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Problem I - Cratering. 2. The Earth


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#601 indydave

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Posted 17 October 2016 - 02:30 PM

 

 

Note:  This is a pretty big concession on my part as in my "Fire and Brimstone" analysis, I suggested rocky fragments all the way to the size of Manhatten.

 

 

The ability of something to be launched upward is a factor of its surface area to weight ratio.  This means that larger stuff would never make it up the 60mi "chasm" until its size was small enough to allow it to be launched.  It would keep being battered by other stuff, which would lessen its size and weight. 



#602 indydave

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Posted 17 October 2016 - 03:47 PM

Pi>>

Indy's proposal for Brown's Hydroplates as an explanation for the lack of 100 mile craters on Earth is rejected due to absence of reasonably expected supporting evidence.>>

Looks to me like more Indian-giving.  The discussion showed how low density objects of great size would be obliterated by Earth's atm. so that they would NOT make 100mi craters on Earth.  Pi even as ADMITTED that, and had to "downgrade" his argument from the absence of 800  100mi craters to now being the absence of 240 1 km craters.  (Or maybe that would be 14.4x that number because he suggested at the end that there MUST be at least 14.4 of those per cloud...which is NOT what Brown suggests at all). 



#603 indydave

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Posted 17 October 2016 - 03:48 PM

Pi>>We even had a physics PhD (DaveB) provide us with an equation that allows us to compute how many impacts we should expect.  It turned out the number depends on the relative velocity of the approaching body and ranges from a minimum of about 13.5x for objects that are going so fast they will exit the solar system and never return to 200x for those that are just barely approaching Earth at very, very slow relative velocities.>>

I have accepted this 13.5x factor for comets or asteroids RETURNING from outside of 1 AU so as to collide with Earth.  That is not what most of the collisions would be, however.  The Brown model has stuff being pushed outward SLOWLY due to radiometer and Yarkovsky effects (RY).  What this means to any sensible mind is that ONLY stuff being pushed outward from INSIDE OF 1 AU could possibly hit Earth.  You don't LIKE this FACT (if Brown is right) but you have no ANSWER for it yet.  This means that of all the stuff being pushed slowly outward, the amount that could hit the Moon is twice that which could hit Earth...because the Moon can be hit by stuff coming from inside 1 AU as well as stuff slightly outside of 1 AU.  So even ACCEPTING your 19x factor, this takes your number from 800 to 400...before a couple of other FAIR adjustments...one of which YOU AGREE with. 

 

>>Sorry, Indy.... the number of such encounters is an essential point in this topic.>>

Not if the ALL are obliterated high in our atmosphere.  There could be 10,000 of them and we still may see very few that would make it entact to hit land. 

 

>>At 17 km/sec, there will be no impacts during the flood year. >>

Once again, you don't even understand the HPT model.  Stuff launched (both pro and retro) will have varying speeds.  Half would be pro, so these would be immediately outside 1AU.  As they are SLOWLY pushed outward for weeks by RY, these COULD hit the Moon, but not Earth.  Some stuff would be in a prograde orbit around the Sun, but slightly slower than Earth and some of this COULD hit Earth during the Flood year.  There also would be stuff launched fast enough retrograde that Earth could encounter it 6 months later, and at a speed much higher than 17km/s.  This stuff also could hit Moon at greater speed too, so it could take a much smaller object to make the larger Moon craters.  I don't know if 50% hitting Earth during the Flood year is the correct number, but SOME would.  Again, it is not that relevant if the stuff is obliterated by the atm. 

 

>>While the model is not exact, it does clearly show there will be no impacts during the flood year.>>

That model does not have a way to show stuff being pushed outward. 

 

>>

indydave, on 15 Oct 2016 - 4:39 PM, said:snapback.png

indydave, on 15 Oct 2016 - 4:39 PM, said:snapback.png

You don't like the 400 part because you don't understand the model, and some of that is due to how I expressed it.  I do not mean stuff shot directly AT the Sun, which you say would then cause it to "return to its point of departure".  I mean stuff shot rearward (retrograde) so that its Sun-orbit-speed is reduced so that it falls into a lower orbit around the Sun.

