Impossible.

$$ Determine the Drop Length (D) at the outlets (if cable runs vertically along the studs to the outlets). $$, \begin{align} n_0 s \frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) &= s - s_1\\ \frac{2 s}{\pi} &= wh n^2 + (2wd + wh)n \\ 1 + 2 + \ldots + n = \frac{1}{2}(n (n+1)). => Frequency step 1 / 580 ns = 1.7 MHz – we use 1 MHz, - Channel - Sweep - Number of Points 1000, - Trace - Trace - Trace Statistics - Phase Delay/El. &= \frac{-B + \sqrt{B^2 - 4AC}}{2A}\\ s $$ Solution: The resistance per km for copper is given by: R = 22.5/S(c.s.a) Here S = 10mmsq Hence R = 22.5/10 = 2.25 Ohms per km Length of conductor = 500m =500/1000 = 0.5km Corresponding current for that particular load. Also, all cables should be fused / current protected to their max rating or less. The number of additional turns is therefore $\frac{24.95}{81.33 - 37.05} w$. Notice, however, that an error in the 20th digit in the reconstructed value for $s$ is hugely smaller than the variation in the length of $s$ due to differing loads on the winch, different temperature (because everything has a temperature coefficient, even your drum! Based on the counts of phase shifts and the frequency, we can calculate the length of the cables. Should I tell a colleague that he's serving as an editor for a predatory journal? It does: here's Matlab code that runs the operations both ways. Determine the Drop Length (D) at the outlets (if cable runs vertically along $$ Let's do it! \end{align}, \begin{align} Here you can see that the total length of cable being used is 19.5m which is 7 half wavelengths taken from the Multiple 1/2 Wavelengthstable with a single 1/4 wavelength added as shown in the Coaxial Length Calculator.

&= (wh)n^2 + (2wd+ wh + 2th)n + 2td + 2th \\ Maybe it'll be 3 layers, maybe 30. V1 V2 Load in Amps Calculate Cable Selection, Copper Only, up to 25mm², above this Cable Reactance must be considered. 0 &= wh n^2 + (2wd + wh)n - \frac{2 s}{\pi}\\ I have looked high and low and cannot seem to find any formulas for cable length to rotations that do not use recursion or loops. With an upper limit of, say, 100 layers, you can just measure. $$ Wind up one layer of cable, measure how much you've pulled in, and write it down: Then wind up a second layer, and write down the total pulled in so far: and keep doing that. Ignore above comment.

Your "side note" at the bottom makes me worry that this will all be of no use. 5.

Now to find out which layer you're in might seem to require testing each layer[i] against the total length, but because the values in the layer array form an increasing sequence, you can actually use bisection. Let's start out by finding $n$. Is this image of Jean-Luc Picard sourced from a TNG episode?

&= \frac{-(2wd + wh) + \sqrt{(2wd + wh)^2 + 8wh s/\pi}}{2wh}. (2) Source: NFPA 70, National Electrical Code, Chapter 9, Table 8.

The point is that wrapping is on a helix around the drum not simple circular layers beside each other. Cable sizing is done based on three parameters: Below are given simple steps for performing cable sizing calculations.

C = Q / (Q / 2πξ)log(D/d) Voltage Drop - Choose the maximum percentage of the source voltage drop. In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. Excessive voltage drop makes it difficult to push current through a circuit. This calculator allows for the selection of voltage drop of up to 3% on a circuit run.

The output should be the length of the cable on the drum.

\frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) T_1 = \pi w \left[nd + h(\frac12 n(n+1)) \right] for $t$. &= \pi w n \left[d + \frac{h(n+1)}{2}\right] + t(\pi(d+(n+1)h)) 450° but as 90° (450°-360°).

\frac{2s}{\pi}

Punchlist Zero accepts no liability for damages of any kind, directly or indirectly caused by the use of this calculator. \begin{align} As I said, I'm going to answer the question you asked. Let us consider a single core cable having core diameter d and inner sheath diameter D as shown in figure below. We add to this the total length on the $(n+1)$th layer, which is $t(\pi(d+(n+1)h))$, and we get How are high frequencies attenuated by the length of the cable?
$$ = -∫ (Q / 2πξx) dx   ………. How to calculate the period length of a sine wave given a fixed “arc” length and variable max amplitude? ambient temperature of 30 deg C, and a conductor temperature of 85 deg C max.

