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Problem I - Cratering. 2. The Earth


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#541 piasan

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Posted 17 July 2016 - 01:16 PM

I've clearly stated more times than I can remember going back to the very first reply on this topic that heat is NOT an issue when we discuss these events.  Your constant attempts at diversion notwithstanding.

I know that this is only tricky wording from you because you have not tried to make it an issue YET but as soon as we decide that cratering would not happen on the Earth then you will quickly change your tune to say that all of this energy would be converted to heat in the upper atmosphere and that would cook the planet. You are not fooling anybody.

You are only fooling yourself.  I have openly declared, since the first reply to this topic in post #2 two years ago, that these impactors will not cause significant atmospheric heating.

 

So we should get on with that part of the discussion ....

That part of the discussion is   :off_topic:

 

Future attempts at diversion will be greeted by reminders that atmospheric heating is not relevant to this issue.



#542 indydave

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Posted 17 July 2016 - 02:51 PM

BTW, I do get what you say about how a less dense object would have more tendency to not spread outward against the high pressure of the hypersonic slipstream, because it can compress easier within the diameter of the original object.  BUT that would be only up to a certain point, which is when the pressure would exceed that of the slipstream.  So as the .6 object gets compressed then at a certain point its pressure would exceed the pressure of the .001 atm, even if it is rushing by at high speed....so then it would expand outward some, causing pancaking.  In the case of SL-9 there had to have been SOME of that happening even with the .6 porous ice object...because there was an opening made in the atm which allowed for 40x atm gases and 1x comet material rto be launched upward in a plume.  The comet did not just cut a hole in the atm exactly equal to its own diameter.  Having a diameter of 6.33x the original comet size means that hole in the atm would be 40x larger...so it seems pretty close to the 7x factor for pancaking which they assumed. 

 

Also BTW, I hope you will remember this concept when we talk again about how a superfast stream of gases (Brown's upward jets) would have the effect of not allowing Earth's atm to encroach into the hole which was cut into the atm by the first blast of the jets.  Though they are at a low pressure (and therefore are very cold, due to expansion) they are traveling so very fast that they act like a solid wall, blocking out any of the atm molecules from entering that space.



#543 piasan

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Posted 17 July 2016 - 10:32 PM

OK... my estimate of the depth of penetration for a density 0.1, 70 km diameter impactor based on the Korycansky Shoemaker-Levy 9 research .... 
 

Ha! Looks like we have some backpedaling going on now! Previously you said the amount of penetration of a .1 object going 61 km/s would be to 1.67 bars. Now you're going to change that!

What you have is a more thorough review including factors (such as Figure 8) that had not been considered before.
 

Indy claims a "pancake factor" of 6.  As we can see in figures 4,5, and 6, the pancaking of the bolide is already factored into the model.  Notice, in Figure 6, basalt with a density of 2.7 actually pancakes and fragments worse than ice with a density of 1.0.  If you think about it a bit, it makes sense.  As the leading edge of the object slows, the less dense object can compress front-to-back without needing to spread against a hypersonic slipstream.  Since the model shows lower densities have less pancaking it clearly includes pancaking in its analysis, this variable is already factored into the results.  There is no justification for double dipping.

I may need to go back, but I don't recall using ANY "pancake factor" in my calcs, however, I might have referred to the idea at some point.  I actually would think that it WOULD be very very fair to have that ADDED because it INCREASES the cross-sectional area and therefore there is more mass to stop the object.  The 1/6 factor applies to the objects WHILE STILL IN SPACE ABOVE THE ATM.  Then AFTER hitting the atm, there could indeed be more spreading outward...to cause more contact with atm molecules.  This would not reduce the density, however, because the volume of the sphere would be changed into a cylinder (of sorts) which is wider than it is tall.  Maybe a flat ellipsoid is a better term than cylinder. 

Well, you were using 6x something times 13x for velocity to get 78x.  Regardless, Fig. 6 in particular establishes that the less dense object actually spreads less than the more dense.  So, whether you "actually would think it would be very very fair to have that ADDED" or not .... the DATA shows it would not be correct at all. 

 

Your proposed bolide has a density of 0.1.  Crystalline rock is over 2.7 and water is 1.0.  This means, there is a LOT of empty space inside it.... a minimum of 90%.  With all that empty space, the density of the object MUST increase as the leading edge slows down while the trailing edge does not.  In other words, it has a lot of packing down to do before it's going to start spreading out.