Pi>>No.  I don't like the 400 part because I do understand the physics.  I say it will return to its point of departure because of the Law of Conservation of Energy and orbital mechanics.  As demonstrated by the "solar system" model you found.>>

 

I shouldn't have to keep repeating this...the spinning Earth would shoot stuff outward and some of that would give it speed so it immediately would be beyond 1 AU.  And about half would immediately fall inward to a lower orbit...UNTIL it gets pushed outward by RY.  Do you or don't you acknowledge the reality of RY?

 

>>You get 400 by simply dividing the number of expected impactors by 2 to allow for those that impacted the moon during the launch phase.  That is not the correct approach.  >>

Nope.  Those hitting Moon could happen during just a few hours after launch, BUT those would be VERY few.  The Moon covers such a tiny part of the sky, so less than (guessing) .01% would be direct hits.  The rest (of what Brown suggests) would come from encounters of stuff being slowly pushed outward from 1AU-400,000 km to 1AU+400,000 km. 

 

>>In order to find out how many struck the moon during launch, the best way is to find out how many craters there are on the back side of the moon because there are very few trajectories that will cause an object to narrowly miss the moon, hook around, and strike the moon on the back side during launch.>>

 

Perhaps that approach could tell us how many were direct hits.  Almost none of the large maria are on the backside and THOSE could have been on the front too but the Moon slowly rotated afterwards into a new tidal lock due to a change in its mass, so that the more massive side faces us today.

 

>>We know there have been impacts since the flood and those impacts will be random because they are objects returning from solar orbit.  Since it would be very unusual for an object launched from Earth to hit the back side of the moon, we can consider them to be after the launch phase.  We also know that after launch, objects will be approaching the Earth/moon from random directions.>>

HPT has a MUCH better way to explain the fairly even distribution of craters EVEN AT THE POLES of the Moon.  Your idea does NOT.  It is rather silly for you to argue about objects returning "since the Flood" since the rate is almost NIL so hardly any would be expected to hit the Moon in the past 4500 years.  Even in Brown's model...there would be hardly any to hit either Earth or Moon on a return since the Flood.

 

>>Unfortunately, according to this list the majority of 100 mile (160 km) craters are on the far side of the moon.  This very strongly suggests few, if any, impacts during Brown's launch period.

>>

Nope.  As I have shown, the THREE crater size sites show us (with no invalidation for .1 density objects) that a cloud of water vapor, ice particles, rocky particles and small rocks (under 200m) could indeed make 100 mi craters.  You have given NO reason to NOT accept that EXCEPT your own incredulity.  I have a professional astronomer (Faulkner) the instruction video, and THREE crater size sites, which agree with me.  I also just recently saw THIS:  LINK

 

Equation to determine Crater Size:

 

 

 

 

D=0.07Cf(ge/g)1/6(W pa/pt)1/3.4

 

Where:

D = Crater Diameter

Cf = Crater Collapse Factor ( this is equal to 1.3 for craters >4km on Earth)

ge = Gravitational Acceleration at the surface of Earth

g = Acceleration at the surface of the body on which the crater is formed

W = Kinetic Energy of the impacting body (in kilotons TNT equivalent)

pa = Density of the impactor (ranging from 1.8g/cm3 for a comet to 7.3g/cm3 for an iron meteorite).

pt = Density of the target rock

 

 

If you look closely D is the result and all the others are the type of INPUTS we found on ALL the crater sites.  However, this "jumps to the chase" by not asking about size or density or even speed of the object...but it just wants the kinetic energy "W".  That is exactly the same as what the instructor's video said.  This formula says by implication that an object of differing size or density but which has the same mass and speed as another object...WILL MAKE THE SAME SIZE CRATER.  So this means that if you scale down the density from 1 to .1, and then you scale up the volume so it is 10x...you will get the same size crater.  That is what the THREE sites showed.  And now this equation says the same thing.  I cannot explain why one site said a 95km object was needed, another said 70km and a third said 28km...but I have indeed shown scientific support that a .1 density 70km crater (or even a 95km one, which could also fit what Brown describes) could indeed make a 100mi crater on the Moon.   