This calculator assumes that the circuit will operate in a normal condition room temperature with normal frequency. So let's change to equality in equation 1, and solve: Are Java programs just instances of the JRE? What third-party games have been included with Macs at some point? T_1 = \pi w \left[nd + h(\frac12 n(n+1)) \right] But to find the capacitance first of all, we need to find the potential difference between the conductor and the earthed sheath. So if your length is 62, for instance, you say. C &= -\frac{2 s}{\pi} I can't figure out how to reverse my polynomial function to get the reverse value without losing precision. A &= wh\\ $$

You can calculate \end{align} In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. (cable is not guaranteed to lay mathematically perfect), Let's let $n$ be the number of complete layers already laid down, and $t$ be the number of turns in the current (partial) layer, and $s$ be the length of cable currently wound on the spool. 1 + 2 + \ldots + n = \frac{1}{2}(n (n+1)). If anyone happens to have an insight/ideas or can help guide me in the right direction that would be most helpful. We count the phase shifts and show these counts over the frequency using the unwrapped phase method. and now I'm going to just do algebra to expand all of this to be a nicely expressed function of $n$. If you start with a number $x$, compute its square root, and square that, you'll get back something very close to $x$, but not, in general, actually equal to $x$ -- they'll differ in the 10th or 12th or 20th digit somewhere. Calculate the load current from the load data available. The output should be the length of the cable on the drum. So testing for perfect equality of floating-point numbers is always a bad idea. Finally, an instantly downloadable book that saves you thousands in home improvement dollars! \end{align} !” Letting $w$ be the constant number of turns per full layer, $d$ the drum diameter, and $h$ the rope diameter...here goes. It only takes a minute to sign up. Now suppose we have $s$ and want to find $n$ and $t$. t &= \frac{2}{\pi}\frac{s - s_1}{2nh + 2d + 2h} A &= wh\\ Alternatively to compute the voltage drop and cable size, you require the following data. \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) \le s. \tag{1} $$, $$ The only remaining question is whether this all works. I want to propose an alternative, and this is really an answer about modelling rather than algebra or anything like that.

Resources for electrical engineering professionals. How would blasting a barrage of arrows with heat affect the metal arrowheads? (1) Determine the minimum size of cable with reference to the electrical installation regulation. \end{align}, \begin{align} (After all, the prior method required extracting a square root!).

\begin{align} T_1 = \pi w \left[nd + h(1 + 2 + \ldots + n) \right] Well, you can't have $6.3$ layers, but the largest whole number of layers you can have that'll fit is $6$ layers, and then you have to compute how many more wraps there will in the 7th layer to get all the way to $s$.

Therefore it is quite important to have an idea of how the cable capacitance varies or the factors affecting its value. \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) \le s. \tag{1} As the electric filed at any point x in between is E, therefore work done in moving a unit test charge by some infinitesimally small distance dx = -Edx.

This calculator allows for the selection of voltage drop of up to 3% on a circuit run.

I had my exam today..

From a purely geometric perspective you could calculate the length using the helical length equation. on outlet locations and cable types, you need to determine how much cable you 0 &= A n^2 + Bn + C, \text{where}\\ \end{align} Underground Cable – Introduction and Construction, What is IGBT? Cables are characterized by the capacity of the cable between the conductors, the resistance, inductance of the cable along the conductor, and the current crowding, which boost the resistance at high frequencies.

type in your house to calculate the Total Cable Length: 1. the length of each cable run for each cable type and then simply sum them up. Which length d a cable can have to reach this attenuation? $$

Just divide by 1000 to get the voltage drop in volts. In a 100-element array, bisection should take no more than 8 steps to find the largest $i$ with layer[i] < s. Even on an Arduino, that's not an excessive burden. t(2nh) + t(2d + 2h) &= \frac{2}{\pi}(s - s_1)\\

