Crankbaits and bottom bouncers

[tacklejunkie]

I wouldn't run a crank behind a bouncer and have tried floating rapalas in the past but I think other weighting systems work better.

IMO..BB are made more for trolling spinners

[cylinder]

An 1/8 jig would dive to an unlimited depth if it wasn't for line resistance/line bouyancy. It will always be denser than water, at least in the realm of fresh water fishing.

You just have much better feel for the heavier jig in deep water. And the line resistance has more effect at these depths, thus heavier jigs to counteract this.

I still maintain bouyancy of a crankbait has little effect on depth reached if you are trolling at a minimum speed.

I agree completely.

[TooTallTom]

BobT is right in theory, if the water could get dense enough it would make lead float. But, I don't think that water exists in that state in nature. Also, lead's density is constant, so bigger jigs wouldn't sink any better than little ones in deeper water.

I also think that it's necessary to have a science g2g for the forum members. Does anyone have a lot of lab jackets laying around? BobbyMalone, I'm looking at you... Lab jackets, flipflops, loud sunglasses, and tackleboxes. I can see it, and it looks beautiful.

[BobT]

BobT is right in theory, if the water could get dense enough it would make lead float. But, I don't think that water exists in that state in nature. Also, lead's density is constant, so bigger jigs wouldn't sink any better than little ones in deeper water.

In theory wouldn't this be true if the size of the lead ball grew in direct proportion to the mass? I'm getting into areas where I very likely am using incorrect terminology but what I mean is this.

An object floats or suspends in a liquid when the amount of liquid (in weight) displaced by the object is equal to or greater than the weight of the object. So, if you place a 1/8oz. ball of lead onto the surface of a liquid and it floats, that doesn't mean a 1/4oz ball will also float unless it can displace the equivalent amount of water proportional to its weight. In theory (I think) the 1/4oz. ball would have to be twice the diameter of the 1/8oz. ball but in reality thia is not the case. Because the 1/4oz. jig is not twice the diameter of the 1/8oz. jig it doesn’t displace twice the water and therefore it is less buoyant and will sink at a faster rate.

[TooTallTom]

So, if you place a 1/8oz. ball of lead onto the surface of a liquid and it floats, that doesn't mean a 1/4oz ball will also float unless it can displace the equivalent amount of water proportional to its weight. In theory (I think) the 1/4oz. ball would have to be twice the diameter of the 1/8oz. ball but in reality thia is not the case. Because the 1/4oz. jig is not twice the diameter of the 1/8oz. jig it doesn’t displace twice the water and therefore it is less buoyant and will sink at a faster rate.

BobT, density = mass/ volume. You're thinking of mass/ area. Volume increases faster than area. Basically, size increases in direct proportion to mass, but it increases in all three dimensions, and diameter is only two-dimensional the way you're thinking of it.

Let's do a thought experiment. The equations for rectangles are simpler than for spheres, so let's work with those instead: Take a 2x2 square, and calculate the square footage. (4 square feet.) Take a 2x2x2 cube and multiply the volume. (8 cubic feet.) Now do the same with 3 feet in each direction. (9 square feet versus 27 cubic feet.) The volume of the cube expands much more rapidly than the area of the square with a similar dimensional increase. So, if we're concerned with weight, a 1/4 oz jig doesn't need to have twice the diameter of an 1/8 oz jig.

Since the density of a given material is constant, it doesn't matter how large the object made of that material is. With a given density, a solid object (of lead in this case) will either float or sink in a liquid.

[solbes]

An object floats or suspends in a liquid when the amount of liquid (in weight) displaced by the object is equal to or greater than the weight of the object. So, if you place a 1/8oz. ball of lead onto the surface of a liquid and it floats, that doesn't mean a 1/4oz ball will also float unless it can displace the equivalent amount of water proportional to its weight. In theory (I think) the 1/4oz. ball would have to be twice the diameter of the 1/8oz. ball but in reality thia is not the case. Because the 1/4oz. jig is not twice the diameter of the 1/8oz. jig it doesn’t displace twice the water and therefore it is less buoyant and will sink at a faster rate.

