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512 Hz tank circuit question 
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#1
May614, 08:21 PM

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I'm trying to help a Plummer friend out. I was once a pretty good Navy Tech with good component level trouble shooting skills, but that is long in the past and I have forgotten all of the math required to figure out even the most simple circuits. He has a sewer camera with a locator for it. He has attached a small wire to the camera cable and the wire is connected to a 512 Hz generator called a locator, but it doesn't work under ground. I'm thinking that a tank circuit is needed to develop the signal near the camera head, but I cannot remember how to do the math to figure our what component values to use. I also think that the "wire" he used should be a pair so that I have a feedback path to the generator. This circuit also needs to be very small in size as it could get caught in the pipes and come loose, however it is very low power, so that shouldn't be a problem I think. Can anyone help me out here?



#2
May614, 11:50 PM

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Is he trying to locate the camera underground? Usually this type of a locator is used to locate pipes underground.



#3
May1214, 08:03 PM

P: 5

Sorry for the delay in response, had a hard time figuring out how to reply  I'm new to this. Anyway, he is trying to locate the camera inside the pipe. The camera goes down the pipe and when he finds damage to the pipe, he wants to locate the camera so he can mark where to dig to fix the problem.



#4
May1214, 08:17 PM

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512 Hz tank circuit question
Assuming the robotic camera is tethered back to the controller. There is an easier way .... mark the tether off in metres/feet/yards whichever is your favourite
That's the way I have commonly seen pipe cameras used Dave 


#5
May1214, 09:08 PM

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If the pipe is metal it could be pretty tricky to get an electrical signal out through it.
But an LC tank circuit will resonate at frequency 1/(2pi√LC) http://calculator.tutorvista.com/lc...alculator.html 


#6
May1314, 07:45 PM

P: 5

Dave, The problem with using a simple measurement for distance in the pipe is that pipes take a lot of twists and turns and ups and downs under ground. Jim, thanks for the formula, but as I stated before, I have not used this math in decades. I will follow the link you provided and see if I can figure it out. Also, I think that the reason they use such low frequency in these locators is due to the wave propagation through ground and pipes. I do know that they work because I have seen them work, I just don't know how to retrofit one onto an older camera that doesn't have it built into it. Thanks again.



#7
May1314, 07:52 PM

P: 5

Jim, I just tried the calculator from your link. It is similar to others I have tried. It requires me to know the value of capacitance and inductance to give the resonant frequency. My problem is that I know the frequency, and have no idea what capacitance or inductance to use, or haw to make an educated guess at it. Do you know of another calculator out there that works the problem from the frequency to the required components?



#8
May1314, 07:57 PM

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their locations and depths are mapped and that info is generally held with those that laid the pipes etc. Therefore the distance along the pipe to the problem was the only totally unknown factor and for the ease of installation, they are generally laid as level and straight as possible ( yes ... OK I realise plans may have been lost or accidently destroyed ) Dave 


#9
May1414, 07:01 AM

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with three terms it takes two of them to solve for the third one. And you only know frequency. So, it'll be necessary to pick either a capacitance or an inductance . If I plug in to that calculator 512 for frequency And 10e6 for capacitance (10uf) i get out 0.00966 henry, or 9.66 millihenry. So, a 9.66 millihenry choke would resonate at 512 hz with a 10 uf capacitor. It's probably more convenient to pick an available choke and make the necessary capacitance by paralleling smaller caps. That's because capacitors are cheaper than chokes. Especially in audio range where you need low resistance to give decent Q. Remember your basics from Navy : Q=X/R Something like this might work. http://www.digikey.com/productdetai...8386ND/774926 http://www.bourns.com/data/global/pdfs/1140_series.pdf it's ten millihenries which resonates at 512hz with 9.66 microfarads, real close to a 5.0 and a 4.7 in parallel. It's about an inch across so should fit down a pipe. The capacitors could be some distance away connected by a twisted pair, preferably shielded. At that frequency you'll have X = about 31 ohms, the choke is 2.76Ω so Q is only a little over 10, not great but you might get away with it. Higher current chokes (or audio crossover type from a speaker store) will do better but cost more. ..... You've not said how your friend's gizmo excites the tank circuit ...but there's how to use the calculator, which was your question. Play with it a while or use your pocket calculator. Good luck , and keep us posted. EDIT Hold on a second  Pocket calculator? What was i thinking  you're an old Navy guy. Dust off that Slide Rule !!!! old jim 


#10
May1414, 09:13 PM

P: 5

Jim, it is nice to know what the more expensive of the two components would be. I will work the problem from that angle. Your help is much appreciated! I will let you know what I come up with and how it works. Thanks again.



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