Answers I DIDN'T get - Noah's Flood covering the mountains

[Magnetic]

I've heard that one before, though no clue where he got the 1/3rd number. Fails to mention the friction of that much water rushing out of the ground that fast should melt large parts of the Earth's crust.

He pulled the 1/3 number out of his ass. He wanted to use that as a guesstimate of how much of the Flood would fall as rain, vs how much would just gush up from these magic springs. He also somehow thinks that the water launching into the upper atmosphere or even space would vent heat into space to kick off the Ice Age.

I've disabused him with math, but he's been having a lot of trouble with the laws of thermodynamics. Apparently to him, a rainstorm is a rainstorm, and raindrops to not release megatons of energy.

How would one demonstrate this release of energy in the megatons from water falling from the sky? I can see his point, in a way. Even the heaviest rains I've seen only managed to cool the outside air.

I no longer believe in a world wide flood. As I heard it on a TV program (Discovery Channel, Science Channel, or whatever), it would take 5 times more water than what is currently in the oceans and underground to rise to the height above the mountains. Just too many variables that are unanswered for it to have actually happened. Too many civilizations that don't show much of any variances during the proposed time of the flood, which was supposed to be some 4,000 years ago.

[DPDarkPrimus]

Experience with rainstorms typically makes people feel cold, .

They haven't been in a REAL rainstorm, then. Sure, you're wet and shivering, but it's goddamn HOT out.

[Rahvin]

How would one demonstrate this release of energy in the megatons from water falling from the sky? I can see his point, in a way. Even the heaviest rains I've seen only managed to cool the outside air.

I no longer believe in a world wide flood. As I heard it on a TV program (Discovery Channel, Science Channel, or whatever), it would take 5 times more water than what is currently in the oceans and underground to rise to the height above the mountains. Just too many variables that are unanswered for it to have actually happened. Too many civilizations that don't show much of any variances during the proposed time of the flood, which was supposed to be some 4,000 years ago.

It's easy. We know the mass of the water. If it fell as rain, we know the terminal velocity of raindrops. All we need to calculate kinetic energy is mass and velocity. When the rain hits the Earth, all that energy is released as heat. Simple.

In my calcs I also left out any calculations for the heat generated by friction with air on the way down (the stuff that causes the rain to have a terminal velocity in the first place). I figured the easiest way to make him understand would be to use the speed of the water when it hits the ground, and the total mass of the water.

His problem is that he can't seem to understand that all of the kinetic energy of the raindrops is released as heat. You're right - rainstorms tend to cool down the air through evaporation and the fact that storms form on fronts of cold air. But energy on the scale he proposes easily overcomes that.

If you want, I can copy-paste my entire match post - it might make it more obvious to you.

[GrandMasterTerwynn]

We can do a quick and calculation to derive the energy deposited on the Earth by this global flood.

If we assume that the entire planet was covered in water, then we have to deposit about six miles of water on the surface of the Earth to ensure complete coverage.

This works out to be 4.95E+21 kilograms of water.

If we assume that it fell from a paltry one mile of altitude, we can assume that it has a total potential energy of 7.8E+25 joules.

We know that it rained for "Forty days and forty nights." This works out to about 3.5 million seconds. In order to deposit 4.95E+21 kilograms of water on the Earth's surface from one mile up, we deposit 7.8E+25 joules of energy on the Earth's surface over the course of 3.5 million seconds. This works out to be about 2.25E+19 watts of power, dissipated as waste heat.

Spread this out uniformly over the surface of the Earth, and we find that each square meter of Earth must dissipate 44.1 kilowatts of waste heat for every second of every hour of all 40 days and nights. This is roughly 44 times the influx of solar radiation coming from space (As measured outside the atmosphere) and 62 times the solar radiation striking the Earth's surface.

Thus, from this BOTE calculation, we find that Noah's Flood (as described in the Bible) would rapidly incinerate all life on Earth. Mind you, this calculation assumes I didn't misplace any decimal points in my haste, and ignores the fact that, realistically after a certain point, the rain would temporarily stop falling, because the air temperature has been driven up high enough that the raindrops would vaporize before hitting the ground, swathing the planet in a superheated water vapor shroud. (Eventually, enough heat would be dissipated that the rain would resume falling, and the whole cycle would begin again.)

[Magnetic]

How would one demonstrate this release of energy in the megatons from water falling from the sky? I can see his point, in a way. Even the heaviest rains I've seen only managed to cool the outside air.