 

>>

indydave, on 15 Oct 2016 - 4:39 PM, said:snapback.png

Then the radiometer and Yarkovsky effects would slowly push it outward (adding tangential speed), keeping its orbit circular.

PI:Insignificant.  Of course, we've already discussed these influences extensively without any agreement there either.>>

You have to just deny science to say that RY are not going to affect the orbits of proto-asteroids.  These effects would be many many times STRONGER on a proto-asteroid than on an asteroid...due to the greater surface/mass ratio.  Do you deny RY?  I may agree with you that the number of impacts IN THE FLOOD YEAR would be less than 50%.  I just don't know what would be implied by the model.  But if RY is true, then only 50% could possibly hit Earth while 100% could hit Moon because half of the stuff would be launched prograde and that could NEVER hit Earth because it would immediately be outside 1AU.

 

>>Also, the orbits of ALL objects launched from Earth will be elliptical, not circular>>

That Solar System model has NO way to demonstrate RY effects on orbits.  It would slowly circularize them so that 4500 years later the a-belt would be circular.  If they were in elliptical orbits in the time when they could impact Moon or Earth...SO WHAT?

 

>>Of course, you mean the YEC astronomer who doesn't agree that these things will even get launched in the first place.  Without successful launch, none of the discussion of Brown's proposal matters at all, does it?  How's that for something on *my* side of the ledger?>>

Faulkner's second paper against Brown spoke of the launching as being "a black box"...meaning in his mind Brown did not explain it (he DID), but that he (Faulkner) would take the launching as a given.  THEN he would give his reasoning why he believed IF it happened, then there would be consequences (heat) and different observations in comets than what we see.  In this (and other) threads here, we also BOTH accepted "for sake of argument" that the stuff WOULD get launched.  NOW, you want to express that you DON'T accept the launching.  Fine.  We all knew that from the start.  The question is whether you want to discuss (like DF did) what the consequences or effects would be IF it were launched.  Apparently you have your running-away shoes on again, so now you want to no longer talk about craters...since you failed to make your case.



#604 piasan

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Posted 17 October 2016 - 11:05 PM

You have to just deny science to say that RY are not going to affect the orbits of proto-asteroids.  These effects would be many many times STRONGER on a proto-asteroid than on an asteroid...due to the greater surface/mass ratio.  Do you deny RY? 

I have not denied the science at all.  What I have said is that the effects you're talking about are insignificant.  That was another 80 post or so side discussion of the type you complain about..... in which we were also unable to agree on any significant issues.

 

 The question is whether you want to discuss (like DF did) what the consequences or effects would be IF it were launched. 

That's what we've been doing for most of the 600 odd posts in this topic.

 

Apparently you have your running-away shoes on again, so now you want to no longer talk about craters...

You are confusing frustration with "running-away." 

 

I'm still right here and more than willing to discuss any questions or issues raised by anyone else.

 

...since you failed to make your case.

If anyone believes I have not made my case, they are encouraged to find a person whose knowledge of physics they respect and get an independent opinion.  Any high school physics teacher, engineer, or even first year university engineering student should do.  (Note: only one of us has made that offer, and it isn't Indy.)

 

Every scientific review I've found of Brown's Hydroplates says it's a "no-go."  But that's an issue more suited to the "Hydroplates and Meteorites" topic.



#605 indydave

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Posted 20 October 2016 - 06:08 PM

I noticed that Pi high-tails it just after I show ANOTHER proof he is wrong.  That equation where it shows that the kinetic energy (no matter how dense an object is) is what determines crater size.  He just is too proud to admit he has been shown (again) to be wrong.

