How to create your own wire length tables

This page describes how to create your own Wire length tables.

Copper wire is considered in this analysis at Wire length tables, but it can easily be used for aluminum wire (or other wire materials) simply by changing the wire materials resistivity (11.2 for copper, in Eqn. 6 on Table 2 below). The Wire Size Section contains Table 1, which can be used to determine wire size based on a few assumptions. The Calculating Wire Sizes Section explains how the values in Table 2 were calculated and will be useful for anyone attempting to remake Table 1 from Wire length tables using a different wire material.

Calculating Wire Sizes[edit | edit source]

The following assumptions, variable declarations and calculation steps explain how the values in Table 1 were calculated. Table 2 shows each equation used, how it was rearranged, and what to do with each equation to obtain the values in Table 1.

Assumptions made:

1. copper wire resistivity = 11.2^[1]
2. 3% loss from wires
3. allowable voltage drop = 0.36V
   • 12V system, 3% (0.03) wire loss -> 12V x 0.03 = 0.36V

Variable Declarations:

• A_n: circular mil area
• n: AWG
• m: AWG
• Conductor resistivity: constant, 11.2
• Current: current through wire in Amps, acquired from table
• Roundtrip (RT) wire length: complete length of wire, ft, acquired from table
• Allowable voltage drop: voltage drop through wires, V, 0.36V in this analysis

Table 2. Calculating Wire Sizes

Eqn. #	Equation	Step/Explanation
#1	${\displaystyle A_{n}=(5*92^{\frac {36-n}{39}})^{2}}$	Eqn. 1 was used, rearranged, and iterated on to produce Table 1. [1].
#2	${\displaystyle n=36-{\frac {39*log{\sqrt {\frac {A_{n}}{5}}}}{log(92)}}}$	Algebraically rearrange Eqn. 1 to solve for n. The resulting equation is Eqn. 2 to the left.
#3	${\displaystyle n=1-m}$	If n<0, replace n with 1-m in Eqn. 1 [2] .
#4	${\displaystyle A_{n}=(5*92^{\frac {36-(1-m)}{39}})^{2}}$	Substituting n equals 1-m into Eqn. 1 yields this equation (Eqn. 3). Use Eqn. 3 if n<0.
#5	${\displaystyle m={\frac {39*log{\frac {\sqrt {A_{n}}}{5}}}{log(92)}}-35}$	Solve Eqn. 3 for m, the resulting equation (Eqn. 4) is to the left.
#6	${\displaystyle A_{n}={\frac {(conductor-resistivity)*(current[A])*)roundtrip-wire-length[ft])}{allowable-voltage-drop[V]}}}$	Eqn. 3 is also a rearrangement of Eqn. 1. Use Eqn. 4 to solve for An
	FINAL STEP	Plug all values into Eqn. 6 to obtain a value for An. Plug this value of An into Eqn. 2 for n>0 or into Eqn. 5 for n<0. Solve for n (of m). This value is the required gauge (size) for the wire necessary for your application.

Example Problem[edit | edit source]

This section will outline an example problem in which the steps in the table above are carried through in more depth.

Example: Assume an allowable voltage drop for a system of 0.36 V and a copper wire resistivity of 11.2. If this system must be able to handle 10 A of current and it’s determined that the roundtrip wire length for the system is 70 feet, what size wire must be used (in AWG)?

Steps: Start with the following governing equation: Equation 1: ${\displaystyle A_{n}=(5*92^{\frac {36-n}{39}})^{2}}$ , Equation 2: ${\displaystyle A_{n}={\frac {(conductor-resistivity)*(current[A])*(roundtrip-wire-length[ft])}{allowable-voltage-drop[V]}}}$ Rearrange Equation 1 to solve for n by:

1. Taking the square root of both sides of Equation 1 to go from ${\displaystyle A_{n}=(5*92^{\frac {36-n}{39}})^{2}}$ to ${\displaystyle {\sqrt {A_{n}}}=(5)(92)^{\frac {36-n}{39}}}$
2. Divide both sides of the equation by 5 to go from ${\displaystyle {\sqrt {A_{n}}}=(5)(92)^{\frac {36-n}{39}}}$ to ${\displaystyle {\frac {\sqrt {A_{n}}}{5}}=92^{\frac {36-n}{39}}}$
3. Use log rules on both sides to undo the exponent to go from ${\displaystyle {\frac {\sqrt {A_{n}}}{5}}=92^{\frac {36-n}{39}}}$ to ${\displaystyle log({\frac {\sqrt {A_{n}}}{5}})={\frac {36-n}{39}}*log(92)}$
4. Divide both sides by log(92) to go from ${\displaystyle log({\frac {\sqrt {A_{n}}}{5}})={\frac {36-n}{39}}*log(92)}$ to ${\displaystyle {\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}={\frac {36-n}{39}}}$
5. Multiply both sides by 39 to go from ${\displaystyle {\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}={\frac {36-n}{39}}}$ to ${\displaystyle {\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}*39=36-n}$
6. Subtract 36 from each side to go from ${\displaystyle {\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}*39=36-n}$ to ${\displaystyle ({\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}*39)-36=-n}$
7. Multiply each side by -1 to go from ${\displaystyle ({\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}*39)-36=-n}$ to ${\displaystyle -({\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}*39)+36=n}$
8. Final, rearranged equation: ${\displaystyle n=36-[39*{\frac {log({\frac {\sqrt {A_{n}}}{5}})}{log(92)}}]}$
9. Plug the given values into Equation 2 ( ${\displaystyle A_{n}={\frac {(conductor-resistivity)*(current[A])*)roundtrip-wire-length[ft])}{allowable-voltage-drop[V]}}}$ ) to solve for ${\displaystyle A_{n}}$: ${\displaystyle A_{n}={\frac {(11.2)*(10A)*(70ft)}{0.36V}}}$, ${\displaystyle A_{n}=21,777.7cmil}$
10. Plug the above value of ${\displaystyle A_{n}}$ into the rearranged equation to solve numerically for n: ${\displaystyle n=36-[39*{\frac {log({\frac {\sqrt {21,777.7cmil}}{5}})}{log(92)}}]}$, n=6.8
11. If n is calculated to be less than 0, plug 1-m in for n in Equation 1 and rearrange to solve for n, then proceed to calculate the wire size as the value of m instead
12. If the above step doesn't apply to the problem you are working on (i.e. n is calculated to be greater than 0), round n down to the nearest even number. In this case, n=6.8. If we round this value down to the nearest even number n=6 AWG

References[edit | edit source]

• https://www.engineeringtoolbox.com/amps-wire-gauge-d_730.html

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Authors	Lonny Grafman
License	CC-BY-SA-4.0
Language	English (en)
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Created	March 5, 2021 by Lonny Grafman
Modified	June 9, 2023 by StandardWikitext bot
Cite as	Lonny Grafman (2021–2023). "How to create your own wire length tables". Appropedia. Retrieved April 18, 2024.
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