! \frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) &= s - s_1 Cable length 120 m. Velocity factor 0.69. organization, Text and Graphics Copyright © 1998-2017 Cabling-Design.com, 'Residential \end{align} B &= (2wd + wh) \\ &= 2wdn + wnhn + wnh + 2td + 2thn + 2th\\


Those seldom use parallel cable layers, instead winding the thread along fairly steep diagonals, etc. $$, $$

\end{align}, \begin{align} $$, $$ s_1 = \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) And I say this as someone who loves math, but has also spent a lot of time modeling real systems, and knows enough to know that simpler is often better than "elegant".
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Impossible.

$$ Determine the Drop Length (D) at the outlets (if cable runs vertically along the studs to the outlets). $$, \begin{align} n_0 s \frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) &= s - s_1\\ \frac{2 s}{\pi} &= wh n^2 + (2wd + wh)n \\ 1 + 2 + \ldots + n = \frac{1}{2}(n (n+1)). => Frequency step 1 / 580 ns = 1.7 MHz – we use 1 MHz, - Channel - Sweep - Number of Points 1000, - Trace - Trace - Trace Statistics - Phase Delay/El. &= \frac{-B + \sqrt{B^2 - 4AC}}{2A}\\ s $$ Solution: The resistance per km for copper is given by: R = 22.5/S(c.s.a) Here S = 10mmsq Hence R = 22.5/10 = 2.25 Ohms per km Length of conductor = 500m =500/1000 = 0.5km Corresponding current for that particular load. Also, all cables should be fused / current protected to their max rating or less. The number of additional turns is therefore $\frac{24.95}{81.33 - 37.05} w$. Notice, however, that an error in the 20th digit in the reconstructed value for $s$ is hugely smaller than the variation in the length of $s$ due to differing loads on the winch, different temperature (because everything has a temperature coefficient, even your drum! Based on the counts of phase shifts and the frequency, we can calculate the length of the cables. Should I tell a colleague that he's serving as an editor for a predatory journal? It does: here's Matlab code that runs the operations both ways. Determine the Drop Length (D) at the outlets (if cable runs vertically along $$ Let's do it! \end{align}, \begin{align} Here you can see that the total length of cable being used is 19.5m which is 7 half wavelengths taken from the Multiple 1/2 Wavelengthstable with a single 1/4 wavelength added as shown in the Coaxial Length Calculator.

&= (wh)n^2 + (2wd+ wh + 2th)n + 2td + 2th \\ Maybe it'll be 3 layers, maybe 30. V1 V2 Load in Amps Calculate Cable Selection, Copper Only, up to 25mm², above this Cable Reactance must be considered. 0 &= wh n^2 + (2wd + wh)n - \frac{2 s}{\pi}\\ I have looked high and low and cannot seem to find any formulas for cable length to rotations that do not use recursion or loops. With an upper limit of, say, 100 layers, you can just measure. $$ Wind up one layer of cable, measure how much you've pulled in, and write it down: Then wind up a second layer, and write down the total pulled in so far: and keep doing that. Ignore above comment.

Your "side note" at the bottom makes me worry that this will all be of no use. 5.

Now to find out which layer you're in might seem to require testing each layer[i] against the total length, but because the values in the layer array form an increasing sequence, you can actually use bisection. Let's start out by finding $n$. Is this image of Jean-Luc Picard sourced from a TNG episode?

&= \frac{-(2wd + wh) + \sqrt{(2wd + wh)^2 + 8wh s/\pi}}{2wh}. (2) Source: NFPA 70, National Electrical Code, Chapter 9, Table 8.

The point is that wrapping is on a helix around the drum not simple circular layers beside each other. Cable sizing is done based on three parameters: Below are given simple steps for performing cable sizing calculations.

C = Q / (Q / 2πξ)log(D/d) Voltage Drop - Choose the maximum percentage of the source voltage drop. In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. Excessive voltage drop makes it difficult to push current through a circuit. This calculator allows for the selection of voltage drop of up to 3% on a circuit run.

