Finding Neutral Buoyancy for 1-oz Fishing Lure

http://mackinawfishing.com/downrigging.htm

jgranger567,

You say you have a 1 ounce floating fishing lure. I presume that is the weight of the lure. What is it's inherent buoyancy? (i.e., how much "lift" does it have? )

Evaluating your question empirically, I do not think that there is a single isolated solution to your question, and offhand, I do not know the formula you would need to use. However, I do know that you need to determine a few things, which are probably not going to be "constants":

The specific gravity of the water (varies with salinity & other dissolved minerals);

The water temperature at your desired depth (15 feet)...

As a "quick and dirty" method, how about hooking the lure to a small spring scale, and submerging the rig to your desired depth; the upward buoyant force should be displayed on the scale (which should equal the necessary weight to "stay in the zone") ...

================================================================

Just my $0.02...

As far as I understand your question there is a problem:

the lure has a constant buoyancy when fully in water what you need is a variable buoyancy which will be compensated by a weight when the depth of 15 ft is reached.

Think how a submarine works and use same principle of a volume which shrinks when pressure increases. I presume you should apply this principle.

I see a problem with what you ask. If you find the point of neutral buoyancy for your lure. How would you get it to the 15 foot depth? If it sinks to 15 foot then you have no neutral buoyancy. If as a sub does it would have to have a propulsion system and controls. Which would be a pretty pricey fishing lure.

A better approach for me is to ask what you fishing for and how you are pressenting this lure trolling, casting or jigging?