 

I'll address the 13x energy due to the velocity difference in a separate post.



#544 indydave

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Posted 18 July 2016 - 01:18 PM

You are only fooling yourself.  I have openly declared, since the first reply to this topic in post #2 two years ago, that these impactors will not cause significant atmospheric heating.

 

That part of the discussion is   :off_topic:

 

Future attempts at diversion will be greeted by reminders that atmospheric heating is not relevant to this issue.

 

It is not off topic for this forum...just for this THREAD.  So finish this thread, by conceding that you are wrong to say that a .1 impactor going 17km/s would penetrate to form a crater at the bottom of the ocean.  (Other than tiny ones by the larger rock inclusions...which would NOT be lethal at all).



#545 indydave

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Posted 18 July 2016 - 02:16 PM

>>So, whether you "actually would think it would be very very fair to have that ADDED" or not .... the DATA shows it would not be correct at all. >>

That's just fine.  K's data shows depths which include pancaking for the basalt and not any for the porous ice.  That's what I have been using.  I believe some pancaking would be likely, however.  Dr, Frank, the guy studying possible small comets, DID seem to observe massive spreading out very very high up...but that may not have been real I guess. 

 

Pi>>Well, you were using 6x something times 13x for velocity to get 78x.>>

Don't play dumb.  (Or maybe you AREN'T playing...hmmm).  I did not use a 6x factor for pancaking at all.  It was not because of ANY effect of contact with the atm.  It was THE ORIGINAL DENSITY IN SPACE.  That is, 6x less dense.  Any effect of pancaking would be ON TOP of the density factor AND the speed factor.  You have agreed to the speed factor being 13x (with some weaselly wiggle-room I suppose, but you finally DID agree, because you THOUGHT it wouldn't ruin your position).  If an object is 6x more massive, then it has 6x more energy.  So if there is 13x more energy due to speed and 6x more energy due to mass, it will go 78x deeper.  Or so you have agreed re. the speed factor...I guess you have doubt about the mass factor.  When you look at K's Fig 7, the 2.7 (basalt) object expends 90% energy at (eyeballing the solid black line) at 150 bars.  (The bars are logarithmic...KM is not, it seems, because the atm gets thicker with depth).  The .6 object which is 22% as dense (or you could say the basalt is 4.5 times denser) releases 90% at only 15 bars or so...10% the depth of basalt going the same speed.  And that is WITHOUT any additional pancaking factor.  That is LESS than 1 to 1 density to depth.  So my estimate of 1/78 as deep is GENEROUS to you.   It SHOULD be 1/130.  10x for density and 13x for speed.  So if a .6 object at 61km/s penetrates to 15 bars, then the .1 object going 17km/s would go between .192 bars (if it is 1/78th) and .115 bars (if it is 1/130th). 
HIGH in the atm!  NO CRATER (other than small ones for the rock inclusions...and they would not be lethal at all).

 

You need to man up...and concede.  And don't just fly off from the chess board either.  CONCEDE.  Then if you have the courage, we can continue to discuss heat...but I doubt you will have it. 



#546 indydave

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Posted 18 July 2016 - 02:24 PM


 

I'll address the 13x energy due to the velocity difference in a separate post.

 

Before you waste a lot of your time on that, take a look at the various density objects in Fig. 7.  That tells you a lot about how depth is affected by energy.  A 1km object with 4.5x the energy (basalt vs. porous ice...going the same speed) went 10 times as deep.  That argues that the energy factor is 2.22 to 1.  So if something has 78x the energy, then we would expect it (acc. to Fig 7) to travel 173x deeper.  And the inverse of that is an object with 1/78th the energy would go 1/173 as deep.  15 bars/173 is .087 bars....WAY up in the atm of Earth!!!



#547 indydave

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Posted 18 July 2016 - 06:55 PM

Pi>> You are only fooling yourself. I have openly declared, since the first reply to this topic in post #2 two years ago, that these impactors will not cause significant atmospheric heating.<<

Yes I get that but that is only because you have believed that 90% of the energy is saved until impact with the surface so that would leave very little to heat up the atmosphere. However I am showing to fair minded readers that there would be no substantial impact and therefore the energy would be lost to the atmosphere. And so you would be forced to try to say that the heat would be lethal. It is doubtful of course that you will ever concede that there would be no craters but if there were none then you would have to say there would be lots of heat. But as I have shown, especially by the recent information about plumes that were observed on Jupiter, heat would never reside in the atmosphere in any large amounts. No craters. No heat.

