What is the distribution of the length of fibers in a 1cm square of nonwoven fabric? You may assume that all the fibers are straight and that the orientations of the fibers are uniformly randomly distributed.



What is the average length of a fiber?

Let me see if I am understanding this correctly. Would an accurate way of describing this problem be "solve for the average length of a randomly oriented straight line that intersects a square 1 cm by 1 cm in 2 places"? EDIT: Put more succinctly, a fiber cannot terminate in the middle of the square, correct?

Assuming that is what you meant, then it seems to me that the brute force way to approach this is to assume without loss of generality that the first intersection is on the x axis at some point (x,0). From there, you can determine the average length of all segments that pass through that point and a different side of the square; it will involve integrals from 0 to 1 of integrands that will look something like [ (1-x)^2 + y^2 ] ^ (1/2) over y from 0 to 1, and you have to do it for each side. (The normalizing integral is 1 because of the choice of units and so I am not mentioning it.) I believe any point on the three sides is equally likely, so to get the average for all orientations you'd add these three integrals up and divide by 3, and get the average length for fibers coming through (x,0). This leaves you with one final average to do. EDIT: Since I know you like numbers, doing this all numerically gives an answer of about .87 cm, which seems somewhat plausible to me. I'll try to think of a more elegant way.

EDIT AGAIN: You could probably retrieve something proportional to the distribution of fiber lengths with an approach as follows. Say you're interested in how many fibers there are of length L. Consider a circle of radius L, and scan it along the x-axis. Every time that circle intersects the square on a different side, that represents a possible thread of length L between the point on the x-axis and that side. For each side, I think you could make a map of the region on the x-axis of where you can get intersections, and how many (for the vertical sides it should be at most 1 intersection, you can get 2 for the top one.) Weighting the length of these regions by the number of intersections should give you a number proportional to the probability of finding a fiber of length L.

Have to love ill-posed questions. I can imagine several different answers with a justification for each. Here is the one I find most reasonable:

Choose a point at random in the unit square.
From that point, choose an angle at random and see how long the resulting line segment is, on average. That is, find integral(length of line segment) as angle ranges from 0 to pi. That happens to be.... the area of the square... 1.

If you choose an angle at random, then choose a position for a line with that orientation with probability proportional to its length, you will get 1 again.

If, however, you choose a point on the edge at random and an angle at random, or something like that, you will get too few lines across the middle of the square and too many congregating in the corners, and get an answer less than 1.

And, of course, we tack "cm" onto our answer as an afterthought so as not to offend any physicists who wander into the thread.

about


exactly


Don't take my word for it though. I didn't check.

FWIW the integral I did was

I let you figure out why.

Technically, knitted garments are non-woven (weaving and knitting are the two major modes of textile manufacturing), so all you have to do is find a cheap sweater, determine its area, and unwind it to find how long the string is.

The problem is not ill posed. There is only one sensible interpretation. I even provided an illustration.

And the average length of a fiber is in fact 1cm. How about the distribution?

Hmmm. I'm just gonna go ahead and call BS.

Boro, very interesting problem. Did this come up in some modeling work you are doing? I love your posts in this forum and am looking forward to the solution.

I'm having trouble believing it's 1. I don't have an answer yet, but here's why 1 seems too high.

First, I tried to see if the answer would be greater to or less than 1.
-All lines cross through the square at 2 points.
-At each point where it intersects, it can leave at some angle between 1-179 degrees (approximately)
-At/Near the corners (for example, bottom edge near left corner) the length will be <1 if it goes through the left edge, >1 if it goes through top or right edge. These are both equally likely as angles 0<angle<90 will go through right or top, 90<angle<180 will go through left and be less than 1.
-As you move towards the center of the edge, your odds of getting a length >1 go DOWN. at .2 along the axis, your length is <1 for 78deg in 1 direction, 36deg in the other for only 66deg of >1 length.
>At the center of an edge, .5, there is only a 60 degree window looking straight up that gives a length greater than 1. This means 120deg will have lengths less than 1.

> So, a quick mental guess would say that, longest case has 50% of lines >1, shortest case has 33% of lines >1. So approximately 40% of lines will have lengths longer than 1. I expect the average to be <1, maybe in the .8-.9 range.

