Table 2.

Summary of parameters used in Eqs. 1 and 3 to find the deformed length

Transect	Heading	Relation to slope	l x (m)	\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ \sum\limits_X {\frac{{{w_{\text{S}}}}}{\alpha }} (m) $$\end{document}		\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ {l_X} - \sum\limits_X {\frac{{{W_{\text{S}}}}}{\alpha }} (m) $$\end{document}	e
				α = 3	α = 12
Scenario 1: assume deposit expansion (l x = l t)						l o	l x = l t
A–C	322	Parallel	390	8.01	2.00	385 ± 3	0.013 ± 0.008
D–E	330	Parallel	230	4.04	1.01	227.5 ± 1.5	0.011 ± 0.007
F–G	240	Perpendicular	190	4.55	1.14	187 ± 2	0.015 ± 0.009
Scenario 2: assume deposit contraction (l x = l o)						l t	l x = l o
A–C	322	Parallel	390	8.01	2.00	385 ± 3	−0.013 ± 0.008
D–E	330	Parallel	230	4.04	1.01	227.5 ± 1.5	−0.011 ± 0.007
F–G	240	Perpendicular	190	4.55	1.14	187 ± 2	−0.015 ± 0.009