A reflection coefficient of 0.0455 as you show below is 5% reflectivity. Meaning that approximately 5% of the incident energy is reflected at that boundary. The remaining 95% will continue travelling through the material. Hope that answers your question.
Nice webinar. In your computer simulations at the beginning the pulse is travelling out in a spherical wavefront, leading to large geometrical spreading losses. Why not simulate it more like a flashlight beam?
The animation of a wavefront is how GPR waves propagate into the subsurface. We don’t try to narrow the beam width like a flashlight because that causes all kinds of other issues in the data.
The round object maybe because it was active meaning it has emision or vibration from transportation. A calibration is needed before confirm a other measurment.
Great lecture Thank you! Can GPR be used to indicate corroded rebar or pipe? for instance would there be a difference in reflectivity between good steel and corroded steel- ferrous oxide?
Have you watched part 2 of this series ua-cam.com/video/2DG3Z65O6yw/v-deo.html Glad you liked it. Do let us know if you need more information. www.sensoft.ca/contact-us-gpr/
Very good seminar, as the others in this channel. But, if you allow me, I disagree with what you said Quiz 4, because I think that if the pipe is located in a high attenuation zone, the reflector just below this zone shouldn't be visible, but a strong reflector just below this zone is still visible. Actually, in quiz 9 you confirm this situation, when an attenuation zone begins, no reflector should be visible deeper.
Thank you for your video, much appreciated if you can provide a link for he presentation as well!
Great Webinar for basic interpreting GPR data
One of the best webinars about GBR, thanks
R=−0.0455 represents approximately 0.207% of reflected power, how can I evaluate this value to 5% of reflectivity? can u please explain
A reflection coefficient of 0.0455 as you show below is 5% reflectivity. Meaning that approximately 5% of the incident energy is reflected at that boundary. The remaining 95% will continue travelling through the material. Hope that answers your question.
Fantastic webinar....now on to part 2 :)
Thank you for the interesting explanation, Can the water table be determined in the GPR profile based on the amplitude and how?
Nice webinar. In your computer simulations at the beginning the pulse is travelling out in a spherical wavefront, leading to large geometrical spreading losses. Why not simulate it more like a flashlight beam?
The animation of a wavefront is how GPR waves propagate into the subsurface. We don’t try to narrow the beam width like a flashlight because that causes all kinds of other issues in the data.
@@sensoftgpr thanks!
The round object maybe because it was active meaning it has emision or vibration from transportation. A calibration is needed before confirm a other measurment.
Very very very! Helpful and informative thank you thank you ..hoping for more videos like this
Thanks for your kind words. We are glad that you found it helpful. Be sure to watch the Interpretation Part 2 video!
Sr pls help me how to do interpretation of underdug caches ( metal items) ...not able to understand the graph
Thank you!
Contact us here if you need more information - www.sensoft.ca/contact-us-gpr
Great Presentation...
Great lecture Thank you! Can GPR be used to indicate corroded rebar or pipe? for instance would there be a difference in reflectivity between good steel and corroded steel- ferrous oxide?
Thank you for posting your presentation, it helped me a lot
Have you watched part 2 of this series ua-cam.com/video/2DG3Z65O6yw/v-deo.html
Glad you liked it. Do let us know if you need more information. www.sensoft.ca/contact-us-gpr/
Hello what 00212629582038 WhatsApp thanks
This is Great! Congratz
Very helpful, thanks for sharing
Very good seminar, as the others in this channel. But, if you allow me, I disagree with what you said Quiz 4, because I think that if the pipe is located in a high attenuation zone, the reflector just below this zone shouldn't be visible, but a strong reflector just below this zone is still visible. Actually, in quiz 9 you confirm this situation, when an attenuation zone begins, no reflector should be visible deeper.
measurement. Reflection of heaven is more when it rain than dried season because universal frequency had more chance to go back from under
vivid explanation
Very interesting thank you
Thanks for posting this, it's very informative.
Check out part 2 of this series - ua-cam.com/video/2DG3Z65O6yw/v-deo.html
Thanks for sharing
Thanks for this amizing video.
Can I have conversation whit you?
Thank you!
Contact us here if you need more information - www.sensoft.ca/contact-us-gpr
Thanks.
Tremendous
The expert can fool the sensor if the know technical data from sensor
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