[carpholeo]

Then the material you remove to make room for the weight is not enough to worry about?

[jim45498]

Originally Posted by carpholeo

Then the material you remove to make room for the weight is not enough to worry about?

The material removed in hardly anything in weight. At least nothing to worry about. It works for me.(My crankbaits are for Bass and may be small compared to some on here.)

[Vodkaman]

Jim. It is not just about the weight of the balsa that you drilled out, which is very small. I'll try to explain. Don't worry about the sums, it's the answers that tell the story, but I cannot think of a simpler way to explain about external and internal ballasting.

Consider a balsa body 80mm in length.

By suspending in a jug of water, it is found that 19.4gm of external lead was required to achieve suspension.

The body weighs 3.0gm.

The density of lead is 11.389gm/cm³.

From the above known information, it is now possible to calculate the volume of the body:

Volume of lead = weight / density = 19.4 / 11.389 = 1.7034cm³.

Total weight of body and external lead = 3.0 + 19.4 = 22.4gm.

Because this weight resulted in suspension, then the weight of water displaced by this combination of body and external lead (22.4gm).

1gm of water has a volume of 1cm³.

Therefore, the body + external lead has a volume of 22.4cm³.

Therefore the volume of the body alone = (volume of body + external lead) – (volume of external lead)

Volume of body = 22.4 – 1.7034 = 20.6966 = rounding off to 20.7cm³.

This means that the volume of the body and internal lead has a volume of 20.7cm³ and when suspended will weigh 20.7gm.

But by fitting the external lead to the inside of the lure, the ballasted body now weighs 19.4 + 3.0 = 22.4gm. This is 1.7gm too heavy.

This has not taken into consideration that you have drilled out a cavity for the lead and in doing so, removed buoyant material. This makes the situation slightly worse, as there is less balsa to support the lead.

In addition to this problem, we then slap a coat of D2T over it all, which has a density of 1.17gm/cm³, which is heavier than water.

If this method works for you, I have to assume that you are doing something extra or different, making an allowance somewhere along the line, or I am making a huge error in my thinking somewhere. No one can argue if you say it works for you, but these figures are pretty hard to argue against too.

The purpose of attaining neutral buoyancy is not to demonstrate to the fish how much effort we have put into catching him, but to stop the lure bobbing to the surface or sinking like a stone, when we pause the retrieve. If the lure rises or sinks slowly, surely we have achieved our goal.

The tolerance required for close to true suspension is approximately 1/100 of the weight of the lure. A lure weighing 20gm needs to be built to 0.2gm accuracy. With some planning, this is just about possible. To control suspension to a depth is probably ten times tighter, 1/1000 of the weight of the lure. This is just not possible without specialist equipment, in any case, you would be at the mercy of temperature fluctuations and water purity.

If the tolerance is widened still further to say ±0.5gm. the result would be a very slow rise or fall, which is surely acceptable for the purpose.

I don’t do these calculations every time I build a lure, I only do them once, in a spread sheet. Next time I just do the external lead suspension thingy in a jug of water, weigh the body and lead and type the weights into the spread sheet, which tells me how much lead and how deep to drill my hole. This gets me well within a tolerance of ±0.5gm and works for me.

As a post script. With the information gathered above, it is now possible to calculate the volume of the body alone, by subtracting the volume of external lead from the volume of body + external lead. Simple, no jars or pitchers AND much more accurate.

[mark poulson]

I would use cold water when trying to achieve suspension. For me, it's in winter/cold water, with suspended jerk bait fishing, that a lure that is actually neutrally buoyant is much more critical. Cold water is more dense, and any lure that suspends in cold water will slowly sink in warmer water. Generally that's not a problem for me, since I use a much more active jerk retrieve in warmer water, as the fish are much more active due to their increased body temperature.
A lure that suspends in warm water will slowly rise in colder water.
I really wouldn't sweat a small amount of rise or sink, though. Unless I'm fishing over flooded brush, where the lure can get snagged, a lure that sinks or rises slowly is still effective.
I guess we'll have to modify that old saying to read, "Close only counts in horseshoes, handgrenades, and suspending baits".

[clemmy]

I may have missed someone mentioning this, but couldn't you simply take the mass of lead of given volume, subtract the mass of same volume of applicable wood density to determine effectiveweight of ballast? Ie lead at 11340 kg/m^3 vs say, 14 lbs balsa at 224 kg/m^3?

[mark poulson]

Clemmy/Vodkaman,
I call foul! No fair talking in engineer talk. My computer doesn't have an automatic translate feature, and neither does my brain.
Are you saying you can take the volume of the lure, calculate weight of the water it would displace, figure out the weight of the lure minus the ballast holes, calculate how much lead (along with an allowance for the finish) is needed to reach the weight of the displaced water, to achieve neutral buoyancy, and go from there?

[jim45498]

Vodkaman, I read and re-read that explanation. I still don't know what I read. But, I have come to a conclusion. Your water in Liverpool is different from my water in Kentucky. I think your test tank has some Vodka mixed in the water. jk

[mark poulson]

Instead of Bourbon.

[Vodkaman]

I will try really hard to come up with a simple explanation. You should know by now that explaining is not my strong point.

Mark, correct.

[Shortlite]

VodkaMan, nothing immmersed in water will displace the same WEIGHT of water. It will displace the same VOLUME of water (Archimedes' principle: Eureka!!), so your calculations are a bit off, when you say that 22.4 g of lure displaces 22.4 g of water. Incorrect. You have to measure the volume of water displaced (using change in depth of water in the jug), to find the volume of the lure and external weight. Also, because the lure has a volume of X cubic cm, it will not weigh X g unless it it composed of water. Come on!! And volume is determined by dimensions: the lure has the same volume after you put the lead into it, it just now weighs more due to different density. It is a composite of two materials, in the same physical space. Volume of lure is unchanged.

A very buoyant lure (ie. floating) displaces less water because of its buoyancy. A sinking lure displaces more because it enters the water, and immerses its full body volume. So the principle does not apply perfectly to all lures. A suspender that floats very slowly is still a buoyant object, so your volumetric calculations have to be take with regard to the amount of body IMMERSED in the water. This is the only part of the lure that can displace water. Buoyancy counteracts weight, so only physical displacement can change water level.

The second part of reasoning with regards to internal vs external, removal of material, and changes in buoyancy due to this and paint and topcoating holds water (pardon the pun).