How do I connect a variable capacitor to an inductor in a crystal radio circuit?

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mearvk
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Typical VC: http://goo.gl/Z4Dbg

I have that guy there or one very similar. My question is how does one hook this up to the inductor? Forget, for now, the rest of the circuit. How do these two piece fit together?

The bottom feet (there are two on each side) I assume take the positive terminal of the inductor and the ground gets soldered to the chassis of the variable capacitor? Then would I connect any power source the same way? Positive to a free foot and negative to the chassis? Or would I hook up the positive to one side of the inductor and negative to the other side? And then to get output from the circuit what would I do?

I've watched a ton of videos and they all neglect how to wire these two guys together.

Thanks.

Crystal radio diagram: http://goo.gl/bZlpM
 
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There are a number of connection points on each side of the capacitor.

To make a parallel tuned circuit, the inductor has to go across the capacitor. The actual point of connection is mostly related to convenience.

You could have one side of the inductor and one side of the capacitor grounded because then the two would be joined together.

You would not normally put a power source directly across a variable capacitor.
 
I'm not sure if this world is intentionally obtuse or what.

Very clearly the inductor goes across the VC. This is clear. What I don't know is how. You say it's a matter of convenience. So I can connect any old way?

I can connect the inductor across the feet, neglecting the VC chassis completely.

I can connect the inductor solely to the chassis of the VC, ignoring the feet connections.

I can connect the inductor from the feet to the chassis.

Now that I've gone out of my way to enumerate the possibilities which is correct? Likely that they're all not correct. And again, once this part is correct what does the output look like? Where is it connected?
 
None of those is correct.

The capacitor has two sets of plates. One set is stationary and the other set is movable,

The movable set is connected to the body of the the capacitor and also to the feet. So, one end of the inductor can connect to any of these.

The other end of the inductor has to connect to the stationary plates of the capacitor which are the top ones in the picture in the first post.
 
Your first link is begin flagged as a known malicious website by my Norton antivirus.
Can you attach the image directly?
 
DQa1C.jpg


The stationary plates appear connected to the feet via soldering. The wood serves to isolate the stationary metal from the chassis which appears to connect to the moving piece via the bronze looking rod. So we're clear.
 
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Looks good. So for future reference it looks like any of the feet will do for your positive terminal and the chassis will do as your negative (ground) terminal.