Another quick mental check is - how much greater than or less than 1 are these lines?
- Well, the longest possible is sqrt(2) = 1.41.
- Shortest possible is essentially 0 near the corner.
-> This again leads me to believe that the answer will be less than 1. Not only are there more shorter lines than longer ones, the shorter lines can be MUCH shorter, and the longer lines only slightly longer.

This now leads me to expect the answer to be in the .7 to .8 range.

Again, these methods are NOT exact, just quick mental gym to see whether the answer arrived at by other means makes sense. And looking at the numbers, some posted do make sense, but not the one by op.

RollWave,

A side of the square is a "typical" transit of the area (though at its very margin). In this way an average fiber length of 1 cm seems intuitively correct.

Going for an exact solution, i'm kinda stuck, but i think i'm on the right track...

3 random variables, uniformly distributed (not gaussian):
0<=x<1
0<=y<1
0<=angle<pi

Note the closed bound on the bottom left, open bound on the top right of the x,y range, this prevents selecting a point on the right boundary with an angle zero which wouldn't make sense. And obviously, 0 and pi give the same line, so based on the x,y ranges, we keep zero, omit pi.

I'm stuck at the trigonometry part. How to calculate the line distance given an x,y,angle. This seems like it should be easy, but i'm blanking.

y = ax + b , line equation
a = tan(angle)
b = y-x*tan(angle)

You could solve for x setting y=0or1 and for y setting x=0or1 to determine the box intercept points. with these 2 points, it would be a simple pythagorean theorem to determine the hypotonuse - which is your length.

Ok, I don't think I'm on the right track anymore...This of course will not give an elegant solution. Its more of a way to plug stuff into matlab and run 100,000 samples and see what it says.

but a side is NOT typical. Look at it this way. A line can veer left, go straight, or veer right. If it goes straight, it will have length ~1, but if it veers to the side, it will either be shorter or longer. And since there is a much wider angle range that result in shorter - shorter paths, ie cutting the corner, are actually 'typical'.

If you enter the bottom, you can exit out the left, top or right. Most paths out the left door are shorter than 1. All out the top are >= 1. Most out the right are shorter than 1. "most of 2/3" is "slightly but noticeably more than 50%". How's that for fuzzy logic, lol

(EDIT: forgot to add that your aiming window to get out the top is MUCH smaller (ie, <1/3) than your window of angles that result in getting out the left or right)

Yes, in fact it did.

RollWave,

I like the cut of your jib, but the average length is 1cm.

Shortest length of a fiber is 0cm
Longest possible length of a fiber is square root of 2 (1+1=c^2)

Average would be square root of 2 divided by 2 (0.707) or no?

RollWave,

A side of the square is indeed typical, because two of them at right angles define the area in the fist place.

If you took a scissors and cut the square into two equal areas with a single straight cut, what would the minimum length of that slice be? 1 cm.

Only if they're uniformly distributed

edit: and by that i mean the lengths

I don't think it is 1... My money is on ~0.8686.

Of these 500,000 samples, the mean of each block of 50,000 samples varied between 0.8663 to 0.8700, so the mean isn't varying much and 500,000 samples is more than adequate. (Hoping I didn't make some stupid mistake or my RNG is shyte.)

Algorithm:

Random would have the same result. Think about flipping a coin.

edit: i guess that graph explains it better, you're right.

Hmmmm, I'm wondering if setting the intersection at each side as a uniformly distributed random number rather than the angle as uniformly distributed random number is what makes my answer differ from the true answer (if it does)... I'm too tired now to think about it...

I think any random distribution of `fibers' to be considered should satisfy the following. If you divide the square into n times n smaller equal squares, then the average length of `fibers' in each smaller square is 1/n times the average length of `fibers' in the whole square (for every n=1,2,3,...).

OK, just came back from dog walk thinking about this... stupid mistake was after picking a random side to start, I then selected the other side of intersection randomly. This isn't the case. Should be easy to fix: pick a random side, pick a random point on the side, pick a random angle and find other intersection. But hitting the bed now...

Equally distributed intersection points doesn't imply equally distributed points and angles. If you take one point fixed at the midpoint of a side, and consider lines going only from there to the opposite side, it's clear that uniform angle density will result in fewer intersections near the corners, because they're further away, but your method has equal intersection rates all along the side, so it can't be right.