Well we are having fun, and I agree we should be out on the water with a rod, reel, and a beer debating these things instead. The density of lead at the same conditions does not vary between a 1/8 oz jig and a 1/4 oz jig. It gets so much easier when we think of density, mass over volume. I wouldn't make a comparison based on diameter, but rather volume. The volume of a sphere (simplifying the jig shape here) is 4/3 * pi * r^3. Lets say a jig has a radius of 1/8" or .125". Volume of that jig is 0.0082 in^3. Double the volume for the 1/4 oz jig and the resultant radius is only 0.157".

Water will never get dense enough to make lead float. Density of lead is 11340 kg/m3. See pic below for density of water vs temp and pressure. Even at 200 bars (or approx atm), it's 1/10 the density of lead. Lead on ice does not count either, that's a phase change

[fivebucks]

I can get pretty dense sometime

[Greg Clusiau]

I've tried pulling cranks with bottom-bouncers and have had very limited success this way (other than the Rainy River where plenty of current is present).

[TravP]

Skip the bottom bouncers and buy some snap weights.

[Greg Clusiau]

"ditto" what Travis said, or else use lead core.

[TravP]

Leadcore is my #1 choice. The reason I say snapweights is because there cheap. Not everybody wants to spend the money on LC.

[goose89]

Leadcore is my #1 choice. The reason I say snapweights is because there cheap. Not everybody wants to spend the money on LC.

Leaving new rods and reels out of the equation,(you could argue you'd need both if trolling lead core or snap weights) lead core is way cheaper than snap weights. $35 (for a set, I haven't seen individuals sold) vs $15-20.

BB's for sure the cheapest route, and they do work. Probably more so on rivers. I'd go with a 3 way (more versatile) if open water cranking and leadcore / snapweights weren't in my arsenal.

[TravP]

You can get 2 offshore releases for like ten bucks. Put on a bell sinker and your snapweight is created.

[hanson]

You can get 2 offshore releases for like ten bucks. Put on a bell sinker and your snapweight is created.

Thats exactly what I was going to say.

[goose89]

You can get 2 offshore releases for like ten bucks. Put on a bell sinker and your snapweight is created.

good point! I stand corrected.

[buzbunni]

Well we are having fun, and I agree we should be out on the water with a rod, reel, and a beer debating these things instead. The density of lead at the same conditions does not vary between a 1/8 oz jig and a 1/4 oz jig. It gets so much easier when we think of density, mass over volume. I wouldn't make a comparison based on diameter, but rather volume. The volume of a sphere (simplifying the jig shape here) is 4/3 * pi * r^3. Lets say a jig has a radius of 1/8" or .125". Volume of that jig is 0.0082 in^3. Double the volume for the 1/4 oz jig and the resultant radius is only 0.157".

Water will never get dense enough to make lead float. Density of lead is 11340 kg/m3. See pic below for density of water vs temp and pressure. Even at 200 bars (or approx atm), it's 1/10 the density of lead. Lead on ice does not count either, that's a phase change

I don't quite get the part where...

I mean...if you ignore the variable in the equation, then you.....

Wait a minute.

What I'm tryin to say is this....

Uh.

OK. I admit it. I'm dumb.

[solbes]

Sorry, got a little carried away there. Suggests that all the beer I drank in college didn't completely wipe out brain cells used while taking math and engineering classes

Bottom line, I think experience has shown that a crankbait will continue to dive beneath a bottom bouncer. Even if it is a bouyant floater style lure.

[chris63]

If you switch to Whiskey instead of Beer I bet you make lead float in water at any pressure!c63

[bigbucks]

The presentation will still work. Just like other weights you don't have to let a bottom bouncer drag the bottom. Having said that there are undoubtedly easier ways to do it, as getting the amount of line right can certainly be a tricky situation...