 

D=0.07Cf(ge/g)1/6(W pa/pt)1/3.4

 

Where:

D = Crater Diameter

Cf = Crater Collapse Factor ( this is equal to 1.3 for craters >4km on Earth)

ge = Gravitational Acceleration at the surface of Earth

g = Acceleration at the surface of the body on which the crater is formed

W = Kinetic Energy of the impacting body (in kilotons TNT equivalent)

pa = Density of the impactor (ranging from 1.8g/cm3 for a comet to 7.3g/cm3 for an iron meteorite).

pt = Density of the target rock

 

 

He should be expected to explain why this equation MUST BE WRONG.  Why, Pi???  If it is RIGHT, then your position (saying the low-density objects at high velocity couldn't make 100mi craters on the Moon) HAS TO BE WRONG.

 

>>You are confusing frustration with "running-away.">>

Always having Indydave show you to be wrong MUST be very frustrating for sure!



#606 indydave

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Posted 21 October 2016 - 03:03 PM

I wrote>>>You need to man up...and concede. And don't just fly off from the chess board either. CONCEDE<<

The term pigeon chess was used by Pi when he was unhappy with how his opponent on another thread had behaved and yet he does the same thing here. He wastes everyone's time with our having the belief that if he were presented some good solid arguments that his position is wrong that he would have the integrity to admit to that. And yet when that was done to the EXTREME, he did not show the character to admit he was wrong. He will still try to appeal to Dr. Danny Faulkner as someone I should listen to regarding heat in HPT, and yet Faulkner says to me that these low-density objects traveling at such high speed would indeed cause massive cratering on the moon up to and beyond 100 miles diameter, and yet that carries no weight at all in the high school science teacher's mind. He may wish to have discussion about Brown's theory further in this forum, but he will have a hard time getting much interaction from me when he refuses to show "good form", as they say in England, in the manner in which he participates on the field of polemic battle.

#607 indydave

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Posted 22 October 2016 - 08:35 PM

This crater size calc site has a paper describing how they figure the sizes.  Click on "The Theory" at the top of the page.

 

>>As the primary example, the volume V of a crater formed by a given impact can be
expected to depend on the impactor radius a, velocity U, and mass density d. Note that
those 3 variables also defined the kinetic energy, momentum, and mass of the impactor, so
equally those could be used, as well as any other choice of three independent variables
containing the three independent units of mass, time and length.>>

 

By "length" I assume he means "distance" which would (along with time) give you the velocity. 

 

Later in the paper:: (a chart showing various densities)  then 

 

>>A user may pick other values in the app are made by choosing “Other” in the pull-down
menu for the impactor type.>>

 

Nothing is said about refraining from using low densities.  The author seems to have written papers on this topic since 1980.

 

When I ran it just now, the result was a 37km .1 object would produce a 160km rim diameter.  Pi knows he is wrong to say these objects could NOT produce such craters, and so he has decided to disappear...rather than face the music as a courageous and fair polemicist would. 



#608 indydave

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Posted 23 October 2016 - 03:37 PM

Pi has decided to leave this topic but he really should return to address what I am going to post here. He conceded that the low-density objects would not cause devastation on Earth which was half of the equation. The other half which he challenged was whether those same objects could produce large craters on the moon in the range of 100 miles diameter. I have shown from many different sources and angles that the experts agree with my view that they would but he has decided that there is something special that happens if you reduce the density from 1 down to .1. We both found reference to a document produced by the crater effects site that said that they should not use their calculator on objects below 1. That's fine and I accept that but I found three other sites and and an instructor's video and an equation as well as Dr. Danny Faulkner who all suggested that these objects would indeed make such large craters. The site that I referenced in the previous post shows that the object needed to produce a 100 mile diameter crater would be only 37 kilometers. I decided I would attempt to get a reply from the originator of that site who has about 40 years experience in this field. I was rather surprised that he quickly replied to me. Because I mentioned that we were involved in a discussion I don't feel at all hesitant to post the response from him.


