## University GPR data help - how to find velocity

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Peter
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### University GPR data help - how to find velocity

Hello everyone!

I need a bit of help, for uni I´m trying to analyse some GPR data using CMP but I´m stuck finding the velocity

I´ve placed the data collected from the GPR using 3 different antenna positions in a excell graph (just like the attached image) and I´m trying to verify that the velocity from the first layer is 300m/nanoseconds (speed of the electromagentic wave on air)

Making some research online I´ve found the second attached image with a mathmatical expression to return the velocity from a hyperbola is this correct?

I´ve found very little stuff to help me on this matter, can anyone give me some tips and directions

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geophix
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### Re: University GPR data help - how to find velocity

The formula is correct. Suppose the depth to the target is d, and the distance the wave travel from an antenna to the target is l, then you have x^2+d^2=l^2 based on the right angle triangle formula. Since d=v*t0/2 and l=v*t/2, you get v^2=4x^2/(t^2-t0^2).

Peter
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### Re: University GPR data help - how to find velocity

Thanks so much for the reply really helped me out now I can find out the velocity of a hyperbola.

But I´m still lost in one thing, I´ve analysed my data and I´ve got these two tables (attached)

Left Collumn ( time in nanoseconds of a peak) Right Collumn (distance of the antenna in meters)

I created a excell graph that suposedly should indicate points from the two hyperbola that define both the surface and the second ground layer.

but in excell It´s nearly impossible to approximate a hyperbola to each set of 3 points..... and I need it to find the velocity and therefore thickness of the layers

Any tips on how to move from here?
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data.jpg (18.43 KiB) Viewed 4202 times

geophix
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### Re: University GPR data help - how to find velocity

I think you are using the wrong curve fitting formula here. Hyperbola is defined as y^2=d^2+x^2, not as the formula format shown in your Excel graph.

Also, the formula v^2=4x^2/(t^2-t0^2) is only good for the first layer, not for the second layer. It is because the wave path will bend when crossing the interface of the two layers due to wave velocity difference of the two layers. So the wave path is not straight any more, you can not use this formula directly from the ground surface to the second layer. I think you need to use the refraction formula v1/v2=sin(incidence_angle)/sin(refraction_angle) for the calculation.However, I am not sure how to do it exactly.

Peter
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Joined: Thu Feb 28, 2019 8:47 am

### Re: University GPR data help - how to find velocity

geophix wrote:
Fri Mar 01, 2019 9:25 pm
I think you are using the wrong curve fitting formula here. Hyperbola is defined as y^2=d^2+x^2, not as the formula format shown in your Excel graph.

Also, the formula v^2=4x^2/(t^2-t0^2) is only good for the first layer, not for the second layer. It is because the wave path will bend when crossing the interface of the two layers due to wave velocity difference of the two layers. So the wave path is not straight any more, you can not use this formula directly from the ground surface to the second layer. I think you need to use the refraction formula v1/v2=sin(incidence_angle)/sin(refraction_angle) for the calculation.However, I am not sure how to do it exactly.
Any software to bypass this step that you´d recommend me?
I can´t use reflexw or roaddoctor but besides those, any other that helps me find those hyperbolas and velocities is fair game

geophix
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### Re: University GPR data help - how to find velocity

Try Reflexw -> Modules -> CMP(1D) -> Velocity Analysis. I use Reflexw for 2D analysis, but never used its module for 1D velocity analysis. Maybe you can find out how it works or whether it works for your problem. Good luck!

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