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.