#548 piasan

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Posted 18 July 2016 - 07:36 PM

Pi>> You are only fooling yourself. I have openly declared, since the first reply to this topic in post #2 two years ago, that these impactors will not cause significant atmospheric heating.<<

Yes I get that but that is only because you have believed that 90% of the energy is saved until impact with the surface so that would leave very little to heat up the atmosphere. However I am showing to fair minded readers that there would be no substantial impact and therefore the energy would be lost to the atmosphere. And so you would be forced to try to say that the heat would be lethal. It is doubtful of course that you will ever concede that there would be no craters but if there were none then you would have to say there would be lots of heat. But as I have shown, especially by the recent information about plumes that were observed on Jupiter, heat would never reside in the atmosphere in any large amounts. No craters. No heat.

:off_topic:



#549 piasan

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Posted 19 July 2016 - 02:07 AM

With regard to the 13x factor for energy used by Indy due to the difference in velocity between a 17km/sec object and one moving at 60 km/sec.  We completely agree that because (kinetic) energy is a function of velocity squared, the proper way to determine the energy difference is (60/17)2.

 

Now to what I said about 13x:

Indy correctly claims his proposed 17 km/sec bolide would have 1/13 the energy of the 60 km/sec SL-9 fragments.  He asserts this justifies a 13x reduction in penetration depth.  I disagree with the 13x adjustment mainly because all the plots we've been looking at have been logarithmic.  This suggests to me that a more appropriate adjustment might be log(13) or about 1.1.  In addition, the faster objects approaching Jupiter will experience 13x the braking force with the same mass.  This means the faster object will slow down much more quickly so a linear relationship between energy and penetration is inappropriate.  However, I'm uncertain what the exact adjustment should be  and rather than drag this out, we can apply the 13x factor.  (Pi: Emphasis added.)

So this was what your said ABOUT OBJECTS GOING 61KM/S.  And you have now agreed that there would be 1/13th the energy for objects going 17km/s.  But ASTONISHINGLY you have used poor and/or tricky logic to say the object would NOT penetrate 1/13th as deep.  I'll try to unravel the tricky logic below. 

 

>>Pi: I disagree with the 13x adjustment mainly because all the plots we've been looking at have been logarithmic. >>

That has nothing to do with it because we both have eyeballed to get the actual data for the amounts.  That is part of why discussing a "curve" shown on a logarithmic chart is bogus.

......

All of this is very confusing.  You need to elucidate better.  

One of the things I can always rely on with Indy is accusations of deception. 

 

How about this for why the 13x factor for penetration isn't appropriate.....

There are, IIRC, 14 equations in the SL-9 paper.  Each and every one of them is integration (calculus); has an exponent; or uses an variable as an exponent.  One does not use calculus when a simple linear relationship will do.  Exponents and variables as exponents yield non-linear results.  Further, as the energy is a function of velocity squared a plot of energy versus penetration will also yield non-linear results.  Most likely, the solution you're looking for would require calculus..... but it is definitely NOT a linear relationship in which 13x the energy = 13x the penetration.

 

There's your "trickery."

 

You want me to elucidate better.  Notice the portion I emphasized (above).  Here's what it means:

"I'm uncertain what the exact adjustment should be..." means I don't know how to determine what the difference would be.

"... rather than drag this out...." means I really don't want to get into another 20 post discussion over this point.

" .... we can apply the 13x factor."  means I was willing to just use what you claimed.

 

Pi, do you believe that a diver who jumps from a platform so that he is traveling at a speed which is 3.6 times his speed from a lower platform will penetrate deeper than he would from the lower platform? According to you he would have more energy but there would be more braking power.

Also I guess I should ask you if a comet which is traveling 61 km/s were to hit Earth's atmosphere is it your belief that it would not travel as deep as a comet which travels 17 km/s... because of the increased braking power?

>>This suggests to me that a more appropriate adjustment might be log(13) or about 1.1. <<

Does that mean that you no longer want to eyeball the charts? Are you trying to tell me that an object that has 13 times more energy is going to only penetrate 1.1 times the depth? You must be joking.

