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erobz
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I saw on a home improvement show some "shortcut" for measuring out a series of fixed measurements over a distance (it pertained to spacing baluster for porch railing). The contractor places a series of lines of some small incremental distance (1 or 2 inches say) on an elastic fabric band, and the stretches the band until the desired spacing is achieved on the last post. I thought, "is that going to yield increments of uniform spacing over the entire length or is this just a close enough technique?"
So, I start with:
$$ \frac{x}{l} = \frac{x + dx}{ l + dl} $$
Where ## x ## is a marked position on the band, and ## l ## is the length of the band.
This yields
$$ dx = \frac{x}{l}dl \tag{1} $$
I am going to say that ##x = fl ##, where ##0 \leq f \leq 1##
From (1)
$$ \begin{align} \int_{{{x_a}}}^{ {x_a}'} dx &= f_a \int_{L_o}^{L} dl \tag*{} \\ \quad \tag*{} \\ {x_a'} &= f_a \Delta l + x_a \tag*{} \end{align}$$
## {x_a}' ## is the new coordinate of ## x_a ## after the band undergoes a change of length ## \Delta l = L - L_o ##
So next imagine ## x_2 - x_1## defines some increment. After a stretch of ## \Delta l ## the new coordinates of the markings are given by:
$$ {x_2}' - {x_1}' = \left( f_2 - f_1 \right) \Delta l + x_2 - x_1 = \left( f_2 - f_1 \right) \left( \Delta l + L_o \right) $$
Similarly let ## x_4 - x_3## define an increment at some other location on the band.
$$ {x_4}' - {x_3}' = \left( f_4 - f_3 \right) \Delta l + x_4 - x_3 = \left( f_4 - f_3 \right) \left( \Delta l + L_o \right) $$
We can now see given some arbitrary ## \Delta l ## as long as ## f_4 - f_3 = f_2 - f_1## (that is to say the increments initially made are of uniform length), we have that:
$$ {x_4}' - {x_3}' = {x_2}' - {x_1}' $$
and the "shortcut" the contractor used should, indeed create increments of uniform length on the band after stretched to its new length.
I think I have it right? This ignores the width of the marks (assumed very thin), and length contraction in the band in orthogonal directions.
So, I start with:
$$ \frac{x}{l} = \frac{x + dx}{ l + dl} $$
Where ## x ## is a marked position on the band, and ## l ## is the length of the band.
This yields
$$ dx = \frac{x}{l}dl \tag{1} $$
I am going to say that ##x = fl ##, where ##0 \leq f \leq 1##
From (1)
$$ \begin{align} \int_{{{x_a}}}^{ {x_a}'} dx &= f_a \int_{L_o}^{L} dl \tag*{} \\ \quad \tag*{} \\ {x_a'} &= f_a \Delta l + x_a \tag*{} \end{align}$$
## {x_a}' ## is the new coordinate of ## x_a ## after the band undergoes a change of length ## \Delta l = L - L_o ##
So next imagine ## x_2 - x_1## defines some increment. After a stretch of ## \Delta l ## the new coordinates of the markings are given by:
$$ {x_2}' - {x_1}' = \left( f_2 - f_1 \right) \Delta l + x_2 - x_1 = \left( f_2 - f_1 \right) \left( \Delta l + L_o \right) $$
Similarly let ## x_4 - x_3## define an increment at some other location on the band.
$$ {x_4}' - {x_3}' = \left( f_4 - f_3 \right) \Delta l + x_4 - x_3 = \left( f_4 - f_3 \right) \left( \Delta l + L_o \right) $$
We can now see given some arbitrary ## \Delta l ## as long as ## f_4 - f_3 = f_2 - f_1## (that is to say the increments initially made are of uniform length), we have that:
$$ {x_4}' - {x_3}' = {x_2}' - {x_1}' $$
and the "shortcut" the contractor used should, indeed create increments of uniform length on the band after stretched to its new length.
I think I have it right? This ignores the width of the marks (assumed very thin), and length contraction in the band in orthogonal directions.
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