[Yosemitest]

"The peak particle size on that chart is about 1 cm or less."

If your statement is true, what would particles sizes be from figure 4 oftheir paper that state:

"Mean = 3.20971"

"Mode = 2.93205"

I don't believe they use the letter "m" to mean meter, but I could be wrong, as I am not schooled in this field.

Explain their statement to me that states:

"An equal amount of dust is assumed to be released between the sizes of 10−9 to 10−1 m in radius,
though only particles larger than 10−5 m are simulated:
given our assumed size distribution,
this gives a fraction > 0.99999 of the mass and 4×10−7 of the number of particles simulated. "

[messierhunter]

It’s a simple log function. Log(Radius(m)) = -3 would be 1 cm. 1*10^-3 m = 0.1 cm. At Log(Radius(m)) = -2 where most of the density ends, the radius would be 1 cm.

Their statement is pretty self-explanatory. If you don’t get it then you need to go back and take some remedial science and math courses. Your constant confusion with log functions suggests you probably should do so anyway. Please, drop this nonsense, it’s making free republic look ridiculous. There are plenty of forums on the internet for pseudoscientific discussions. I’m begging you, this is embarrassing.

[Errant]

I don't believe they use the letter "m" to mean meter...

Now you're getting somewhere... Don't feel bad about confusing metric symbols. Even NASA gets symbols wrong sometimes. They once crashed into Mars because of confusion not between metric units, but between the two systems .

The great thing about metrics is once you know what you're dealing with, it's easy to convert between the units.

:-)

Still, the model you're using only predicts dust particles. It isn't going to help much in predicting how many large rocks are trailing a fragmented comet nucleus. Best to use observation/radio returns for that.