One.   See the words "suggests" and "might?"  Those words reflect the uncertainty previously mentioned.

 

It is hard to know how apropos this site LINK  is, but I tried to use it to figure what to expect for two objects, of the same size and density, traveling at 2 speeds.  One is at 500 ft/s (500x500 = 250,000) and the other is 354 ft/s (354 x 354 = 125,000, which is half the energy).  The first one travels through the atm for 600 yds before hitting the ground. http://www.shootersc....php?t=0cb5cdf1   And the slower one travels 310 yds. http://www.shootersc....php?t=9b7775d9   It is pretty close to double the distance when there is double the speed energy.  The maximum range is not 2x but that is hard to figure IMO because I believe it is a calc to determine when the bullet would be falling straight downward if it were shot over a cliff of infinite height.  I am not sure it is fair because you have acceleration of gravity involved...which pulls harder on the faster bullet because it is in forward flight longer.  The maximum range of the faster one over the slower one is 1514/906...which is 1.67.  It is also hard to know if you scale it up to being 17km/s or 61km/s if you would see the same ratios.  I don't think the calculator will go that high.  The faster one stays in the air 3.86seconds but the slower one is 2.67 seconds, so gravity gets 1 second longer to pull down on the bullet.  If that factor were removed, I believe the faster bullet...travelling through the same density of air...would go twice as far.  This supports my view that speed energy is directly proportional to penetration ability. 

 

I think the distance factor for speed energy is WAY larger than 1.1...which seems to be what Pi says.  I am not sure how he would apply that number though.  It appears he would say that if there is 13x more energy it would go only 1.1 times as deep.  That is ABSURD!  He should stop stalling.  Show how the number 1.1 is supposed to be used!  My guess is he has already talked to his physics PhD buddies and didn't get the answer he hoped to hear!

Two.  I tried the link, and it can be useful, but I'm not sure they're directly applicable to hypersonic velocities.  When I used the model, I input different speeds.  Since we've been having problems with using models outside their (apparent) limitations and the model was for firearm ballistics, I scaled the numerical values of the velocities (17 and 60) by 50x to get 850 and 3000.  In feet per second, this is well within the typical velocities of bullets. Say a .22 and a 30-06.  IIRC, the ratio for a 25 foot (or was it inch) drop was somewhere around 2.3-2.5.  (I used the amount of drop so we were speaking of equal time intervals.)

 

If nothing else, the calculator shows Indy's 1:1 relationship between energy and penetration depth (13x the energy = 13x the depth) is, as Indy would say.... "bogus."

 

BTW, I haven't been in direct contact with DaveB for a year or so.  We are both on a couple YahooGroup lists and occasionally exchange comments there.

 

Pi >>....

Indy correctly claims his proposed 17 km/sec bolide would have 1/13 the energy of the 60 km/sec SL-9 fragments.  He asserts this justifies a 13x reduction in penetration depth.  I disagree with the 13x adjustment mainly because all the plots we've been looking at have been logarithmic.>>

 

It is bogus to suggest that a logarithmic chart means something about the values themselves.  It is just a means of showing a much larger group of plots on a smaller size of paper (or smaller screen).  It is easy to convert the numbers and use the actual numbers.

 

>>This suggests to me that a more appropriate adjustment might be log(13) or about 1.1.>>

 

RIDICULOUS!

 

>>  In addition, the faster objects approaching Jupiter will experience 13x the braking force with the same mass.  This means the faster object will slow down much more quickly so a linear relationship between energy and penetration is inappropriate. >>

 

the ballistics calculator disproved this.

Three.  I think my "bogus" and "tricky" reservations with Indy's straight-line analysis have been confirmed by the calculator he presented.  Which, incidentally, was after my comments.

 

So, I told Indy I was willing to use his 13x factor in order to avoid a lengthy argument about it and he then made 3 posts challenging a proposal that I had already said I wasn't going to use.  Then, after he finds a calculator that disproves his 13x claim, he uses 13x in several calculations.  Ya can't have it both ways, Indy.  You have disproven 13x and have produced data indicating a more realistic number may be more like 2.4x or so.  My previous offer to accept your claim of 13x is withdrawn on the basis of better evidence (provided by you).