Dr. Holsapple,
I found your email address at the size calculator site here: http://keith.aa.wash...aling/index.htm

I wonder if you would answer a simple question. Our discussion group is considering if perhaps the 100+mile diameter craters on the Moon could be caused by a very large diameter 100kg/m3 (.1 density) object traveling at 17km/s. When we used the site to calculate it, the result was a 37km object.

Does this sound right to you or is there something additional to consider if the density is very low? If you have an object that is 1/10th the density (eg. 100kg/m3) but its volume is 10x, would it produce the same crater size as one that is 1000kg/m3 and the volume is 1x? We are considering an object about the density of loose snow. Many thanks!


Hi David:

In fact, we have very little information on the outcomes of very high speed, low density objects (e.g. comets), so the guesses are in fact extrapolations with little data.

But the bottom line is that for any high speed impact, the crater is much larger than the impactor. In those cases, the details of the impactor composition do not matter much, the answer is determined by some combination ot the impactors mass and velocity, not its mass density or size.

So, I think the answer of 37 km is as good of a guess as possible.

Cheers,

Keith Holsapple

#609 piasan

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Posted 24 October 2016 - 08:38 AM

Pi knows he is wrong to say these objects could NOT produce such craters, and so he has decided to disappear...rather than face the music as a courageous and fair polemicist would. 

Again, Pi has not disappeared.  Pi simply declines to continue :icon_deadhorse: with Indy after over 600 posts on this topic alone plus dozens of other posts in associated threads without achieving a point of agreement on even the most mundane of topics.

 

Pi is still right here.

 

Pi has decided to leave this topic but he really should return to address what I am going to post here.

As this is something new, I'll comment on it.....

 

I decided I would attempt to get a reply from the originator of that site who has about 40 years experience in this field. I was rather surprised that he quickly replied to me. Because I mentioned that we were involved in a discussion I don't feel at all hesitant to post the response from him.

Dr. Holsapple wrote.......In fact, we have very little information on the outcomes of very high speed, low density objects (e.g. comets), so the guesses are in fact extrapolations with little data.

But the bottom line is that for any high speed impact, the crater is much larger than the impactor. In those cases, the details of the impactor composition do not matter much, the answer is determined by some combination ot the impactors mass and velocity, not its mass density or size.

So, I think the answer of 37 km is as good of a guess as possible.

I notice Indy didn't include that these objects contain some pretty significant rock inclusions.  That said.....

 

I will agree that based on Holsapple's statement that these "guesses are in fact extrapolations with little data..." and he feels "37 km is as good of a guess as possible,"  These things could potentially leave a 100 mi diameter crater on the moon.

 

This, of course, does not address whether your suggested impactor should be treated as a single body or one that is highly fragmented.

 

There are other issues that have not even been mentioned yet......

For example, a 17 km/sec launch velocity from Earth will not give a 17km/sec impact velocity on the moon.  Impact on the moon will be much, much slower than that.

 

Then there is the "target size" of the moon.  As observed from Earth, the moon occupies only 0.5 degrees of arc.  That's 1/720 of a circle.  However, we are talking about 3d space.  From physicist DaveB:

In case you are interested (so you won't have to guess about it any more), if there is a round object in the sky (such as the Moon or Sun) whose diameter subtends an angle of θ then that object occupies a solid angle of 4*π*sin2(θ/4) steradians or a sin2(θ/4)th of the whole celestial sphere.  This means that if the distance between the center of the object and the center of the Earth is D and the radius of the round object is R then the fraction of the whole celestial sphere occupied/blocked by the object is (1/2)*(R/D)2/(1 + sqrt(1 - (R/D)2)).  In the case of the Moon, if we take its average distance as its orbit's semi-major axis length D = 384399 km, and take the mean lunar radius as R = 1737.4 km then the Moon typically occupies a 1/195804 fraction of the whole celestial sphere.

So, there is only about a one in 200,000 chance of an object launched from Earth in a random direction even striking the moon in the first place.  (Note:  We've been over this before, but I don't see any need for readers to wade thru 3 years of this stuff to find it.)

