Severe plastic deformation

In surveying, tape correction(s) refer(s) to specific mathematical technique(s) used to apply corrections to a taping operation. Tape correction is applied to systematic or instrument errors or combination of both.

Correction due to incorrect tape length

Manufacturers of measuring tapes do not usually guarantee the exact length of tapes, and standardization is a process where a standard temperature and tension are determined at which the tape is the exact length. The nominal length of tapes can be affected by physical imperfections, stretching or wear. Constant use of tapes cause wear, tapes can become kinked and may be improperly repaired when breaks occur.

The correction due to tape length is given by:

$C_L=M_L\pm Corr\times \frac{M_L}{N_L}$

Where:

C_L is the corrected length of the line to be measured or laid out;

M_L is the measured length or length to be laid out;

N_L is the nominal length of the tape as specified by its mark.

Note that incorrect tape length introduces a systematic error that must be calibrated periodically.

Correction due to slope

Correction due to slope. $C_h$ is the correction of height due to slope, $\theta$ is the angle formed by the slope line oriented from the horizontal ground, s is the measured slope distance between two points on the slope line, h is the height of the slope.

When distances are measured along the slope, then the equivalent horizontal distance may be determined by applying a slope correction.

When applying corrections due to slope, it is necessary to determine the slope angle of the length measured. (Refer to the figure on the other side) Thus,

• For gentle slopes, $m<20\%$

$C_h=\frac{h^2}{2s}$

• For steep slopes, $20\%\le m\le 30\%$

$C_h=\frac{h^2}{2s}+\frac{h^4}{8s^3}$

• For very steep slopes, $m>30\%$

$C_h=s(1-\cos \theta )$

Where:

$C_h$ is the correction of measured slope distance due to slope;

$\theta$ is the angle between the slope line and horizontal line;

s is the measured slope distance.

The correction $C_h$ is subtracted from $s$ to obtain the equivalent horizontal distance on the slope line:

$d=s-C_h$

Correction due to temperature

When measuring or laying out distances, there is always a change in temperature especially when the taping operation requires time to do so. Usually, to avoid circumstances where there is an introduced error due to temperature, tapes were standardized as a response to such factor, and a standard temperature for the tape determined.

The correction of the tape length due to change in temperature is given by:

$C_f=C\cdot L(T-T_s)$

Where:

$C_f$ is the correction to be applied to the tape due to temperature;

T is the observed temperature or average observed temperature at the time of measurement;

$T_s$ is the standard temperature, the temperature at which the tape was standardized;

C is the coefficient of thermal expansion of the tape;

L is the length of the tape or length of the line measured.

The correction $C_f$ is added to $L$ to obtain the corrected distance:

$d=L+C_f$

Usually, for common tape measurements, the tape used is a steel tape with coefficient of thermal expansion C equal to 0.0000116 units per unit length per degree Celsius change. This means that the tape changes length by 1.16 mm per 10 m tape per 10°C change from the standard temperature of the tape.

Correction due to tension

Tension introduces error when the tape is pulled at a force that differs from the standard tension used at standardization. It will stretch less than its standard length when an insufficient pull is applied making the tape too short.

The tape stretches in an elastic manner (up until it reaches its elastic limit where it will deform permanently, essentially ruining the tape).

The correction due to tension is given by:

$C_p=\frac{(P_m-P_s)L}{AE}$

Where:

$C_p$ is the total elongation in tape length due to pull; or the correction to be applied due to incorrect pull applied on the tape; meters;

$P_m$ is the pull applied to the tape during measurement; kilograms;

$P_s$ is the standard tension, it is the pull applied to the tape during standardization; kilograms;

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

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

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

The correction $C_p$ is added to $L$ to obtain the corrected distance:

$d=L+C_p$

The value for A is given by:

$A=\frac{W}{(L)(U_w)}$

Where:

W is the total weight of the tape; kilograms;

$U_w$ is the unit weight of the tape; kilogram per cubic centimeter.

For steel tapes, the value for $U_w$ is given by $7.866\times 10^{-3}kg/c{m}^3$.

Correction due to sag

A tape not supported along its length will sag and form a catenary between supports. The correction due to sag must be calculated for each unsupported stretch separately and is given by:

$C_s=\frac{{\omega }^2{L}^3}{24P^2}$

Where:

$C_s$ is the correction applied to the tape due to sag; meters;

$\omega$ is the weight of the tape per unit length; kilogram per meters;

L is the length between two ends of the catenary; meters;

P is the tension or pull applied to the tape; kilogram.

The correction $C_s$ is subtracted from $L$ to obtain the corrected distance:

$d=L-C_s$

Note that the weight of the tape per unit length is equal to the weight of the tape divided by the length of the tape:

$\omega =\frac{W}{L}$

so: $W=\omega L$

Therefore, we can rewrite the formula for correction due to sag by:

$C_s=\frac{W^{2}L}{24P^2}$

Derivation (sag)

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

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

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

$L=\underset{-k/2}{\overset{+k/2}{\int }}\sqrt{1+{(dy/dx)}^2}dx$

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

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

The tape length L is then found by integrating:

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

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

A Taylor series expansion of $\delta$ in terms of the quantity 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 L and the next nonvanishing term is to the fifth power of L. Thus a series expansion for $\delta$ is reasonable. To this end we need to find an expression for $\delta$ that contains L but not k. We already have an expression for L in terms of k, but now need to find the inverse function (for k in terms of L):

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

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

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

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

Evaluating delta at 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:

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

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

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

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

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

Thus the first surviving term in the Taylor series is:

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

Notice that the variable P here is the tension on the cable, whereas above 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 P/g here and equate it to 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.

References

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