 

 

On this; the comment by Faulkner, and the heat "issue" Indy reminds me of my wife.  I've been known to tell her "for the fourth time YES."  In Indy's case, sometimes it takes a lot more than four.



#550 piasan

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Posted 19 July 2016 - 02:59 AM

You need to man up...and concede.  And don't just fly off from the chess board either.  CONCEDE.  Then if you have the courage, we can continue to discuss heat...but I doubt you will have it. 

Indy wants a concession..... OK..... here it is:

1)  I agree there are objects that could produce a significant crater on the moon while making no detectable sign on the Earth that there had been an encounter with them.

2)  Based on the research of Dr. Frank provided by Indy, I agree the Earth could encounter a 0.1 density object made entirely of water (or water with really small particulate) and no large blocks of ice and the planet would show no surface effects.  (Of course that same research shows those objects wouldn't have any (detectable) surface effects on the moon either.)

 

Note:  The proposed 0.1 density, 70 km diameter cloud of water and rock will make notable surface damage on either the Earth or moon.   More on that in another topic I'll open up.  This one is beyond exhausted.

 

One final comment here.....  Indy wondered why the SL-9 researchers would make multiple runs at density 0.6 with different masses.  The answer is probably that the purpose of the research was to investigate impact results with respect to comet SL-9.  SL-9 was thought to have a density of 0.6 and the comet consisted of 20+ fragments of varying size, it makes sense they would evaluate penetration due to the different sizes of the various pieces of SL-9.



#551 indydave

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Posted 19 July 2016 - 02:24 PM

Pi>>2)  Based on the research of Dr. Frank provided by Indy, I agree the Earth could encounter a 0.1 density object made entirely of water (or water with really small particulate) and no large blocks of ice and the planet would show no surface effects.  (Of course that same research shows those objects wouldn't have any (detectable) surface effects on the moon either.)>>


I won't try to comment on most of what you wrote...yet...but this part is intriguing.  Of course as I believe you pointed out, the breakup/expansion of the supposed small comets happened something like 300 or maybe it was 600 miles up.  Way above what is thought of as E's atm.  The breakup was from electrostatic forces, IIRC.  The Moon has no (significant) atm so that could not cause breakup, so we might expect to see impacts today on the Moon...UNLESS the objects get expanded by means of the same sort of electrostatic forces.  I seem to recall speaking with one of Frank's associates (he had died) to ask about the quandry of absence of visible impacts on the Moon, and he had an answer for it, but I just can't recall what it was.  It could be that electrostatic forces can obliterate a smaller .1 object but have little effect on a larger one...capable of making larger craters shortly after or during the Flood.  

 

BTW, if you tell your wife "yes" the first 3 times, I hope you manage to say it clearer than you supposedly said yes to me re. the 13x.  It seemed a bit weaselly...or if you prefer, "ambiguous."  I can't help but doubt that you would not have gone along with 13x for very long...especially when you got a "good" number for you only when you FORGOT THE LOWER DENSITY FACTOR.  I will look at the bullet energy stuff to see if you blundered (maybe you used a blunderbuss as your firearm!).  In the meantime, you should look at the point I made using the various densities and their corresponding depths...taken from Fig 7, which were all the SAME diameter.  There was LESS than a 1/1 energy to depth relationship.  As I wrote in #545:

 

If an object is 6x more massive, then it has 6x more energy.  So if there is 13x more energy due to speed and 6x more energy due to mass, it will go 78x deeper.  Or so you have agreed re. the speed factor...I guess you have doubt about the mass factor.  When you look at K's Fig 7, the 2.7 (basalt) object expends 90% energy at (eyeballing the solid black line) at 150 bars.  (The bars are logarithmic...KM is not, it seems, because the atm gets thicker with depth).  The .6 object which is 22% as dense (or you could say the basalt is 4.5 times denser) releases 90% at only 15 bars or so...10% the depth of basalt going the same speed.  And that is WITHOUT any additional pancaking factor.  That is LESS than 1 to 1 density to depth.  So my estimate of 1/78 as deep is GENEROUS to you.   It SHOULD be 1/130.  10x for density and 13x for speed.  So if a .6 object at 61km/s penetrates to 15 bars, then the .1 object going 17km/s would go between .192 bars (if it is 1/78th) and .115 bars (if it is 1/130th).  HIGH in the atm! 



