M: Please explain to me the scalability issues and just general details of what is involved in using a tidal harness for power besides the basic idea that I've got of the moon lifts the water and then gravity turns a turbine as it drains out. Because on the surface, it seems like that should be an easy one to pull off in terms of small scale, but that's without having any numbers to work with.
G: TL;DR:
Tidal power is predictable but energy-sparse per square
meter, mechanically hostile, and hard to scale without either
huge civil works (barrages/lagoons) or lots of
underwater hardware (tidal-stream turbines). Small scale is
possible, but it’s usually educational or niche, not
economically self-sustaining.
（日本語）要点（ようてん）：潮汐（ちょうせき）発電（はつでん）は予測（よそく）できるが、単位面積（たんいめんせき）あたりのエネルギーが小さい。— Key point: tidal power is predictable, but low in energy density.
This is closest to your mental model: fill a basin at high tide, drain it through turbines at low tide.
image_group{"query":["tidal barrage diagram","tidal lagoon power plant","La Rance tidal power plant France"]}
Energy available scales with:
A simplified energy estimate per tide: [ E \approx \tfrac{1}{2},\rho,g,A,h^2 ]
Why this matters:
Doubling tidal range gives 4× energy. That’s why only a
few locations are viable.
Real example:
Scalability wall:
You can’t “miniaturize” geography. Small basins → trivial power.
Think underwater wind turbines driven by fast tidal currents.
image_group{"query":["tidal stream turbine underwater","tidal current turbine array","underwater tidal turbine close up"]}
Power scales with: [ P \propto \rho,A,v^3 ]
Where v = water speed.
Key reality check:
Pros:
Cons:
Even with seawater’s high density, slow movement kills power output.
You don’t get the cost curve benefits solar/wind enjoy.
Saltwater causes:
Every maintenance trip may require:
Tides are:
To be useful at grid scale, you still need:
So tidal doesn’t replace baseload cleanly—it supplements.
Barrages especially:
This leads to:
Notably, tidal power succeeds only where geography is doing most of the work for you.
（日本語）結論（けつろん）：地形（ちけい）が強力（きょうりょく）でないと、潮汐発電は小規模では割（わ）に合わない。— Conclusion: without strong geography, small-scale tidal isn’t worth it.
Your intuition is right conceptually, but the numbers are unforgiving:
Tidal is elegant, reliable, and real—but only when nature already stacked the deck.
M: When I was talking small scale, I was talking very small scale. Like, personal solar panel scale. Maybe a little bit larger, given the situation, but... Like, let's say I had a house on Virginia Beach, and... on the water, in a way that allowed me to work with it. I know that the beach is supposed to be public property. So, just roll with me on the legality of it. I'm using it for an example, not as a legal argument. Let's say that I had that situation, and... I don't know, let's say it was a... area 20 feet across. And I wanted to just basically... dig a trough across half of it. In a way that let water into some sort of... marine plastic... uh... marine plastic trough. Let it escape like a dam, and capture energy... when the tide goes out. Now, let's go with the perfect version of that. Like, there's no mechanical issues. The materials are perfect, and every ounce of energy can be captured. How would that scale to... a single solar panel? In terms of energy produced, if everything is optimal for the tidal harness.
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G: Let’s do the perfect-physics upper bound for your “tiny tidal trough” and compare it to “one solar panel.”
If your geometry is different (e.g., 20-ft diameter semicircle, or 20×20 full basin), the result scales linearly with area.
（日本語）前提（ぜんてい）：潮（しお）の干満差（かんまんさ）と、ため池（いけ）の面積（めんせき）で決（き）まる。— Assumption: it’s set by tidal range and basin area.
A standard idealized estimate for tidal-range energy per tide is:
[ E \approx \tfrac{1}{2}\rho g A h^2 ]
Where:
Per tide:
Per day (2 drains/day):
That’s the perfect upper bound for that footprint, at mean tidal range.
Virginia Beach average solar resource is around ~5.2
kWh/m²/day (a “peak sun hours” style figure).
citeturn0search2turn0search13
A common “single panel” reference is ~400 W. Under
~5–5.5 peak sun hours, that’s roughly **~2.0–2.2 kWh/day** in good
conditions. citeturn0search3
So:
Ratio:
Result: your ideal tiny tidal setup is about ~40× less daily energy than a single modern solar panel.
（日本語）結論（けつろん）：この大（おお）きさだと、太陽光（たいようこう）パネル1枚（まい）のほうが、だいたい40倍（ばい）くらい強（つよ）い。— Conclusion: at this scale, one solar panel is roughly ~40× stronger.
