Here’s a defensible theory model for how long a physoclistous black bass would need to “equalize enough” to get back down to 10 m (33 ft) on its own.
The physics first (sets the size of the problem)
At 10 m, absolute pressure is ~2 ATA. At the surface it’s ~1 ATA.
By Boyle’s law (good approximation here), if the bass is reeled from 10 m → surface and cannot vent quickly:
- Swim bladder volume at surface ≈ 2× its volume at 10 m
- The “excess” gas volume relative to being neutral at 10 m is roughly one 10 m-equivalent bladder volume (call it ΔV ≈ V10)
So the question becomes: How long to remove enough gas (ΔV) that the fish is no longer too buoyant to initiate descent?
What “enough” equalization actually means
A bass does not need perfect neutral buoyancy at the surface to go back down. It needs buoyancy low enough that it can kick down a few meters. Once it gets even to 2–3 m, pressure increases and the bladder compresses, which rapidly reduces buoyancy, making further descent easier (a positive feedback). So the time-to-return is usually dominated by one of two regimes:
Regime A — “Immediate descent” (seconds)
If the fish can simply overpower the excess buoyancy and get down a couple meters:
- Time to get back down: seconds to <1 minute
- Equalization required: essentially none (recompression does the work)
This is why many bass released from ~10 m look okay: they just swim down and pressure helps them.
Regime B — “Needs gas resorption first” (tens of minutes to hours)
If the fish is so buoyant it can’t initiate descent (or keeps popping back up), then it must resorb gas via blood → gills before it can dive. That time depends mainly on:
- How big ΔV is (how much “extra” gas)
- How fast resorption is (species/temperature/stress dependent)
A rough quantitative estimate (order-of-magnitude)
We need plausible swim bladder volumes. Many physoclists have swim bladder volumes on the order of ~2–7% of body volume (varies a lot by species and condition). For a bass:
- 1.0 kg bass ≈ 1.0 L body volume
- If bladder at 10 m is ~3–6% of body volume → V10 ≈ 30–60 mL
- At surface it doubles → 60–120 mL
- “Excess” to lose to behave like it did at 10 m: ΔV ≈ 30–60 mL
Now, what resorption rate? We don’t have black-bass-specific numbers here, but physoclist resorption rates reported in other teleosts tend to be on the order of ~0.05 to 0.3 mL/kg/min (ballpark range, varies widely with physiology/temperature/stress). For a 1 kg bass, that’s 0.05–0.3 mL/min.
To remove 30–60 mL:
- Fast end: 30 mL / 0.3 ≈ 100 min (~1.5 hr)
- Slow end: 60 mL / 0.05 ≈ 1200 min (~20 hr)
That’s full correction to the original 10 m gas load equivalent. But to merely be able to dive, it likely needs to remove only a fraction of ΔV (enough that sustained tail-kicks beat buoyancy), so divide by ~2–5:
Practical “able to get down” estimate (if it can’t descend immediately)
- ~20 minutes to ~4 hours is a reasonable theoretical window for many fish
- Could be longer (6–24+ hours) in colder water, exhausted fish, or if the fish is badly overinflated/injured
The key behavioral discriminator
- If it can get its head down and reach ~2–3 m, it will often “zip” the rest of the way down via recompression. That looks like instant recovery.
- If it can’t initiate descent at all, you’re likely in the hours regime.