#552 indydave

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Posted 19 July 2016 - 03:23 PM

Pi>>Say a .22 and a 30-06.  IIRC, the ratio for a 25 foot (or was it inch) drop was somewhere around 2.3-2.5.  (I used the amount of drop so we were speaking of equal time intervals.)>>

I just went there and when I did it (using the links I gave for my 2 speed charts, then adjusting those).  I first changed the sight height to zero.  This was because there was an upward-then-downward trajectory for the slow one, but an only upward one for the fast one...meaning that gravity was working against it the whole time.  With a flat sight angle both had gravity for the whole time.  Then I did it using your numbers.  3000 and 850.  In one second,  fast traveled: 850yds  slow: 275  (3.09x)  In 3 seconds, fast: 1975   slow: 800 (2.47x).  I then tried some other speeds, again using 13xenergy.  In one second, fast (610fps): 200  slow (170fps): 60 (3.33x).  Three seconds: fast: 595  slow: 170 (3.5x)   Using 6100 and 1700fps, in one second it travels 1540 and 505 (3.05x). 

 

I did another one, this time with a sight angle of .5in.  Using 850fps...I got 575yds before it hit the ground and using 3000fps, it reached apex at 1760, meaning it would travel 2x that (3520) before it hit the ground.  That is a 6.1x factor all based on an adjustment in sight angle.  I believe it is not right to use a set amount of time (eg. 1 sec. or 3 sec.) because we are not interested in what is the distance in a certain TIME...but rather what distance would it have all its energy expended, regardless of how long that takes. 

 

Because there is quite a bit of variance using different inputs (besides just 2 speeds) it is hard to say if the ballistics calculator gives good info about what the effects of various speed inputs are. 



#553 indydave

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Posted 19 July 2016 - 07:45 PM

Pi>Indy wants a concession..... OK..... here it is:
1) I agree there are objects that could produce a significant crater on the moon while making no detectable sign on the Earth that there had been an encounter with them.<<

Okay, I guess I have to ask what is the tricky part? Is it in the definition of the word significant? Please elucidate better. What you have said is impossible is an object that would create a 100 mile diameter crater on the moon would NOT cause a huge crater on the earth...so I am assuming that the word "significant" must be a lot smaller than that. And if so that is not much of a concession.

#554 indydave

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Posted 19 July 2016 - 07:53 PM

Pi, please before you go through the gun ballistics stuff I would like for you to address the point I made to you based on the various masses of objects which have different amounts of energy and travelled different depths according to figure 7 in the SL9 article. It shows clearly that with the same sized objects with different masses because of density, there are different amounts of energy and it shows the different depths for the different energies. That should give us the answer regarding energy and depth of penetration. If not explain why not.

#555 piasan

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Posted 23 July 2016 - 04:51 AM

Pi>Indy wants a concession..... OK..... here it is:
1) I agree there are objects that could produce a significant crater on the moon while making no detectable sign on the Earth that there had been an encounter with them.<<

Okay, I guess I have to ask what is the tricky part? Is it in the definition of the word significant? Please elucidate better. What you have said is impossible is an object that would create a 100 mile diameter crater on the moon would NOT cause a huge crater on the earth...so I am assuming that the word "significant" must be a lot smaller than that. And if so that is not much of a concession.

You are correct that "significant" does not mean the kind of 100 mile diameter lunar craters that are the observational evidence on which this discussion is based.  By "significant" I mean craters that are miles to a few tens of miles across.... something I could probably easily see in my little 4.5 inch telescope.

 

Until I pointed out this proposed bolide will have rocks big enough to survive reentry and leave a lot of really noticeable craters on Earth, your claim was that the proposed object would basically be shredded during entry and leave NO crater.  Not just a "huge crater."

 

I think we have a point of agreement that Brown's model would produce impactors that will, in fact, leave some craters on Earth that would be pretty hard to miss.  (Note:   I'm not claiming these proposed bolides would be sufficient to cause a lot of damage more than a few miles from the area of impact.)  This is worthy of a new topic.

 

Pi, please before you go through the gun ballistics stuff I would like for you to address the point I made to you based on the various masses of objects which have different amounts of energy and travelled different depths according to figure 7 in the SL9 article. It shows clearly that with the same sized objects with different masses because of density, there are different amounts of energy and it shows the different depths for the different energies. That should give us the answer regarding energy and depth of penetration. If not explain why not.