Example: if you somehow got 6 ft range instead of ~3.3 ft, energy goes up by about ((6/3.3)^2 \approx 3.3×). That would turn 52 Wh/day → ~170 Wh/day, still far under ~2 kWh/day.
At “personal solar panel scale,” the limiting factor is simple:
there just isn’t much gravitational potential energy in a few
tens of square meters of water moving up/down a meter or so.
You need either:
M: Ouch. The only mitigating factor at that absolute point, the optimized point, is that you might be able to wash things in the trough. No, there are a couple other things, but your point is taken. Let's try and put it into real-world terms, though. What would be able to run off of that 0.05 kilowatt-hours a day?
G: 0.05 kWh/day is 50 Wh/day.
That’s not “run a thing continuously” energy. It’s “tiny gadget budget.” In real-world terms, after storage/inverter losses you’d probably net more like 25–40 Wh/day, but I’ll answer using the ideal 50 Wh/day you asked about.
（日本語）これは小（ちい）さい電気（でんき）だ。— This is a small amount of electricity.
To run something 24/7, divide by 24:
So: anything that averages ~2 W can run all day.
Examples:
This is where it’s more intuitive.
If an item draws P watts, then:
Examples:
If you insisted on building something like this, the “makes sense” category is:
If you tell me what you want it to feel like (lighting, charging, “keep a network node alive,” run a small pump, etc.), I can map it to an actual battery/inverter/storage setup that matches this budget.
M: So if someone were to use this for some sort of hypothetical energy solution in something fictional, you know, near-future fiction, it's not like someone would ever plausibly be able to spend millions and set up a small house with a tidal basin harness. But you could conceivably have someone who lived off the land and whose only connection to the outside world was a phone set up something like this. No, I'm not trying to come up with anything. I'm just, that's where my mind went as I tried to file away this little bit of information. And that's in optimal conditions, right? So there's no way to hand wave away a significant... There's no way to hand wave away something that's basically a water slide scale trough somehow powering an entire neighborhood or something like that, no matter how you play with the physics of the situation. The energy has an upper bound.
G: Yep. Your read is basically correct.
If the character’s electrical life is:
Even then, the realistic version is “keeps small electronics topped up,” not “always plenty of power.”
For tidal-range harvesting (your basin/trough model), there’s a hard ceiling set by:
[ E \approx \tfrac{1}{2}\rho g A h^2 ]
You can’t “clever engineering” your way past that without changing A (area) or h (head/tidal range). So a tiny footprint can’t yield big energy. That’s not an engineering limitation; it’s the available gravitational potential energy.
If you want “tidal power that matters,” you have to change the premise:
So you can write “tidal powers a town,” but only if the setting includes one of those macro-scale enabling conditions. The “toy basin in the sand” can’t be scaled up by vibes.
（日本語）上限（じょうげん）は変（か）わらない。面積（めんせき）か高（たか）さを増（ふ）やさないと、電気（でんき）は増（ふ）えない。— The upper bound doesn’t change: without more area or more head, you don’t get more power.
“Tides are reliable, but they’re not dense. You don’t defeat that with cleverness—you defeat it with geography.”
If you want, give me the fictional coastline constraints (tidal range, whether there’s an inlet/estuary, what “neighborhood” means), and I’ll tell you what size of basin/turbine array would be barely plausible.
M: Yeah, when I say hand wave, I'm thinking about how, like, in places that try to be plausibly futuristic like Star Trek, they'll sometimes take the absolute potential energy of something and act as though they managed to harness the entirety of it. With a tidal harness, there's really nothing to do like that. It's a limited resource. Okay, related but separate idea. Say, hypothetically, a natural tidal harness setup formed. Like something about the geography created a perfect situation where, you know, I'm going to say a city block of coastal area in terms of, well, area, created a situation where the walls are something secure enough that we can treat it as though it were that marine-safe plastic, and the tide fills it up somehow by like five meters every day. Something about the setup creates a situation in which you've got a city block five meters deep of water. But with the catch being that there is actually no physical escape for the water. So, it gets refilled every day, but doesn't actually drain. Now, that's just the setup. Assume all that to be correct. Now, let's say you managed to drill a hole to allow drainage and allow you to capture this as a tidal harness. You said before that the energy was limited by that equation involving gravity. In this situation though, if the hole weren't very large, wouldn't that become moot? And it would become related to the actual size of the hole and the pressure involved in all that, right? If so, what would the equation turn into for how much energy you could get out of that setup? And if not, what am I thinking wrong about or being unclear about in this situation?
