# Landscape Fabric Roll

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• erobz
erobz
Homework Helper
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I was trying to figure out how much landscape fabric I have left on a roll, given a known thickness ##T## of the fabric and the radius ##r## of that roll. I obviously don't want to do an integral to calculate it while I'm working, but was looking for some approximation that can be readily determined by comparing it to the integral. I'm proposing a simple Archimedean Spiral ##r = k \theta##

So I start with the arclength formula in Cartesian coordinates:

$$ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } dx$$

With ##x= r \cos \theta, y = r \sin \theta ##

I start by making a change of variables to ##r##:

$$\left( \frac{dy}{dx} \right)^2 = \left( \tan \theta \right)^2$$

for the differential I get:

$$dx = \left[ k \cos \theta - r \sin \theta \right] d\theta$$

this goes to:

$$ds = \sqrt{1 + \left( \tan \theta \right)^2 } \left[ k \cos \theta - r \sin \theta \right] d\theta$$

If Iv'e done this correctly up to this point I see that it is integrable by researching it and I can get ##s## as a funtion of ##\theta##. But I realize its not really what I want... I want to measure the radius in the field.

I guess its true by definition of the Archimedean spiral that ##\theta## ( the upper limit of the integration ) is just the ##\frac{r}{T }2 \pi ##?

So if that is true, Is there a reasonable measure of some radius I can take in the field that would approximate the integral?

I guess since the fabric is very thin I can probably just get away with concentric circles, hence:

$$s \approx 2 \pi r + 2 \pi ( r+T) +2 \pi ( r+2T) \cdots 2 \pi ( r+nT) = 2 \pi ( n+1) \left( r + \frac{Tn}{2} \right)$$

Where ##n = \frac{r_o}{T}##

and ##r## is the initial radius ( from which it spirals out) near the center of the roll.

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Even better perhaps, since the rolls start from a very small initial radius:

$$s \approx 2\pi T + 2\pi 2T + \cdots 2\pi nT = \pi T n ( n+1)$$

Where ##n = \frac{r_o}{T}##

$$s \approx \pi \frac{r_o}{T} ( r_o + T )$$

Furthermore if we neglect ## T \ll r_o##, we get a rather simple expression to work with:

$$s \approx \pi \frac{r_o^2}{T}$$

I guess I solved my own problem!

erobz said:
I guess I solved my own problem!
Well that's a good thing!

But my first thought was to maybe use weighing instead of measuring the diameter of the roll. Do you know what the tare is for a roll (like the supporting core, whatever that is)? Do you have a scale handy that is accurate over the range of weights from the starting weight down to the tare?

Maybe compare the measuring technique with the weighing technique to see which is more accurate and easier for you to use.

berkeman said:
Well that's a good thing!

But my first thought was to maybe use weighing instead of measuring the diameter of the roll. Do you know what the tare is for a roll (like the supporting core, whatever that is)? Do you have a scale handy that is accurate over the range of weights from the starting weight down to the tare?

Maybe compare the measuring technique with the weighing technique to see which is more accurate and easier for you to use.
The rolls weigh about 10 lbs. It’s not critical enough for me to have a precision scale, but it surely would be accurate. Usually landscaping you might have a tape measure handy. It’s more of a “should i stop where I’m at and run to Lowe’s, or should I try to finish the job, and still end up running to Lowe’s” question!

Oh, sorry, I think I totally misinterpreted the term "landscape fabric roll". My wife is an avid seamstress, so there is a of sewing stuff and sewing projects around our house. And I assumed that your fabric was involved in some sort of sewing that was termed "landscaping".

But now I'm seeing that it is some sort of fabric used in actual outside landscaping, so yeah, measuring with a tape would be lots more practical than lugging out a scale. What exactly is landscape fabric used for? Just curious.

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erobz
berkeman said:
Oh, sorry, I think I totally misinterpreted the term "landscape fabric roll". My wife is an avid seamstress, so there are lots of sewing stuff and projects around our house. And I assumed that your fabric was involved in some sort of sewing that was termed landscaping.

But now I'm seeing that it is some sort of fabric used in actual outside landscaping, so yeah, measuring with a tape would be lots more practical than lugging out a scale. What exactly is landscape fabric used for? Just curious.
It's a permeable weed barrier that goes over a bed (on the soil), underneath the decorative mulch, or landscape stone. If applied well it minimizes weeding and spraying for many years.

berkeman
erobz said:
$$s \approx \pi \frac{r_o^2}{T}$$
Might be obvious, but just in case...

##r_i## is internal radius and ##r_e## is external radius.

Area of an end-face of the roll is ##\pi (r_e^2 - r_i^2)##.
Area of the corresponding edge of the roll ##sT##.

Assuming a tightly wound circular roll and ##T \ll s##, the above 2 quantities are the same, giving ##s = \frac {\pi (r_e^2 – r_i^2)}T## with virtually no maths needed.

erobz
Steve4Physics said:
Might be obvious, but just in case...
I didn't see it. It's obvious now though, thanks!

Steve4Physics

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