CURVGRAV-GUI: a graphical user interface to interpret gravity data using curvature technique

Abstract

CURVGRAV-GUI is an open source software that was developed to interpret gridded gravity data by using curvature technique. It was developed using C# language with Microsoft.NET Framework 4.0. This program calculates the critical and extreme points, and estimates the depths of source bodies at this points. Besides, CURVGRAV-GUI processes gravity data by using minimum curvature, one of the attributes of curvature, and detects the subsurface lineaments. It is a user-friendly application that can display obtained solutions and gravity data thanks to image and scatter maps. CURVGRAV-GUI was designed to develop both synthetic and field applications. Additionally, the β constant, a parameter related to the source geometry, was examined for different source types such as sphere, horizontal and vertical cylinder and thin vertical fault. This program was tested by using two synthetic model applications. In the first synthetic model application, it was used a complex synthetic model consisting of three sphere and a horizontal cylinder located at the different depths. In the second synthetic model application, a graben model consisting of two thin vertical fault was used. Finally, the performance of the CURVGRAV-GUI was tested with using a real gravity data belonging to Kozakli-Central Anatolia region, Turkey. Very successful results were obtained for both synthetic and field data. Earth scientist can use CURVGRAV-GUI for educational experiments.

Appendix: Calculation of the β values for some source bodies

Sphere

$$ g\left( x, z\right)=\frac{Az}{{\left({x}^2+{z}^2\right)}^{3/2}} $$

(18)

$$ \frac{\partial g}{\partial x}=\frac{-3 Axz}{{\left({x}^2+{z}^2\right)}^{5/2}}\kern0.75em ,\kern0.5em \frac{\partial^2 g}{{\partial x}^2}=\frac{15 A{x}^2 z}{{\left({x}^2+{z}^2\right)}^{7/2}}-\frac{3 Az}{{\left({x}^2+{z}^2\right)}^{5/2}} $$

(19)

$$ K(x)=\frac{\frac{\partial^2 g}{{\partial x}^2}}{{\left(1+{\left(\frac{\partial g}{\partial x}\right)}^2\right)}^{\frac{3}{2}}}\Rightarrow \kern1.25em K(x)=\frac{\frac{15 A{x}^2 z}{{\left({x}^2+{z}^2\right)}^{7/2}}-\frac{3 A z}{{\left({x}^2+{z}^2\right)}^{5/2}}}{{\left(1+{\left(\frac{-3 A xz}{{\left({x}^2+{z}^2\right)}^{5/2}}\right)}^2\right)}^{\frac{3}{2}}},\kern0.5em K(0)=\frac{-3 A}{z^4}\kern0.5em $$

(20)

$$ K(0)=-\frac{2\beta g(0)}{z^2}\Rightarrow \kern1.25em \frac{-3 A}{z^4}=\frac{-2\beta \frac{A}{z^2}}{z^2},\kern0.5em \beta =\frac{3}{2} $$

(21)

Horizontal cylinder

$$ g\left( x, z\right)=\frac{Az}{x^2+{z}^2} $$

(22)

$$ \frac{\partial g}{\partial x}=\frac{-2 Axz}{{\left({x}^2+{z}^2\right)}^2}\kern0.75em ,\kern0.5em \frac{\partial^2 g}{{\partial x}^2}=\frac{8 A{x}^2 z}{{\left({x}^2+{z}^2\right)}^3}-\frac{2 Az}{{\left({x}^2+{z}^2\right)}^2} $$

(23)

$$ K(x)=\frac{\frac{\partial^2 g}{{\partial x}^2}}{{\left(1+{\left(\frac{\partial g}{\partial x}\right)}^2\right)}^{\frac{3}{2}}},\kern0.5em K(x)=\frac{\frac{8 A{x}^2 z}{{\left({x}^2+{z}^2\right)}^3}-\frac{2 A z}{{\left({x}^2+{z}^2\right)}^2}}{{\left(1+{\left(\frac{-2 A xz}{{\left({x}^2+{z}^2\right)}^2}\right)}^2\right)}^{\frac{3}{2}}},\kern0.5em K(0)=\frac{-2 A}{z^3}\kern0.5em $$

(24)

$$ K(0)=-\frac{2\beta g(0)}{z^2},\kern0.5em \frac{-2 A}{z^3}=\frac{-2\beta \frac{A}{z}}{z^2},\kern0.5em \beta =1 $$

(25)

Vertical cylinder

$$ g\left( x, z\right)=\frac{A}{{\left({x}^2+{z}^2\right)}^{1/2}} $$

(26)

$$ \frac{\partial g}{\partial x}=\frac{- Ax}{{\left({x}^2+{z}^2\right)}^{3/2}}\kern0.75em ,\kern0.5em \frac{\partial^2 g}{{\partial x}^2}=\frac{3 A{x}^2}{{\left({x}^2+{z}^2\right)}^{5/2}}-\frac{A}{{\left({x}^2+{z}^2\right)}^{3/2}} $$

(27)

$$ K(x)=\frac{\frac{\partial^2 g}{{\partial x}^2}}{{\left(1+{\left(\frac{\partial g}{\partial x}\right)}^2\right)}^{\frac{3}{2}}},\kern0.5em K(x)=\frac{\frac{3 A{x}^2}{{\left({x}^2+{z}^2\right)}^{5/2}}-\frac{A}{{\left({x}^2+{z}^2\right)}^{3/2}}}{{\left(1+{\left(\frac{- A x}{{\left({x}^2+{z}^2\right)}^{3/2}}\right)}^2\right)}^{\frac{3}{2}}},\kern0.5em K(0)=\frac{- A}{z^3}\kern0.5em $$

(28)

$$ K(0)=-\frac{2\beta g(0)}{z^2},\kern0.5em \frac{- A}{z^3}=\frac{-2\beta \frac{A}{z}}{z^2},\kern0.5em \beta =\frac{1}{2} $$

(29)

Vertical thin fault

$$ g\left( x, z\right)= A\left(\frac{\pi}{2}+{tan}^{-1}\left(\frac{x}{z}\right)\right) $$

(30)

$$ \frac{\partial g}{\partial x}=\frac{Az}{x^2+{z}^2} $$

(31)

The horizontal gradient of the vertical thin fault anomaly is equal to the horizontal cylinder anomaly. In this study, the curvature method was applied on horizontal gradient of vertical thin fault anomaly, therefore, the β parameter is 1.