G: You’re mixing up energy vs power. The hole size changes how fast you can get the energy out (power), not how much you can get total (energy), as long as you eventually let the basin drop.
Even with a tiny hole, the total extractable energy per fill-and-drain cycle is still capped by the gravitational potential energy of the water that drops.
That cap doesn’t care whether you drain it in 10 minutes or 10 hours.
You described:
Two cases:
Then total energy per cycle is still:
[ E \approx \tfrac{1}{2}\rho g A h^2 ]
Same equation as before. Hole size only changes how quickly you realize that (E).
Then you don’t get a continuing energy source from “stored water,” because once pressures equalize (or the basin stops changing level), the system stops doing work. To keep producing, you need ongoing change: inflow at high tide and outflow at low tide (or some other continuously maintained head difference).
That’s the power equation, not the energy equation.
If the water surface is (H(t)) meters above the outlet/turbine, then the ideal hydraulic power is:
[ P(t) = \rho g Q(t) H(t) ]
Where (Q(t)) is the volumetric flow rate through the hole/turbine.
Ideal-ish form:
[ Q(t) = C_d,a,\sqrt{2gH(t)} ]
Combine them:
[ P(t) = \rho g \Big(C_d a \sqrt{2gH(t)}\Big) H(t) = \rho g C_d a \sqrt{2g},H(t)^{3/2} ]
So yes: power scales with hole area and (H^{3/2}).
But the total energy you can extract as it drains from (H=h) down to 0 is the time integral of power, and it collapses back to the same gravitational-energy cap:
[ E = \int P(t),dt = \tfrac{1}{2}\rho g A h^2 ]
(as long as you’re extracting through a turbine with no losses and the basin surface area is roughly constant with depth).
A tiny hole gives you a weak trickle for a long time. A big hole gives you a gush for a short time. Same “battery,” different “discharge rate.”
If it truly refills every day while never draining, that implies something physically odd (it would overflow or reach equilibrium with the sea). What you probably mean is:
That creates a one-way ratchet until it hits a new equilibrium (overflow, seepage, or equalization). Once it stops changing, power stops.
To have continuous daily harvesting, you need a daily head difference to cycle: fill → drain → fill → drain.
If you want to pin it down with numbers, tell me:
…and I can give you (1) energy per cycle, (2) peak power, and (3) how long it would take to drain.
M: Okay, that's sort of why I asked it the way I did. It's because I knew that I would probably phrase something badly. Yeah, I get what you mean about the power versus energy difference. What I was trying to point out, I used specifics in my example just so I had something tangible instead of just using vague terminology. But the point is supposed to be that once the amount of liquid becomes a irrelevant point, in this situation a naturally occurring reservoir that cannot physically empty during a single cycle of the tides, I'm aware that the potential energy still would be the same in terms of what could be extracted. I was looking for the idea encompassed by the right terminology that says that if you've got a hole that's like one foot by one foot in the appropriate point, what is the maximum energy that could be extracted, and what equation would relate to that situation. And again, we're talking about an idealized hypothetical. So yeah, I looked at the idea that if it didn't drain, there would be a collection of sediment or something like that. For this hypothetical, we are working with the same concept as the optimized trough, where that's beside the point in terms of answering my question. In practical terms, it would be every point, because eventually it would have gotten filled up and there wouldn't be something like that there anyway. I mean, I could come up with a scenario like, I don't know. The sediment had filled it, but the people trying to get the energy out realized that what it contained was nothing but sediment. And so they cleared it out and drilled a hole so that it's functioning the way I described. The story isn't the point, except in that it sets up this hypothetical connection to how much energy could be extracted from the situation in a single tidal cycle. This is the pain in the ass with asking questions even with you. I can't ask you to give me less information because often what I need in order to ask a question correctly is about halfway through the entire response. But that means that the other half either goes unread or it turns into a giant fork conversation where three or four ideas are being dealt with at once. When I've got the energy for it, it's kind of fun. When I'm tired but it's mandatory, then it becomes a chore.
G: Got it. Thank you for the patience note — I’ll stay tight and surgical here.
You were not asking for more story or more cases. You were asking:
“If I have a fixed hole (say 1 ft × 1 ft) at a given head, what is the maximum energy I can extract per tidal cycle, and what is the right concept / equation that governs that?”
Here is the clean answer.
What you want is hydraulic power through an orifice under a fixed head, and its time-integrated energy.
The key terms are:
The hole size never sets the energy limit.
It sets the power limit (rate of extraction).