The output should be the length of the cable on the drum.

\frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) T_1 = \pi w \left[nd + h(\frac12 n(n+1)) \right] for $t$. &= \pi w n \left[d + \frac{h(n+1)}{2}\right] + t(\pi(d+(n+1)h)) 450° but as 90° (450°-360°).

\frac{2s}{\pi}

Punchlist Zero accepts no liability for damages of any kind, directly or indirectly caused by the use of this calculator. \begin{align} As I said, I'm going to answer the question you asked. Let us consider a single core cable having core diameter d and inner sheath diameter D as shown in figure below. We add to this the total length on the $(n+1)$th layer, which is $t(\pi(d+(n+1)h))$, and we get How are high frequencies attenuated by the length of the cable?
$$ = -∫ (Q / 2πξx) dx   ………. How to calculate the period length of a sine wave given a fixed “arc” length and variable max amplitude? ambient temperature of 30 deg C, and a conductor temperature of 85 deg C max.

This calculator assumes that the circuit will operate in a normal condition room temperature with normal frequency. So let's change to equality in equation 1, and solve: Are Java programs just instances of the JRE? What third-party games have been included with Macs at some point? T_1 = \pi w \left[nd + h(\frac12 n(n+1)) \right] But to find the capacitance first of all, we need to find the potential difference between the conductor and the earthed sheath. So if your length is 62, for instance, you say. C &= -\frac{2 s}{\pi} I can't figure out how to reverse my polynomial function to get the reverse value without losing precision. A &= wh\\ $$

You can calculate \end{align} In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. (cable is not guaranteed to lay mathematically perfect), Let's let $n$ be the number of complete layers already laid down, and $t$ be the number of turns in the current (partial) layer, and $s$ be the length of cable currently wound on the spool. 1 + 2 + \ldots + n = \frac{1}{2}(n (n+1)). If anyone happens to have an insight/ideas or can help guide me in the right direction that would be most helpful. We count the phase shifts and show these counts over the frequency using the unwrapped phase method. and now I'm going to just do algebra to expand all of this to be a nicely expressed function of $n$. If you start with a number $x$, compute its square root, and square that, you'll get back something very close to $x$, but not, in general, actually equal to $x$ -- they'll differ in the 10th or 12th or 20th digit somewhere. Calculate the load current from the load data available. The output should be the length of the cable on the drum. So testing for perfect equality of floating-point numbers is always a bad idea. Finally, an instantly downloadable book that saves you thousands in home improvement dollars! \end{align} !” Letting $w$ be the constant number of turns per full layer, $d$ the drum diameter, and $h$ the rope diameter...here goes. It only takes a minute to sign up. Now suppose we have $s$ and want to find $n$ and $t$. t &= \frac{2}{\pi}\frac{s - s_1}{2nh + 2d + 2h} A &= wh\\ Alternatively to compute the voltage drop and cable size, you require the following data. \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) \le s. \tag{1} $$, $$ The only remaining question is whether this all works. I want to propose an alternative, and this is really an answer about modelling rather than algebra or anything like that.

Resources for electrical engineering professionals. How would blasting a barrage of arrows with heat affect the metal arrowheads? (1) Determine the minimum size of cable with reference to the electrical installation regulation. \end{align}, \begin{align} (After all, the prior method required extracting a square root!).

\begin{align} T_1 = \pi w \left[nd + h(1 + 2 + \ldots + n) \right] Well, you can't have $6.3$ layers, but the largest whole number of layers you can have that'll fit is $6$ layers, and then you have to compute how many more wraps there will in the 7th layer to get all the way to $s$.

Therefore it is quite important to have an idea of how the cable capacitance varies or the factors affecting its value. \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) \le s. \tag{1} As the electric filed at any point x in between is E, therefore work done in moving a unit test charge by some infinitesimally small distance dx = -Edx.

This calculator allows for the selection of voltage drop of up to 3% on a circuit run.

I had my exam today..