I no longer believe in a world wide flood. As I heard it on a TV program (Discovery Channel, Science Channel, or whatever), it would take 5 times more water than what is currently in the oceans and underground to rise to the height above the mountains. Just too many variables that are unanswered for it to have actually happened. Too many civilizations that don't show much of any variances during the proposed time of the flood, which was supposed to be some 4,000 years ago.

It's easy. We know the mass of the water. If it fell as rain, we know the terminal velocity of raindrops. All we need to calculate kinetic energy is mass and velocity. When the rain hits the Earth, all that energy is released as heat. Simple.

In my calcs I also left out any calculations for the heat generated by friction with air on the way down (the stuff that causes the rain to have a terminal velocity in the first place). I figured the easiest way to make him understand would be to use the speed of the water when it hits the ground, and the total mass of the water.

His problem is that he can't seem to understand that all of the kinetic energy of the raindrops is released as heat. You're right - rainstorms tend to cool down the air through evaporation and the fact that storms form on fronts of cold air. But energy on the scale he proposes easily overcomes that.

If you want, I can copy-paste my entire match post - it might make it more obvious to you.

I guess your entire match post isn't necessary, especially if it is too long. Unless you have it on another page that I could link to.

It is just 'funny' to think of rain causing heat. But then the characteristics of water cause things to happen that don't seem right, such as a fall from a bridge into water, if the fall is high enough it can produce injuries serious enough, even death. I guess if you have enough speed thrown into the mix, even something like a liquid can produce such results.

I should ask my pastor about the deal with water causing heat. . . . . . .Now, I know what you'll think/say. However, the pastor of my church is a rather huge oddity as pastors go. He actually went to a major university and has a Ph.D. in aerospace engineering during that time, he felt a call to be a pastor. His doctoral thesis was on a specific area of fluid dynamics.

[Veramocor]

Grand Master K, Alkyesa, Rahvim.

I disagree with your 'boiling' the Earth statements calculations, completely.

Alkyesa you made a random statement and didn't back it up at all.

GMK you tried to back it up but prove yourself wrong with your own numbers.

Rahvin I'd love to see your "special" calculation which make you so much better than the people you are arguing against.
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Lets assume that each rain drop forms at 20,000 ft (a more than 'paltry' 3.8 mi) in the sky. The only energy associated with this droplet is the potential energy and the heat already in the droplet.

Lets use a basis of 1 lb.

All of the heat from potential energy will be transferred to heating the water droplets.

Rain drop starts at 50 deg F

PE = H*M*g

PE= 20,000 ft * 1 lb mass * 32.2 ft/sec2

PE = 20,000 ft * 1 lb force

PE = 20,000 ft lbs

PE = 25.7 BTU's

1 BTU heats up 1 lb water 1 deg F.

25 BTU will heat 1 lb of water 25.7 degrees F.

Enough to heat the water up to 75 deg F hardly boiling.


Grand Master K, look at your own numbers. I'll even trust you didn't make a decimal point error. You prove my point in them. 4.9 E+21 kgs = 4.9 E+24 grams and 7.8E+25 joules = 1.9 E+25 calories.

1 calorie heats up 1 gram water 1 degree Celcius.

1.9E+25/4.9 E+24 = 3.8 degrees C (6.84 deg F)

You used 1 mile in your calculation I used 3.78. 3.78 * 6.84 =

25.86 deg F.

Which is damn close to my easy 1 lb basis calculation where I got 25.7 deg F without having to resort to large numbers to confuse people.

Just because you can show calulation with really big numbers like 10^25 doesn't mean you can incinerate the Earth.

[DPDarkPrimus]

Excuse me, but would you happen to be a fundamentalist?

[wautd]

Just found this pearl on the startrek.com forum about a guy that says the earth is 6000years old:

if the earth was millions of years old, the magnetic field woud have gotten so big that the earth would have gained more mass than anything else int he universe, and melted.

[Dooey Jo]

All of the heat from potential energy will be transferred to heating the water droplets.

Actually wrong. The potential energy will be converted into kinetic energy as the droplets fall. When they hit the ground, much of that kinetic energy will be converted into heating the surface.

Just because you can show calulation with really big numbers like 10^25 doesn't mean you can incinerate the Earth.

No, but dropping 5e21 kg of water over the course of 3.5e6 seconds will. That is in fact just ten times less than the mass of the Moon! Calculations with big numbers show that dropping that much mass from a 1.6 km height over the course of 3.5e6 seconds would be 2.2e19 W. That's 42 kW/m^2. Just like GrandMasterTerwynn said. Fact is, assuming black body properties, that irradiance would heat the surface to about 930 K or about 660 degrees C:
(42e3 / sigma)) ^ 0.25 = ~930 K
And that is enough to boil the water.

