Tape correction (surveying)

The general formula for a catenary formed by a tape supported only at its ends is

${\displaystyle y={\frac {P}{\omega g}}\cosh \left({\frac {x\omega g}{P}}\right)}$.

Here, ${\displaystyle g}$ is the gravitational acceleration. The arc length between two support points at ${\displaystyle x=-k/2}$ and ${\displaystyle x=+k/2}$ is found by usual methods via integration:

${\displaystyle L=\int _{-k/2}^{+k/2}{\sqrt {1+\left(dy/dx\right)^{2}}}\,dx}$

For convenience set ${\displaystyle a={\frac {P}{\omega g}}}$. The integrand is simplified as follows using hyperbolic function identities:

${\displaystyle {\sqrt {1+\left(dy/dx\right)^{2}}}={\sqrt {1+\left({\frac {d}{dx}}\left(a\cosh \left({\frac {x}{a}}\right)\right)\right)^{2}}}={\sqrt {1+\sinh ^{2}\left({\frac {x}{a}}\right)}}=\cosh \left({\frac {x}{a}}\right)}$

The tape length ${\displaystyle L}$ is then found by integrating:

${\displaystyle L=\int _{-k/2}^{+k/2}\cosh \left({\frac {x}{a}}\right)dx=\left[a\sinh \left({\frac {x}{a}}\right)\right]_{x=-k/2}^{x=+k/2}=\left(2a\right)\sinh \left({\frac {k}{2a}}\right)}$

Now the correction for tape sag is the difference between the actual span between the supports, ${\displaystyle k}$, and the arc length of the tape's catenary, ${\displaystyle L}$. Call this correction ${\displaystyle \delta =k-L}$. The absolute value of this ${\displaystyle \delta }$ correction is ${\displaystyle C_{s}}$ above, the amount you would subtract from the tape measurement to get the true span distance.

A Taylor series expansion of ${\displaystyle \delta }$ in terms of the quantity ${\displaystyle L}$ is desired to give a good first approximation to the correction. In fact, the first nonvanishing term in the Taylor series is cubic in ${\displaystyle L}$, and the next nonvanishing term is to the fifth power of L; thus, a series expansion for ${\displaystyle \delta }$ is reasonable. To this end, we need to find an expression for ${\displaystyle \delta }$ that contains ${\displaystyle L}$ but not ${\displaystyle k}$. We already have an expression for ${\displaystyle L}$ in terms of ${\displaystyle k}$, but now need to find the inverse function (for ${\displaystyle k}$ in terms of ${\displaystyle L}$):

${\displaystyle {\frac {L}{2a}}=\sinh \left({\frac {k}{2a}}\right)}$

${\displaystyle \sinh ^{-1}\left({\frac {L}{2a}}\right)={\frac {k}{2a}}}$

${\displaystyle k=\left(2a\right)\sinh ^{-1}\left({\frac {L}{2a}}\right)}$

${\displaystyle \delta =k-L=\left(2a\right)\sinh ^{-1}\left({\frac {L}{2a}}\right)-L}$

Evaluating ${\displaystyle \delta }$ at ${\displaystyle L=0}$ yields zero, so there is no zero-order term in the Taylor series. The first derivative of this function with respect to L is

${\displaystyle {\frac {d\delta }{dL}}={\frac {1}{\sqrt {{\frac {L^{2}}{4a^{2}}}+1}}}-1}$.

Evaluated at L=0, it vanishes and so does not contribute a Taylor series term. The second derivative of ${\displaystyle \delta }$ is

${\displaystyle {\frac {d^{2}\delta }{{dL}^{2}}}=-{\frac {L}{4a^{2}\left({\frac {L^{2}}{4a^{2}}}+1\right)^{3/2}}}}$.

Again, when evaluated at L=0 it vanishes. When evaluated at L=0, the third derivative survives, however.

${\displaystyle {\frac {d^{3}\delta }{{dL}^{3}}}=-{\frac {\left(8a^{3}-4{\text{aL}}^{2}\right)}{\left(4a^{2}+L^{2}\right)^{5/2}}}}$

Thus, the first surviving term in the Taylor series is:

${\displaystyle \delta \cong \left[{\frac {d^{3}\delta }{{dL}^{3}}}\right]_{L=0}{\frac {L^{3}}{3!}}=-{\frac {1}{4a^{2}}}{\frac {L^{3}}{6}}={\frac {-L^{3}}{24a^{2}}}={\frac {-L^{3}\omega ^{2}g^{2}}{24P^{2}}}}$

Notice that the variable ${\displaystyle P}$ here is the tension on the cable, whereas above, ${\displaystyle P}$ is the mass whose gravitational force (mass times gravitational acceleration) equals the tension on the cable. The only conversion necessary then is to take ${\displaystyle P/g}$ here and equate it to ${\displaystyle P}$ above. Also, this formula is the tape sag correction to be added to the measured distance, so the negative sign in front can be removed and the tape sag correction can be made instead by subtracting the absolute value as is done in the preceding section.

Some tapes are already calibrated to account for the sag at a standard tension.[1][2] In this case, errors arise when the tape is pulled at a Tension which differs from the standard tension used at standardization. The tape will pulled less than its standard length when a tension less than the standard tension is applied, making the tape too long.

A tape stretches in an elastic manner until it reaches its elastic limit, when it will deform permanently and ruin the tape.

The correction due to tension is given by:

${\displaystyle C_{p}={\frac {(P_{m}-P_{s})L}{AE}}}$

Where:

${\displaystyle C_{p}}$ is the elongation in tape length due to pull; or the correction to be applied due to applying a tension which differs from standard tension; meters;

${\displaystyle P_{m}}$ is the tension applied to the tape during measurement; newtons;

${\displaystyle P_{s}}$ is the standard tension, when the tape is the correct length, often 50 newtons; newtons;

L is the measured or erroneous length of the line; meters

A is the cross-sectional area of the tape; square centimeters;

E is the modulus of elasticity of the tape material; newtons per square centimeter;

The correction ${\displaystyle C_{p}}$ is added to ${\displaystyle L}$ to obtain the corrected distance:

${\displaystyle d=L+C_{p}}$

The value for A is given by:

${\displaystyle A={\frac {W}{(L)(U_{w})}}}$

Where:

W is the total weight of the tape; kilograms;

${\displaystyle U_{w}}$ is the unit weight of the tape; kilograms per cubic centimeter.

For steel tapes, the value for ${\displaystyle U_{w}}$ is ${\displaystyle 7.866\times 10^{-3}kg/cm^{3}}$.