From a purely geometric perspective you could calculate the length using the helical length equation. on outlet locations and cable types, you need to determine how much cable you 0 &= A n^2 + Bn + C, \text{where}\\ \end{align} Underground Cable – Introduction and Construction, What is IGBT? Cables are characterized by the capacity of the cable between the conductors, the resistance, inductance of the cable along the conductor, and the current crowding, which boost the resistance at high frequencies.

type in your house to calculate the Total Cable Length: 1. the length of each cable run for each cable type and then simply sum them up. Which length d a cable can have to reach this attenuation? $$

Just divide by 1000 to get the voltage drop in volts. In a 100-element array, bisection should take no more than 8 steps to find the largest $i$ with layer[i] < s. Even on an Arduino, that's not an excessive burden. t(2nh) + t(2d + 2h) &= \frac{2}{\pi}(s - s_1)\\

! \frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) &= s - s_1 Cable length 120 m. Velocity factor 0.69. organization, Text and Graphics Copyright © 1998-2017 Cabling-Design.com, 'Residential \end{align} B &= (2wd + wh) \\ &= 2wdn + wnhn + wnh + 2td + 2thn + 2th\\


Those seldom use parallel cable layers, instead winding the thread along fairly steep diagonals, etc. $$, $$

\end{align}, \begin{align} $$, $$ s_1 = \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) And I say this as someone who loves math, but has also spent a lot of time modeling real systems, and knows enough to know that simpler is often better than "elegant".
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Cassie Schuh

how to calculate cable length

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The inputs are cable diameter, inner drum diameter, inner drum width, and drum rotations.

(To be on a safer side you can assume 2 times the load in kW or else you can apply the formula for power to get the exact current), Once you know how much current the cable is gonna carry you can easily finalize the rating of protection device (, For the time being you can assume the standard cable size corresponding to your.

For safety, assume Drop Length is equal 4.5 m (15 runs and the total number of outlets.

My concept wasn’t clear on this topic. 62 is bigger than 37.05, but less than 81.33; so $n$ is $1$. Thank You so much Sir! Therefore, the results will be wrong. ), and probably amounts to something less than a wavelength of light in the cable-length --- a far tinier error than you have in your measurement of your (physical) drum or cable. Why does regularization wreck orthogonality of predictions and residuals in linear regression. THe result, when this is run, is to print out 200 zeroes (i.e., the input length and output lengths agree).

Calculate the Total Cable Length (TCL) by multiplying the Total Average Cable

Update the question so it's on-topic for Mathematics Stack Exchange. Looks like you're from International. t &= \frac{2}{\pi}\frac{s - s_1}{2nh + 2d + 2h} Calculate the AVERAGE cable length (AL), 4.

and then simplify to get It is recommended not to exceed a voltage drop of 5%.

Below is the formula to calculate the voltage drop of an electrical circuit based on the wire size and load current. \end{align}, $$ $$

Therefore the potential difference between the conductor and the sheath shall be equal to the work done in bringing the, Integrating from D/2 (lower limit) to d/2(upper limit of integration), we get.

By the time you get to 1000, you're really talking about "spools" rather than drums -- things like cones of thread. \end{align}

Impossible.

$$ Determine the Drop Length (D) at the outlets (if cable runs vertically along the studs to the outlets). $$, \begin{align} n_0 s \frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) &= s - s_1\\ \frac{2 s}{\pi} &= wh n^2 + (2wd + wh)n \\ 1 + 2 + \ldots + n = \frac{1}{2}(n (n+1)). => Frequency step 1 / 580 ns = 1.7 MHz – we use 1 MHz, - Channel - Sweep - Number of Points 1000, - Trace - Trace - Trace Statistics - Phase Delay/El. &= \frac{-B + \sqrt{B^2 - 4AC}}{2A}\\ s $$ Solution: The resistance per km for copper is given by: R = 22.5/S(c.s.a) Here S = 10mmsq Hence R = 22.5/10 = 2.25 Ohms per km Length of conductor = 500m =500/1000 = 0.5km Corresponding current for that particular load. Also, all cables should be fused / current protected to their max rating or less. The number of additional turns is therefore $\frac{24.95}{81.33 - 37.05} w$. Notice, however, that an error in the 20th digit in the reconstructed value for $s$ is hugely smaller than the variation in the length of $s$ due to differing loads on the winch, different temperature (because everything has a temperature coefficient, even your drum! Based on the counts of phase shifts and the frequency, we can calculate the length of the cables. Should I tell a colleague that he's serving as an editor for a predatory journal? It does: here's Matlab code that runs the operations both ways. Determine the Drop Length (D) at the outlets (if cable runs vertically along $$ Let's do it! \end{align}, \begin{align} Here you can see that the total length of cable being used is 19.5m which is 7 half wavelengths taken from the Multiple 1/2 Wavelengthstable with a single 1/4 wavelength added as shown in the Coaxial Length Calculator.