And how come you managed to misspell the names of all the people you addressed? Was that some sort of joke?

[Veramocor]

DPDP: No not even close. Agnostic, believe in evolution fully, and one of the few liberals in the South it seems. But I happen to be right on this. Does calling me a fundamentalist relieve you of having to work out the equations I provided?

Dooey Jo:

Did you not even look at the calculations I did? You don't have enough energy to heat up the water that you dropped let alone the rest of the planet!

All of the heat from potential energy will be transferred to heating the water droplets.

Actually wrong. The potential energy will be converted into kinetic energy as the droplets fall. When they hit the ground, much of that kinetic energy will be converted into heating the surface.

I think you'll find that If you convert all my potential energy to kinetic energy you will come out with the same number (Conservation of Energy).

So lets follow the logic train. choo, choo

One poung of droplets is created at 20,000 ft, with 20,000 ftlbs of potential energy.

It drops to the surface where the 20,000 ftlbs of PE are converted to 20,000 ftlbs of KE energy. The droplet hits the surface of the water on the ground transferring its energy to 20,000 ftlbs of vibrational energy. Vibrational energy through friction is transfered to 20,000 ftlbs of heat energy or 25.7 BTUs. Enough to heat the water by 25 deg F.

Repeat this same thought exercise for 1000000 lbs you will still only have enough energy to heat the water that you dropped 25 degrees F.

For the rest of the world it takes ~4100 Joules to heat on kg of water 1 deg C. Lets say 8E25 Joules and 5E21 kg water rounde from GMT's numbers. (8E25/5E21)/4100 =3.8 deg C!

So please show me where I'm going wrong with a simple heat balance.


As for spelling I rushed through that part. I apologize.

[Rahvin]

Rahvin I'd love to see your "special" calculation which make you so much better than the people you are arguing against.

Alrighty, here you go. Remember that the scenario proposed was that 1/3 of the water contained in the oceans and icecaps fell as rain, and the rest of the water poured out of the "fountains" like an everyday spring. He also proposed that the water thrown up as rain would first reach an alititude sufficient to bleed off heat into space, lowering the temperature of the planet and setting up for the Ice Age, and then fall down over 40 days.

Okay, let's see how much energy we are talking about with 30% of the world's ocean mass falling as rain. We'll ignore the icecaps, becuase I don't feel like doing THAT much math, and it only makes it worse for you if I do.

According to a previous post, the Earth's oceans contain 1,370,000,000 cubic kilometers of water. That's 1.37e18 cubic meters, so 1.37e18 metric tons. 1/3 of that would be 4.57e17 tons. 4.57e20 kilograms. We already know that rain falls at about 7 m/s. Kinetic energy = (mv^2)/2. That's 2.24e22 Joules of energy, which translates to 5,358,851.67 megatons of energy. That's about 10 kilotons of energy per square kilometer of the planet. Each day, it would be the equivalent of 10 metric tons of TNT exploding over every square kilometer on the planet. Not nearly as impressive or devastating as the previous calculations, but still more than enough to cause changes in the Earth we would still detect a few thousand years later. Not to mention that a wooden boat wouldn't be much protection.

AND we still need explosions several orders of magnitude stronger just to get the water up there, from the "fountains of the deep." 18,000 mph is about 29,000 kph. That's 1.04e11 m/s. The kinetic energy required to make 1/3 of the Earth's oceans reach escape velocity is 2.47e42 Joules, which converts to 5.91e26 megatons, and this one isn't spread over 40 days. It also doesnt take into account the drag force of pushing through several cubic kilometers of ocean, about 62 kilometers of air (if I recall correctly), and STILL retaining a speed of nearly 29,000 kph. These calculations assume that air resistance is not a problem, and that the water is ejected from the surface of the earth, not the ocean's depths, so it's a pretty good lower limit, I think.

It would only take 5.26e8 megatons to melt the entire surface of the Earth into magma to a depth of one meter. This calculation assumes that the entire planet is made up of silicate material (ie, rock), with no oceans. The enrgy required to put 1/3 of the Earths ocean into the upper atmosphere would be enough to melt the surface of the earth to a depth of 1.13e18 meters.

That's my cleaned-up final post in the other forum. Well, the math part anyway.

You'll notice that I handwaved away large portions of the energy involved by only using the terminal velocity of raindrops and the mass of the water. Energy released during the fall itself is not added to my totals, only the impact. I wanted to give as many concessions as humanly possible to the Creationist so that he could see how bad his scenario really was.

