Cultivar plantas en una cueva submarina para lograr oxígeno vía fotosíntesis

AAAAAAAAAAAAAAHHHHHHH!!!!

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Pocos lugares mejores para sobrevivir al caos social que una cueva submarina, siempre y cuando nadie sepa que la habitas. Para ello, no puedes subir a la superficie constantemente para llenar botellas de oxígeno o pilas, lo ideal es habitarla durante largos periodos de tiempo, saliendo únicamente a pescar submarinamente.

¿De cuánto oxígeno se dispone en aquellas cuevas que tienen mucha vegetación? Como me espero que la respuesta a la anterior cuestión sea que no hay apenas vegetación porque no hay luz, me pregunto: ¿Por qué no usar un complejo sistema de espejos concatenados para reflejar la luz de la superficie y conseguir fotosíntesis? ¿Se pierde mucha luz con cada reflejo? Edito: reflejan casi toda la luz que choca contra su superficie.

La tierra, abono y semillas se llevarían al principio, ahora, antes de que estallen los acontecimientos, y se iría cultivando todo. ¿En medio año se podría disponer ya de un buen nivel de oxígeno?


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Y la luz para la fotosíntesis? Con espejos ni de coña vas a iluminarlas para eso... Con lámparas y conexión al vecino como los de la Cañada?
He encotrado esto, aunque está en inglés:

Is it possible to light up a room covered with mirrors with just one bulb in it assuming that no space is left without a mirror?

It’s an interesting thought experiment. Let’s see if we can work an estimate.

Let’s assume a cubical room of side S measured in light-seconds - this unit is convenient as it avoid tedious tracking of units. Suppose a bulb emitting P photons per second. Then while a photon from the bulb is traveling to the wall for the first time, the bulb will put out about PxS photons, which will then all be in the room. Thanks to our choice of units, that time will be S seconds. So if the walls are black, that will be the number of photons in the room at one time, PxS.

Now if the walls are reflective, in the next S seconds, the bulb will emit another PxS photons, and the walls will reflect PxSxR photons, where R is the reflectivity of the mirrors. So in after 2S seconds, there will be PxS + PxSxR photons in the room. In the next time interval, after 3S seconds, there will be PxS + PxSxR + PxSxR2 photons in the room, where R2 is R squared. You can see where this is going. After NxS seconds, there will be:

PxSx(1 + R + R2 + R3 + …….+ RN + ….)

So we have an interesting result. The room will be brighter by a factor equal to the sum of an infinite series:

(1 + R + R2 + R3 + …….+ RN + ….)

Also, we find that the factor does not depend on the size of the room.

As long as R is between 0 and anything a little less than 1.0, this series converges. In fact it is an infinite geometric power series and the sum is 1/(1-R)

The reflectivity of ordinary second surface mirrors is typically 0.8 - 0.85. If we take 0.8, the series sums to 5.0

So, after a very short time, the number of photons in the room with mirrors is 5 times greater than in the room with black walls. Not a bad gain. Also, with the light coming from all directions, the illumination should be ultra uniform.

Fresh, pure silver first surface mirrors have a reflectivity near 0.99, so the room would be 100 times brighter, or the equivalent of having 100 bulbs in a room with black walls. That might get a little warm.

There are some assumptions in here. I approximated that 1 = 1/2 (see if you can find where.) That’s because I treated this as a Fermi problem, where we are interested in a meaningful estimate, not in a precise calculation. And of course we assumed that none of the photons are absorbed by people or things (or the bulb) in the room.

One of the pleasures of physics is the ability to work out “what if” questions for physical phenomena.
 
He encotrado esto, aunque está en inglés:

Is it possible to light up a room covered with mirrors with just one bulb in it assuming that no space is left without a mirror?

It’s an interesting thought experiment. Let’s see if we can work an estimate.

Let’s assume a cubical room of side S measured in light-seconds - this unit is convenient as it avoid tedious tracking of units. Suppose a bulb emitting P photons per second. Then while a photon from the bulb is traveling to the wall for the first time, the bulb will put out about PxS photons, which will then all be in the room. Thanks to our choice of units, that time will be S seconds. So if the walls are black, that will be the number of photons in the room at one time, PxS.

Now if the walls are reflective, in the next S seconds, the bulb will emit another PxS photons, and the walls will reflect PxSxR photons, where R is the reflectivity of the mirrors. So in after 2S seconds, there will be PxS + PxSxR photons in the room. In the next time interval, after 3S seconds, there will be PxS + PxSxR + PxSxR2 photons in the room, where R2 is R squared. You can see where this is going. After NxS seconds, there will be:

PxSx(1 + R + R2 + R3 + …….+ RN + ….)

So we have an interesting result. The room will be brighter by a factor equal to the sum of an infinite series:

(1 + R + R2 + R3 + …….+ RN + ….)