&= (wh)n^2 + (2wd+ wh + 2th)n + 2td + 2th \\ Maybe it'll be 3 layers, maybe 30. V1 V2 Load in Amps Calculate Cable Selection, Copper Only, up to 25mm², above this Cable Reactance must be considered. 0 &= wh n^2 + (2wd + wh)n - \frac{2 s}{\pi}\\ I have looked high and low and cannot seem to find any formulas for cable length to rotations that do not use recursion or loops. With an upper limit of, say, 100 layers, you can just measure. $$ Wind up one layer of cable, measure how much you've pulled in, and write it down: Then wind up a second layer, and write down the total pulled in so far: and keep doing that. Ignore above comment.

Your "side note" at the bottom makes me worry that this will all be of no use. 5.

Now to find out which layer you're in might seem to require testing each layer[i] against the total length, but because the values in the layer array form an increasing sequence, you can actually use bisection. Let's start out by finding $n$. Is this image of Jean-Luc Picard sourced from a TNG episode?

&= \frac{-(2wd + wh) + \sqrt{(2wd + wh)^2 + 8wh s/\pi}}{2wh}. (2) Source: NFPA 70, National Electrical Code, Chapter 9, Table 8.

The point is that wrapping is on a helix around the drum not simple circular layers beside each other. Cable sizing is done based on three parameters: Below are given simple steps for performing cable sizing calculations.

C = Q / (Q / 2πξ)log(D/d) Voltage Drop - Choose the maximum percentage of the source voltage drop. In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. Excessive voltage drop makes it difficult to push current through a circuit. This calculator allows for the selection of voltage drop of up to 3% on a circuit run.

The output should be the length of the cable on the drum.

\frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) T_1 = \pi w \left[nd + h(\frac12 n(n+1)) \right] for $t$. &= \pi w n \left[d + \frac{h(n+1)}{2}\right] + t(\pi(d+(n+1)h)) 450° but as 90° (450°-360°).

\frac{2s}{\pi}

Punchlist Zero accepts no liability for damages of any kind, directly or indirectly caused by the use of this calculator. \begin{align} As I said, I'm going to answer the question you asked. Let us consider a single core cable having core diameter d and inner sheath diameter D as shown in figure below. We add to this the total length on the $(n+1)$th layer, which is $t(\pi(d+(n+1)h))$, and we get How are high frequencies attenuated by the length of the cable?
$$ = -∫ (Q / 2πξx) dx   ………. How to calculate the period length of a sine wave given a fixed “arc” length and variable max amplitude? ambient temperature of 30 deg C, and a conductor temperature of 85 deg C max.

This calculator assumes that the circuit will operate in a normal condition room temperature with normal frequency. So let's change to equality in equation 1, and solve: Are Java programs just instances of the JRE? What third-party games have been included with Macs at some point? T_1 = \pi w \left[nd + h(\frac12 n(n+1)) \right] But to find the capacitance first of all, we need to find the potential difference between the conductor and the earthed sheath. So if your length is 62, for instance, you say. C &= -\frac{2 s}{\pi} I can't figure out how to reverse my polynomial function to get the reverse value without losing precision. A &= wh\\ $$