For the "Fountains of the Deep" calculations, which is where we actually destroy the planet (since his Flood used a tiny fraction of the water that would really have been necessary), I handwaved various issues that I don't know how to calculate that only add MORE energy to the equations. Again, trying to set lower limits and give as many concessions as possible. Oh, and thanks to Mike for the BDZ calculations.

Please, if anyone wants to double-check my math, feel free to do so. I've been out of physics class for a while, and I might be a tad rusty. Besides, some of you have a lot more experience calculating global devastation than I do.

Enough to heat the water up to 75 deg F hardly boiling....Just because you can show calulation with really big numbers like 10^25 doesn't mean you can incinerate the Earth.[/quote[

You're an idiot. We aren't talking about one pound. We aren't talking about boiling the water. We're talking about a few bazillion megatons of energy being converted to heat on impact.

[Dooey Jo]

For the rest of the world it takes ~4100 Joules to heat on kg of water 1 deg C. Lets say 8E25 Joules and 5E21 kg water rounde from GMT's numbers. (8E25/5E21)/4100 =3.8 deg C!

So please show me where I'm going wrong with a simple heat balance.

I'm not saying your calculations are wrong. In fact, here's a quick proof that they are not:
m*g*h = C*m*ΔT
The mass is irrelevant in that equation, so the change in temperature would be ΔT = (g*h)/C. C is the specific heat capacity. Ergo:
T = (9.8*1600)/4190 = ~3.7 K (or deg C if you like, since it's change in temperature).

But Terwynn's calcs are not wrong either, and the fact still remains that the Earth would be hit by 2.25e19 J of kinetic energy every second for 40 full days. The question is wether that energy would be radiated off as waste heat, or if it would go to heating the water itself...


Well anyhow, it doesn't really matter since any possible incineration from falling rain is neither the only nor biggest problem with a global flood. Like, where did the water come from, and where did it go for instance. Or what the atmospheric effects of 5e21 kg of water condensing into rain would be...

[Magnetic]

How soon would plant life bounce back after a half a year or so under water?

[Dooey Jo]

Well, all plant life would be exterminated by the rain, so it would probably need a few hundred million years to evolve back

AND we still need explosions several orders of magnitude stronger just to get the water up there, from the "fountains of the deep." 18,000 mph is about 29,000 kph. That's 1.04e11 m/s. The kinetic energy required to make 1/3 of the Earth's oceans reach escape velocity is 2.47e42 Joules, which converts to 5.91e26 megatons, and this one isn't spread over 40 days. It also doesnt take into account the drag force of pushing through several cubic kilometers of ocean, about 62 kilometers of air (if I recall correctly), and STILL retaining a speed of nearly 29,000 kph. These calculations assume that air resistance is not a problem, and that the water is ejected from the surface of the earth, not the ocean's depths, so it's a pretty good lower limit, I think.

Actually, 1.04e11 m/s is about 300 times faster than the speed of light! In order to reach escape velocity, you need to apply 60 MJ/kg, or 11.2 km/s. It works out to ~2.8e28 J, or 7 trillion megatons. Don't panic; that's still more than enough for quite a mighty BDZ (which were calculated to be from 1e24 to 1e26 J)...

[Veramocor]

No, but dropping 5e21 kg of water over the course of 3.5e6 seconds will. That is in fact just ten times less than the mass of the Moon! Calculations with big numbers show that dropping that much mass from a 1.6 km height over the course of 3.5e6 seconds would be 2.2e19 W. That's 42 kW/m^2. Just like GrandMasterTerwynn said. Fact is, assuming black body properties, that irradiance would heat the surface to about 930 K or about 660 degrees C:
(42e3 / sigma)) ^ 0.25 = ~930 K
And that is enough to boil the water.
?

You can't use that simplistic of a model Dooey. I assumed you got 42 kW/m^2 by dividing 2.2E19W by the surface area of the Earth.

If you concentrated all your heat on 'the same surface' you'd be correct.

Your not doing that. Your model is incorrect. As it rains you are discretly laying down a new surface of water and also doing a lot of mixing at the top surface area of the water.

But Terwynn's calcs are not wrong either, and the fact still remains that the Earth would be hit by 2.25e19 J of kinetic energy every second for 40 full days. The question is wether that energy would be radiated off as waste heat, or if it would go to heating the water itself...