 

And we haven't even begun to mention the number of really, really big craters on the moon (500+ miles (800 km) in diameter).

 

 

He conceded that the low-density objects would not cause devastation on Earth which was half of the equation. The other half which he challenged was whether those same objects could produce large craters on the moon in the range of 100 miles diameter.

Not so fast, Buster.

 

You are making specific claims and it is entirely fair game to see if the reasonably expected result of such low density objects, with their rocky inclusions, would leave evidence they had actually encountered Earth.

 

Holsapple's comments are applicable only to lunar craters.  They do not, in any way, address the crater fields we can reasonably expect to find on Earth if Brown's claims are true.  In other words, you have produced no evidence that gives grounds for a reconsideration of my most recent position.

 

The bottom line remains......

Brown's Hydroplate model is STILL rejected as a source of cratering on the moon due to the absence of supporting evidence on Earth.



#610 indydave

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Posted 24 October 2016 - 02:46 PM

Yeah, it's a dead horse to you now that it's clear you have failed to defend your position. If you felt you had me on the ropes you would be all gung-ho to continue!

It is good that you have finally conceded that my second main point is also true, which is that these low-density objects could indeed make the large craters on the moon just as I stated right at the beginning. But then you want to suddenly switch to discussing how many of them there would be or what about those one or two 500 mile diameter craters on the moon. Anything but to allow it to sink in that you failed to show there would be lethal devastation on the Earth and now you also failed to show that these low-density objects could not produce 100+ mile diameter craters on the moon. So you want to just distract us away from the fact that you failed. I may address more of what you had to say in another post but I DON'T want to distract us away from the fact that you have failed.

The lesson here is that if you quit too soon you may be just a post or two away from winning the argument. I'm glad I hung in there. And I should also say I appreciate that Pi came back to say that he conceded on the second of the two major points. He wants now to move the goal posts and try a new game but at least he admits now that he lost in the first game.

#611 indydave

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Posted 28 October 2016 - 12:18 PM

Even Pi's "save face minor arguments" he has tried to use...AFTER losing on the two MAJOR arguments...are falling by the wayside.  One has been his claim that if a large low density object with rock inclusions were to hit the Moon then it would show up as a field of smaller impact craters (perhaps within a larger one).  So I went back to Dr. Holsapple to ask:

 

"If you care to comment further, I have another question.  If this gigantic "snowball" (.1 density) were to have some denser (3.0) inclusions which were much smaller...a maximum of 200m, is it likely that we could see separate discrete craters from those objects or would they get obliterated by the main excavating blast?  Would it appear like a "crater field" or just one big crater? Thanks in advance for your reply. "

 

Here was his answer, received today: 

 

"the governing principle is that, since the crater is much larger than the impactor, any internal structure of the impactor is not relevant.  What determines the crater is the impactors mass and velocity only.

 
Cheers,
 
Keith H."
 
This was the answer I expected.  So he first says (in prior email) that density is irrelevant...only mass and velocity...to determine size.  Then he also says that internal structure is also irrelevant to affecting the kind of cratering that occurs.  Pi's position now is at odds with DF as well as this (presumed) evolutionary professional astronomer. 


#612 indydave

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Posted 29 October 2016 - 03:19 PM

Pi>>There are other issues that have not even been mentioned yet......

For example, a 17 km/sec launch velocity from Earth will not give a 17km/sec impact velocity on the moon.  Impact on the moon will be much, much slower than that.>>

 

They could be much faster OR slower.  We used 17km/s just as an example.  Brown does not say what speed objects would be which hit the Moon.  If it is faster than 17km/s, then a much smaller diameter object could make the 100mi craters.  And vice versa. 

