“Dark Fall” by David Weber

A 25 km on a side cube has almost double the volume and hence mass and impact energy as a sphere. A prolate spheroid has even less volume hence mass and energy.
I once provoked a fecal storm by citing a female college math professor as an example of an oblate spheroid which would have even more volume, mass and impact energy as a cube of similar height.

TFLYTSNBN wrote:A 25 km on a side cube has almost double the volume and hence mass and impact energy as a sphere. A prolate spheroid has even less volume hence mass and energy.
I once provoked a fecal storm by citing a female college math professor as an example of an oblate spheroid which would have even more volume, mass and impact energy as a cube of similar height.

In calculating the yield of an impact OF THIS SIZE, the shape is essentially meaningless. Mass, velocity and density rule all here.

In the case of an Alpha type body hitting Earth, it will cross the atmosphere from the space/atmosphere boundary to impact in approx 3.5 seconds. The amount of drag of the different shapes does not have time to matter. If the asteroid is an iron type (the densest), is roughly spherical and is at the given 25km diameter, it will weigh roughly 64,386,280,100,000,000 kilograms (if my math was close)

doug941 wrote:

TFLYTSNBN wrote:A 25 km on a side cube has almost double the volume and hence mass and impact energy as a sphere. A prolate spheroid has even less volume hence mass and energy.
I once provoked a fecal storm by citing a female college math professor as an example of an oblate spheroid which would have even more volume, mass and impact energy as a cube of similar height.

In calculating the yield of an impact OF THIS SIZE, the shape is essentially meaningless. Mass, velocity and density rule all here.

In the case of an Alpha type body hitting Earth, it will cross the atmosphere from the space/atmosphere boundary to impact in approx 3.5 seconds. The amount of drag of the different shapes does not have time to matter. If the asteroid is an iron type (the densest), is roughly spherical and is at the given 25km diameter, it will weigh roughly 64,386,280,100,000,000 kilograms (if my math was close)

But shape certainly does matter when translating the dimension across into it's volume and thus it's mass.

doug941 wrote:

TFLYTSNBN wrote:A 25 km on a side cube has almost double the volume and hence mass and impact energy as a sphere. A prolate spheroid has even less volume hence mass and energy.
I once provoked a fecal storm by citing a female college math professor as an example of an oblate spheroid which would have even more volume, mass and impact energy as a cube of similar height.

In calculating the yield of an impact OF THIS SIZE, the shape is essentially meaningless. Mass, velocity and density rule all here.

In the case of an Alpha type body hitting Earth, it will cross the atmosphere from the space/atmosphere boundary to impact in approx 3.5 seconds. The amount of drag of the different shapes does not have time to matter. If the asteroid is an iron type (the densest), is roughly spherical and is at the given 25km diameter, it will weigh roughly 64,386,280,100,000,000 kilograms (if my math was close)

Your math is within one significant didget of mine.

Impact energy at 50 km/s is then 7.8ee25 Joules or

78 YottaJoules.

Weber's number of "almost 700 YottaJoules" seems to be high by an order of magnitude.

I get 18.6 Billion Megatons not Weber's 1.5 Trillion Megatons.

May be the impactor had a neutronium core?

May be some type of weapon?


Oops.

Note to David Weber:
Do not drink and calculate?

TFLYTSNBN wrote:6

Impact energy at 50 km/s is then 7.8ee25 Joules or

78 YottaJoules.

?

That's a lotta joules.

Just on the off chance that David Weber utilized one of the online asteroid impact calculators when he was writing DARKFALL, I perused some of them. Most are not supported by my Samsung tablet which suggests that they might have problems. A few seem to be limited to a maximum asteroid diameter of 10 km which indicates problems. One was ludicrously innaccurate suggesting that someone was guilty of Drinking and Coding, an offense that I myself havebeen guilty of. One calculated the impact energy correctly then FUBARED converting to Megatons equivalent.

If Weber used one of these online calculators then any innaccuracies would not be his fault.

Assuming a 25km diameter sphere of Nickle Iron asteroid with a density of 8,000 kg/m^3, I get:

Volume = 1.6eex13 m^3
Mass = 1.25eex17 Kg
Energy = 1.6eex26Joules
Energy = 156 YottaJoules

Megaton equivalent = 37 Billion

Scaling from the surface burst effects of nuclear weapons, I get a crater radius 2,000 kilometers!!!

Of course with an impact velocity of 50 km/s, I would expect such an asteroid to behave more like a subsurface burst so crater radius would be smaller.

The enormous crater dimensions would multiply the effects of the impactor by literally blowing off a chunk of the planetary crust exposing the mantle below. The crust acts as an insulator for the molten mantle beneath. Heat transfer rate is inversely proportional to the thickness of the solid crust. Remove a few million square kilometers of planetary crust and an enormous quantity of heat will be released until enough mantle solidifies to act as an insulator again.


Anyone else want to play with the numbers before I do?

http://ffden-2.phys.uaf.edu/212_spring2 ... hysics.htm

TFLYTSNBN wrote:Assuming a 25km diameter sphere of Nickle Iron asteroid with a density of 8,000 kg/m^3, I get:

Volume = 1.6eex13 m^3
Mass = 1.25eex17 Kg
Energy = 1.6eex26Joules
Energy = 156 YottaJoules

I got exactly half that at every step.

V= 4/3*pi*r^3
V= 4/3*3.14*(12500 m)^3
V= 8.18 * 10^12 m^3

Energy = 8.18 * 10^25 Joules or 81.8 yottajoules

Galactic Sapper wrote:

TFLYTSNBN wrote:Assuming a 25km diameter sphere of Nickle Iron asteroid with a density of 8,000 kg/m^3, I get:

Volume = 1.6eex13 m^3
Mass = 1.25eex17 Kg
Energy = 1.6eex26Joules
Energy = 156 YottaJoules

I got exactly half that at every step.

V= 4/3*pi*r^3
V= 4/3*3.14*(12500 m)^3
V= 8.18 * 10^12 m^3

Energy = 8.18 * 10^25 Joules or 81.8 yottajoules

You caught my mistake.

I calculated V = D^3 but forgot to multiply by π/6.

Crap!

When are we going to get the story of Calvin's Hope's journey to Refuge, and the years leading through the volcanic eruptions up to the survey team discovering Refuge?

George J. Smith wrote:When are we going to get the story of Calvin's Hope's journey to Refuge, and the years leading through the volcanic eruptions up to the survey team discovering Refuge?

You suffering from FUSIA SHOCK?

I will never forget the story of King Richard the Artichoke Hearted.