Well anyhow, it doesn't really matter since any possible incineration from falling rain is neither the only nor biggest problem with a global flood. Like, where did the water come from, and where did it go for instance. Or what the atmospheric effects of 5e21 kg of water condensing into rain would be...

I don't disagree that the world wouldn't be in very very bad shape. The pressure alone from just a mile of water would be incredible. 5240 ft/2.31 = 2268 psig (source Cameron's Hydraulics).

I do disagree that the world would be incenerated. That is the statement that a bunch of people in this thread have been throwing around. It is simply false. And a bad example of group think by people claiming to be scientific.

The kinetic energy would go into heating the water. In fact I am 100% sure it would. What is happening?

A rain drop falls on a surface of water. It pentrates the surface of the water and friction slows it down. The friction is transferred to heat energy. Which is absorbed by the rest of the water by conductivity/convection.

The temperature will never get hot enough for radiative cooling to take place. Convection and conductivity will dominate at temperatures around 50-75 deg F.

[Veramocor]

You're an idiot.

You get your own post, dumbass.


Dooey stated the equation quite right. Thank him for showing it symbolically.

m*g*h = C*m*ΔT

Look at it for awhile and then recant and admit you are wrong.

We aren't talking about one pound.

Yes we're talking about a lot of pounds. Too bad mass canceled out in the equation.

We aren't talking about boiling the water.

Again we aren't talking about boiling water. We'd never boil the water there isn't enough energy to even heat it up 30 deg F.

We're talking about a few bazillion megatons of energy being converted to heat on impact.

A 'bazillion' tons of energy which has to heat up the bazillion tons of water you dropped from X miles high.

I've shown you you don't have enough actual heat to heat up the water you are dropping.

The only argument left is heat transfer mechanics now which I talked about with Dooey in my last post.

[Rahvin]

Actually, 1.04e11 m/s is about 300 times faster than the speed of light! In order to reach escape velocity, you need to apply 60 MJ/kg, or 11.2 km/s. It works out to ~2.8e28 J, or 7 trillion megatons. Don't panic; that's still more than enough for quite a mighty BDZ (which were calculated to be from 1e24 to 1e26 J)...

Holy shit, THAT was a fucking huge mistake. Thanks for pointing it out. Ill be more careful in the future.

But, hey, as long as the energy release is still on the level of a BDZ, the argument still stands.

[Surlethe]

But, hey, as long as the energy release is still on the level of a BDZ, the argument still stands.

Wattage, in this case, is more important. BDZ wattage is ~1E23 W (main site). Great Flood wattage is ~2E19 W (GMT's calcs.).

That means a Star Destroyer releases energy at 10000 times the rate of the Great Flood. The oceans are unable to dissipate the energy that quickly, and so a BDZ will vaporize oceans, while the Great Flood would merely heat up the ocean several degrees.

[Rahvin]

A 'bazillion' tons of energy which has to heat up the bazillion tons of water you dropped from X miles high.

I've shown you you don't have enough actual heat to heat up the water you are dropping.

The only argument left is heat transfer mechanics now which I talked about with Dooey in my last post.

Dd you read my post? The "world annihilation" came from the energy released to throw the water into the upper atmosphere.

The rain had a lot of energy (10 tons of TNT per square kilometer over the entire planet each day for 40 days), but I never claimed that the rain itself vaporised everything. Not with the amount of water this fundy proposed, anyway (disregarding the fact that it's not nearly enough to flood the planet).

Dooey Jo pointed out a flaw in my calculationd for the water jets, but they still would have resulted in explosions large enough to destroy the planet. Well...fry the surface, anyway.

[Surlethe]

I messed up; didn't read the thread closely enough.

My point stands, but not in the context of Rahvin's quote at the top.

[Dooey Jo]

You can't use that simplistic of a model Dooey. I assumed you got 42 kW/m^2 by dividing 2.2E19W by the surface area of the Earth.

If you concentrated all your heat on 'the same surface' you'd be correct.

Your not doing that. Your model is incorrect. As it rains you are discretly laying down a new surface of water and also doing a lot of mixing at the top surface area of the water.

Yep, it's surface area of Earth, and actually, now that I think about this more closely, your explanation does seem to be the more likely. I think shall revise my position (not that I actually had a position about this particular question before today, but I digress).

[Rahvin]

You're an idiot. We aren't talking about one pound. We aren't talking about boiling the water. We're talking about a few bazillion megatons of energy being converted to heat on impact.

Retracted. You are correct - the energy would heat the water droplets themselves rather than the surrounding environs.

The real point I made, that of the energy required to shove the water into the upper atmosphere, should still stand, however.