 

>>Then there is the "target size" of the moon.  As observed from Earth, the moon occupies only 0.5 degrees of arc.  That's 1/720 of a circle.  However, we are talking about 3d space.  From physicist DaveB:

Quote

In case you are interested (so you won't have to guess about it any more), if there is a round object in the sky (such as the Moon or Sun) whose diameter subtends an angle of θ then that object occupies a solid angle of 4*π*sin2(θ/4) steradians or a sin2(θ/4)th of the whole celestial sphere.  This means that if the distance between the center of the object and the center of the Earth is D and the radius of the round object is R then the fraction of the whole celestial sphere occupied/blocked by the object is (1/2)*(R/D)2/(1 + sqrt(1 - (R/D)2)).  In the case of the Moon, if we take its average distance as its orbit's semi-major axis length D = 384399 km, and take the mean lunar radius as R = 1737.4 km then the Moon typically occupies a 1/195804 fraction of the whole celestial sphere.

So, there is only about a one in 200,000 chance of an object launched from Earth in a random direction even striking the moon in the first place.  (Note:  We've been over this before, but I don't see any need for readers to wade thru 3 years of this stuff to find it.)>>

 

It might surprise Pi that I have no problem at all with this calculation (although I haven't checked it...it sounds fine)...EXCEPT that it is built on a misunderstaaaaaanding of what Brown asserts.  This idea is saying that the Moon is only hit as a DIRECT SHOT from Earth.  If that were the only way the Moon could be hit, then you could indeed need something in the range of 200,000x times the stuff shot outward...or 200,000x the number of Moon craters.  And if you count all the mass Brown says was launched and compare that to the objects which hit the Moon, it may BE about 200,000x.  Of course, if that idea (craters are formed only by a direct shot) were true, you'd have almost NO craters on the Moon's far side!   But that is NOT what Brown says about objects that can cause craters on the Moon.  As Pi SHOULD know, the stuff hitting the Moon could come from all the mass that was inside and just outside of 1 AU which is in orbit around the Sun and gets slowly pushed outward.  The collisions with the Moon could occur over years or decades...maybe even centuries.  We haven't mentioned this much, but it could also be stuff in orbit around the Earth or the Moon...but not in a stable low eccentricity orbit, which would eventually collide with the Moon (including at the poles)...as well as with Earth.  



#613 piasan

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Posted 30 October 2016 - 02:19 PM

 

Even Pi's "save face minor arguments" he has tried to use...AFTER losing on the two MAJOR arguments...are falling by the wayside.  One has been his claim that if a large low density object with rock inclusions were to hit the Moon then it would show up as a field of smaller impact craters (perhaps within a larger one).  So I went back to Dr. Holsapple to ask:

 

"If you care to comment further, I have another question.  If this gigantic "snowball" (.1 density) were to have some denser (3.0) inclusions which were much smaller...a maximum of 200m, is it likely that we could see separate discrete craters from those objects or would they get obliterated by the main excavating blast?  Would it appear like a "crater field" or just one big crater? Thanks in advance for your reply. "

 

Here was his answer, received today: 

 

"the governing principle is that, since the crater is much larger than the impactor, any internal structure of the impactor is not relevant.  What determines the crater is the impactors mass and velocity only.

 
Cheers,
 
Keith H."
 
This was the answer I expected.  So he first says (in prior email) that density is irrelevant...only mass and velocity...to determine size.  Then he also says that internal structure is also irrelevant to affecting the kind of cratering that occurs.  Pi's position now is at odds with DF as well as this (presumed) evolutionary professional astronomer. 

 

Good.

 

Now, go back and ask him one more question......

"What will be the result of a 0.1 density impactor with rocky inclusions up to 200 meters across be when it impacts Earth?"

 

Brown's proposal is still rejected as an explanation due to the absence of reasonably expected evidence on Earth.



#614 indydave

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Posted 30 October 2016 - 07:54 PM

Just another one of your attempts to try to save face again, eh?