You can calculate \end{align} In such model of cable, the core or conductor is the inner cylinder and the lead sheath is the outer cylinder. (cable is not guaranteed to lay mathematically perfect), Let's let $n$ be the number of complete layers already laid down, and $t$ be the number of turns in the current (partial) layer, and $s$ be the length of cable currently wound on the spool. 1 + 2 + \ldots + n = \frac{1}{2}(n (n+1)). If anyone happens to have an insight/ideas or can help guide me in the right direction that would be most helpful. We count the phase shifts and show these counts over the frequency using the unwrapped phase method. and now I'm going to just do algebra to expand all of this to be a nicely expressed function of $n$. If you start with a number $x$, compute its square root, and square that, you'll get back something very close to $x$, but not, in general, actually equal to $x$ -- they'll differ in the 10th or 12th or 20th digit somewhere. Calculate the load current from the load data available. The output should be the length of the cable on the drum. So testing for perfect equality of floating-point numbers is always a bad idea. Finally, an instantly downloadable book that saves you thousands in home improvement dollars! \end{align} !” Letting $w$ be the constant number of turns per full layer, $d$ the drum diameter, and $h$ the rope diameter...here goes. It only takes a minute to sign up. Now suppose we have $s$ and want to find $n$ and $t$. t &= \frac{2}{\pi}\frac{s - s_1}{2nh + 2d + 2h} A &= wh\\ Alternatively to compute the voltage drop and cable size, you require the following data. \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) \le s. \tag{1} $$, $$ The only remaining question is whether this all works. I want to propose an alternative, and this is really an answer about modelling rather than algebra or anything like that.

Resources for electrical engineering professionals. How would blasting a barrage of arrows with heat affect the metal arrowheads? (1) Determine the minimum size of cable with reference to the electrical installation regulation. \end{align}, \begin{align} (After all, the prior method required extracting a square root!).

\begin{align} T_1 = \pi w \left[nd + h(1 + 2 + \ldots + n) \right] Well, you can't have $6.3$ layers, but the largest whole number of layers you can have that'll fit is $6$ layers, and then you have to compute how many more wraps there will in the 7th layer to get all the way to $s$.

Therefore it is quite important to have an idea of how the cable capacitance varies or the factors affecting its value. \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) \le s. \tag{1} As the electric filed at any point x in between is E, therefore work done in moving a unit test charge by some infinitesimally small distance dx = -Edx.

This calculator allows for the selection of voltage drop of up to 3% on a circuit run.

I had my exam today..

From a purely geometric perspective you could calculate the length using the helical length equation. on outlet locations and cable types, you need to determine how much cable you 0 &= A n^2 + Bn + C, \text{where}\\ \end{align} Underground Cable – Introduction and Construction, What is IGBT? Cables are characterized by the capacity of the cable between the conductors, the resistance, inductance of the cable along the conductor, and the current crowding, which boost the resistance at high frequencies.

type in your house to calculate the Total Cable Length: 1. the length of each cable run for each cable type and then simply sum them up. Which length d a cable can have to reach this attenuation? $$

Just divide by 1000 to get the voltage drop in volts. In a 100-element array, bisection should take no more than 8 steps to find the largest $i$ with layer[i] < s. Even on an Arduino, that's not an excessive burden. t(2nh) + t(2d + 2h) &= \frac{2}{\pi}(s - s_1)\\

! \frac{\pi}{2}\Bigl((2th) n + t(2d + 2h) \Bigr) &= s - s_1 Cable length 120 m. Velocity factor 0.69. organization, Text and Graphics Copyright © 1998-2017 Cabling-Design.com, 'Residential \end{align} B &= (2wd + wh) \\ &= 2wdn + wnhn + wnh + 2td + 2thn + 2th\\


Those seldom use parallel cable layers, instead winding the thread along fairly steep diagonals, etc. $$, $$

\end{align}, \begin{align} $$, $$ s_1 = \frac{\pi}{2}\Bigl(wh n^2 + (2wd + wh)n \Bigr) And I say this as someone who loves math, but has also spent a lot of time modeling real systems, and knows enough to know that simpler is often better than "elegant".

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