func(mid)-target >= 0
        if mv == bv:
            bot = mid
        else:
            top = mid
    return (top + bot)*0.5

############

def get_xy(t):
    s = k/math.sqrt((a*math.sin(t))**2 + (b*math.cos(t))**2)
    return [ (a - b*s)*math.cos(t), (b - a*s)*math.sin(t) ]

############

min_t = 0
max_t = math.pi*0.5

# Adjust min_t if there's an intersection
def gtz(t): return (a*math.sin(t))**2 + (b*math.cos(t))**2 -  (a*k/b)**2
if gtz(0) < 0:
    small = 1e-15
    min_t = binarySearch(gtz, 0, 0,math.pi*0.5, small)
    # print("New min_t", min_t)

area = math.pi*a*b # area of the ellipse

print(area - math.pi*(a-k)*(b-k), "<- result based on two ellipses")
inner_area = 0
# Calculate an approximate inner area using Trapezoidal rule
# https://en.wikipedia.org/wiki/Trapezoidal_rule#Non-uniform_grid
[last_x, last_y] = get_xy(max_t)
for i in range(1, n+1):
    t = max_t - (max_t - min_t)*i/n
    [x,y] = get_xy(t)
    # print(x, y, inner_area)
    inner_area += (y + last_y) * 0.5 * (x - last_x)
    last_x = x
    last_y = y
inner_area *= 4 # the above just integrates over one quadrant
# print(area, inner_area)
area -= inner_area

print(area, "<- result from numerical integration")