After you are willing to admit plainly that you have failed in the primary questions of this thread, then maybe we can look at whether or not we should expect more land craters in the range of 1 kilometer or so to be visible today. It might be that if there are not enough then I would agree with you that there could not have been so many 200-meter rocks to land on the Earth. I have already questioned you about whether there should be 14 of them inside each of the objects that would produce a 100-mile crater on the moon. Brown's position is not that there are fourteen or even one in each of the clouds. But more substantially, I have shown you evidence and I presented it to Brown recently, that what he used as evidence for 200 meter solid rocks... the observed rotation rates of rocks in the asteroid belt...is not necessarily valid. A single rock could rotate ten times a day or it could also be a conglomeration of smaller rocks which are adhered together with ice. The source I showed you said that they could have 1000 times less cohesion than a solid rock. That means that the object could be just a conglomeration of 3 meter or 50 meter diameter rocks. If that is the case then there would be hardly any cratering on the Earth, and that is exactly what we see. It is my expectation that Brown will see the point that I made and adjust his view and if you want you can take some minor satisfaction in that, but it certainly is not any kind of disproof of his theory. I guess it will depend on how desperate you are for a minor victory. You got beat soundly in this thread but you just don't want to admit it so you are going to take any minor victory you think you can get. But everyone here knows you got your hat handed to you already.

So do you want to discuss further these supposed smaller craters that you believe we should see on the earth if HPT is true? Or is this just your pot shot you want to take over your shoulder as you are running off the field of conflict?

#615 indydave

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Posted 31 October 2016 - 03:24 PM

I spoke briefly to Brown today and he confirmed that his position is that not every one of the large proto-asteroid objects would have a 200m rock at the center.  That was the MAXIMUM size of any solid "seed rock" for each cloud.  So Pi can hypothesize all he wants about how there MUST be at least 1 millionth of the mass of each cloud that is made up of these, but that is NOT Brown's position at all.  Also, I mentioned again to him about the paper which says the objects seen in the a-belt today which spin up to 10x per day could have ONE THOUSANDTH the cohesion of solid rock....but he said he hasn't had time to read the paper or think about whether he should change what he wrote about that.  He wasn't too concerned about spending time about something that is only of concern to Pi...or even if it was an interest of mine.  He is doing major work on the comet chapter right now regarding other things.

 

The PAPER said that in order to maintain shape while spinning 10x per day, the object could be 1000 times weaker than solid rock.  In fact it said re. such fast spinning object in the a-belt...

 

"While they are sometimes called “monoliths,” their internal
structure can, however, be almost anything except “true”
rubble pile."

 

This means of course that if it is a pile of rocks held together loosely by ice or snow, then they would be easily torn apart high in the atm...and no large craters would occur even if they hit land. 

 

Pi wants to IGNORE the science about this and just stick to his attack on even a MINOR point in what Brown wrote...so he can manage to eek out a claim of a MINOR victory.  For Pi to think that even if he "wins" on this, that he has blown a significant hole in HPT is just ridiculous.  But that's the only scrap that he hopes to take home with him as a trophy after months and months of battle on this subject.



#616 indydave

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Posted 03 November 2016 - 12:07 PM

Pi said the following in the meteorites and hydroplate topic.

>> The claim that the objects you propose are a "solution" to the issue is still rejected due to the absence of reasonably expected evidence such events actually took place.<<

Your position now is that something like 240 objects may have hit over land and each one of these should have fourteen 200 meter objects of solid rock within them, and so we should have easily identifiable crater fields for each one of these. However Brown does not agree that each of these objects would have that size of a solid rock inside it. He only has said that is the maximum size it could be. There could as easily be a collection of smaller rocks that serve as the seed for the proto asteroid. So even if we are going completely by what Brown said in his book, you have not accurately depicted his position. However his idea about the 200-meter solid rocks is apparently wrong and I have expressed that to him. If he is indeed wrong about that then he can easily make that correction without having the slightest effect on his theory. As a reminder, the reason I believe he is wrong about that is that there is good evidence that you can have fast rotating asteroids which are made up of not one single rock but rather they are made of a rock pile that is held together by ice or even snow and that means that when that object hits Earth's atmosphere it would be torn apart and no cratering at all would occur.




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