[ Q = C_d , a , \sqrt{2 g H} ]
Where:
This is what your intuition was circling.
[ P = \rho g Q H ]
Substitute (Q):
[ P = \rho g a \sqrt{2g} , H^{3/2} ]
This is the equation where hole size matters.
Bigger hole → more power right now.
This is the part that does not change, no matter how small the hole is:
[ E = \int P , dt = \tfrac{1}{2} \rho g A h^2 ]
Where:
This is the upper bound you cannot hand-wave past.
Same total energy, as long as the water level drops by the same amount.
The hole never creates energy.
It only controls how fast you spend the stored gravitational
energy.
（日本語）穴（あな）の大（おお）きさは速（はや）さを決（き）める。エネルギーの上限（じょうげん）は変（か）わらない。
— Hole size sets the rate, not the total energy.
“In tidal systems, geometry sets the battery size; the outlet only sets the discharge rate.”
You weren’t wrong — you were just one layer early in the stack.
M: Yes, thanks. Thank you for the clear reframing. Up until now, the potential energy has essentially been equivocated with the power extraction because of the optimal settings of the thought experiment. It's the reason why I had the curiosity, is because that equivocation leads to this kind of exploration of where the differences actually lie. Okay, so what does it do to the equation if the head is consistently dropping, but just never gets to the point where, you know, the flow is zero? Like in the hypothetical, the numbers might be different, but assume the specific numbers I gave might not work this way, but just assume that there is constant flow and that the head never gets below one meter above the hole. Given that a tidal cycle is relatively fixed, we probably could reverse engineer the actual height of the basin just from that one single static number.
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G: Right — now you’re asking about the time-varying head case where you don’t fully drain during the available window, so (H(t)) stays above some minimum (H_{\min}=1,\text{m}).
Two clean pieces: (A) what happens to the energy equation, and (B) how to reverse engineer basin size from “it never drops below 1 m.”
The “battery” equation (E=\tfrac12 \rho g A h^2) assumed you drained from (H=h) down to (0).
If you only drain from (H_0) down to (H_{\min}) (and stop there, or the tide turns and refills), then the maximum extractable energy from the basin for that partial drop is:
[ E_{\text{cycle}} = \tfrac{1}{2}\rho g A\left(H_0^2 - H_{\min}^2\right) ]
So the effect of “never gets below 1 m” is: you harvested less total energy, but you maintained higher power the whole time (because head stayed higher).
（日本語）水位（すいい）が下（さ）がり切（き）らないと、合計（ごうけい）エネルギーは減（へ）る。— If it doesn’t drain all the way down, total energy harvested is smaller.
If we stay in the idealized “hole to atmosphere / tailwater at the outlet” model, the dynamics are:
Orifice flow: [ Q(H)= a\sqrt{2gH} \quad (\text{ideal } C_d=1) ]
Basin level drop: [ A,\frac{dH}{dt} = -Q(H) ]
So: [ \frac{dH}{dt} = -\frac{a}{A}\sqrt{2gH} ]
This integrates to:
[ \sqrt{H(t)}=\sqrt{H_0}-\frac{a\sqrt{2g}}{2A},t ]
Now impose your condition: after one “draining window” of length (T) (roughly the time you allow outflow before the tide turns / refills), the head is still (H(T)=H_{\min}\ge 1,\text{m}).
Solve for basin area (A):
[ A=\frac{a\sqrt{2g},T}{2\left(\sqrt{H_0}-\sqrt{H_{\min}}\right)} ]
That’s the “reverse engineer the basin” equation you were gesturing at.
Then with (g=9.81), (\sqrt{2g}\approx 4.43), so:
[ A \approx \frac{0.0929\cdot 4.43 \cdot T}{2(\sqrt{H_0}-\sqrt{H_{\min}})} = \frac{0.206,T}{(\sqrt{H_0}-\sqrt{H_{\min}})} ]
(with (T) in seconds, (A) in m²)
So if you tell me your notional (H_0) and the window (T) you mean by “a tidal cycle” (often you really mean ~6 hours of “net draining”), you can directly back out a basin area that guarantees it never drops below 1 m.
In real tide-driven setups, the head is usually basin level minus sea level: [ H(t) = H_{\text{basin}}(t) - H_{\text{sea}}(t) ] So “it never drops below 1 m” can happen either because the basin is huge or because the sea level rises again before the basin finishes draining. Same math, but you’re tracking (H(t)) relative to the changing tailwater.
If you want, give me just two numbers—your assumed (H_0) and your drain window (T)—and I’ll compute (A) and the per-cycle energy using the clean formulas above.