I don't think it will work because the data we do have clearly doesn't plot on a straight line and we have no idea how an extension of the (obvious) curve will behave.

 

Further, I think our efforts to adapt the model to a 0.1 impactor are (more likely than not) stretching the model beyond its limitations.... for the same reason.  We do know the researchers did a test using 0.1 density but, unfortunately, they did not publish that data.  We can only speculate why.

 

Finally, the point of agreement mentioned above makes the SL-9 event largely irrelevant.



#556 indydave

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Posted 23 July 2016 - 01:10 PM

Pi>>I think we have a point of agreement that Brown's model would produce impactors that will, in fact, leave some craters on Earth that would be pretty hard to miss.  (Note:   I'm not claiming these proposed bolides would be sufficient to cause a lot of damage more than a few miles from the area of impact.)  This is worthy of a new topic.>>

There has never been a dispute that some sort of craters might be formed...but your whole point has been the impacts would BE LETHAL to any life on the planet.  You seem now to be retracting that, and that is indeed a concession re. what this whole thread has been about.  And besides, the smaller rock inclusions would be unlikely to even cause a crater if the impact happened during the Flood year, and after that, there is a 70% chance the impact would be in the ocean and no discernable crater would happen.  If the Yucatan crater is real, then one very large object did indeed impact, but that was not significant enough to obliterate all the ark occupants.  It may have ended certain species of life however. 

 

>>I don't think it will work because the data we do have clearly doesn't plot on a straight line and we have no idea how an extension of the (obvious) curve will behave.>>

You are dodging.  If you agree that there is a 1/1 relationship between mass and energy, then the chart in Fig 7, which you eyeballed, shows how energy affects depth of penetration.  That can be used to show how various speeds (which have different energies) affect depth.  It is most likely that the non-linear aspect is simply due to the fact that as you get deeper, the atm is thicker.  But of course that is also true on Earth.

 

>>Further, I think our efforts to adapt the model to a 0.1 impactor are (more likely than not) stretching the model beyond its limitations.... for the same reason. >>

The earth impacts site said it should not be used under 1...and I agree.  However the Korykansky paper did NOT say that and indeed in the text they mentioned that they did modeling runs using .1 impactors.  So they seem to have thought it would be valid to use their model for a very low density object. 

 

>>

Finally, the point of agreement mentioned above makes the SL-9 event largely irrelevant.>>

Nope.  We don't agree that the smaller rocky objects which MAY have made smaller craters would cause death to the whole planet.  Do you agree they would NOT?  Let's wrap this up!  If you agree, then you have conceded the essential POINT of this thread.  AND you have also said they would not cause a heat problem...so we are really getting somewhere now!

 

>>You are correct that "significant" does not mean the kind of 100 mile diameter lunar craters that are the observational evidence on which this discussion is based.>>

These are indeed weasel-words.  It is bogus that you for some reason now want to wriggle free of the point that was agreed to long ago...which is that a 70km .1 density object WOULD MAKE A 100 MILE CRATER ON THE MOON.  Previously you agreed it WOULD but that it would NOT make a crater on Earth.  It appears now you want to reverse that, just to avoid conceding.  Do I have to go back to show that to you???  What is your basis now to say it would NOT?  You can't just throw that excrement on the table and then walk off...or fly off.  Defend WHY you say that now...after previously agreeing with me.



#557 indydave

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Posted 23 July 2016 - 01:29 PM

Very simple question for you Pi...if you could prove that the kind of object that would make a 100 mile crater on the Moon would penetrate to 2 bars on Jupiter or on Earth, why is that not just 33 feet under the ocean? That would indeed make a large splash, but no crater...regardless of what the effects site says. This should be easily answered so I hope you don't take too long.

#558 indydave

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Posted 23 July 2016 - 01:42 PM

If a 100mi diameter cloud of water molecules and rocky stuff with an average density of .01 were to impact the Moon at 17km/s, which is the correct answer?:

 

1.  No crater at all would form

2.  A 100mi diameter or larger would form.

3.  A crater smaller than 100 mi diameter would form.

 

If you change that density to .001, what would the answers be?

 

Please note that Brown has not stated what the original density of the proto-asteroids were.  He has only said they started very low-density and then were consolidated by gravity and moved outward so that they are their present densities at their present distances. 






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