Also, we find that the factor does not depend on the size of the room.

As long as R is between 0 and anything a little less than 1.0, this series converges. In fact it is an infinite geometric power series and the sum is 1/(1-R)

The reflectivity of ordinary second surface mirrors is typically 0.8 - 0.85. If we take 0.8, the series sums to 5.0

So, after a very short time, the number of photons in the room with mirrors is 5 times greater than in the room with black walls. Not a bad gain. Also, with the light coming from all directions, the illumination should be ultra uniform.

Fresh, pure silver first surface mirrors have a reflectivity near 0.99, so the room would be 100 times brighter, or the equivalent of having 100 bulbs in a room with black walls. That might get a little warm.

There are some assumptions in here. I approximated that 1 = 1/2 (see if you can find where.) That’s because I treated this as a Fermi problem, where we are interested in a meaningful estimate, not in a precise calculation. And of course we assumed that none of the photons are absorbed by people or things (or the bulb) in the room.

One of the pleasures of physics is the ability to work out “what if” questions for physical phenomena.
No tengo mucha idea de física de la luz, pero ahí no menciona nada si es posible dar la suficiente radiación para que hagan la fotosíntesis, no es sólo que sea más brillante si no que absorben determinada radiación del espectro (que ahora no recuerdo). Yo lo veo chungo así a bote pronto. Luego está el tema de que las plantas respiran por la noche, es decir cuando no hay irradiación te van a consumir oxígeno.

Pero ánimo con el tema. Yo eso sí no me metería ahí, prefiero espacios abiertos
 
No tengo mucha idea de física de la luz, pero ahí no menciona nada si es posible dar la suficiente radiación para que hagan la fotosíntesis. Yo lo veo chungo así a bote pronto. Luego está el tema de que las plantas respiran por la noche, es decir cuando no hay irradiación te van a consumir oxígeno.
¿Qué porcentaje de oxígeno consumen relativo al que producen? Pongamos que tenemos 15 hotas de luz y 9 de oscuridad.

Esto es interesante respecto a cuánta luz hace falta para producir la fotosíntesis:

All light wavelengths in the visible range (400 to 700 nm) are usable for photosynthesis. But some wavelengths are absorbed better than others. Photosythesis is fundamentally photoelectric so a blue photon can do no more photosynthesis than a red photon even though blue photons have more energy. Photons are photons and it does not matter how they are produced.

Where we have trouble is that the human eye is basically logarithmic in its response to light and so what appears as quite bright light for example under a street lamp or in a house is actually extremely low by plant standards.



La longitud de onda ideal para las plantas varía:

main-qimg-1e4038e9077da0d6d8bafe451258d231.webp

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A ver iluso, si tienes suficiente luz como para que crezcan plantas (incluyendo mil espejos y platanodas), es que tienes suficiente agujero hacia la superficie como para tener el oxígeno que necesitas.
 
Es más eficaz emplear esa electricidad en hacer la electrolisis al agua como en los submarinos nucleares,.
Y tienes combustible y oxígeno

Enviado desde mi SM-N975F mediante Tapatalk
 
Si hay sitio para atracar un mini submarino nuclear para usar el reactor como generador eléctrico vale, todo lo demás es pobreza energética.

Lo de usar plantas para generar oxígeno lo veo colateral, una toma de aire fresco, si quieres filtrado, hay que tenerla si o si.
 
Menudo reúma que te vas a pillar y catarrazos...
 
¿Qué porcentaje de oxígeno consumen relativo al que producen? Pongamos que tenemos 15 hotas de luz y 9 de oscuridad.

Esto es interesante respecto a cuánta luz hace falta para producir la fotosíntesis:

All light wavelengths in the visible range (400 to 700 nm) are usable for photosynthesis. But some wavelengths are absorbed better than others. Photosythesis is fundamentally photoelectric so a blue photon can do no more photosynthesis than a red photon even though blue photons have more energy. Photons are photons and it does not matter how they are produced.

Where we have trouble is that the human eye is basically logarithmic in its response to light and so what appears as quite bright light for example under a street lamp or in a house is actually extremely low by plant standards.



La longitud de onda ideal para las plantas varía:

main-qimg-1e4038e9077da0d6d8bafe451258d231.webp

main-qimg-888b253a79217e0840bb3143d989c6a8-lq
Me suena eso de clorofila a y b de cuando lo estudié. Ya te digo que cultivar con lámparas es estropeado, tienes que dar el espectro adecuado, tienes que tenerlo super controlado. Como has puesto tendrías que darle las longitudes de onda más cortas del espectro visible que es donde absorben más, o sea los azules si no me equivoco.

Luego está el tema de la temperatura también. Cultivar plantas en interior no es como ponerse un poto en casa, tienen un montón de variables a controlar.
 
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