Con-Weapons

[Daistallia]

I should really think about how a denser atmosphere [Somewhere around 1.2atm] and a lower gravity ['round 0.80m/s^2] affect weapons development and stuff...

It means that drag is higher and weight lower, so I'd expect projectile weapons would tend to be heavier. More javelins, heavier arrows with less fletching, that sort of thing.

And because of that drag, they'll be slower and shorter ranged or require different propulsion mechanisms.

Also, heavier implies a thicker cross section, but you may want to go thinner to avoid drag.

[bulbaquil]

I should really think about how a denser atmosphere [Somewhere around 1.2atm] and a lower gravity ['round 0.80m/s^2] affect weapons development and stuff

Your planet's gravity is 1/12 that of Earth's? Or did you mean 0.8x Earth's gravity?

But yeah; Mebhara would be similar in that respect, as its atmospheric pressure is 1.43 atm and its gravity is 94% Earth's.

[Torco]

I should really think about how a denser atmosphere [Somewhere around 1.2atm] and a lower gravity ['round 0.80m/s^2] affect weapons development and stuff...

It means that drag is higher and weight lower, so I'd expect projectile weapons would tend to be heavier. More javelins, heavier arrows with less fletching, that sort of thing.

And because of that drag, they'll be slower and shorter ranged or require different propulsion mechanisms.

Also, heavier implies a thicker cross section, but you may want to go thinner to avoid drag.

but heavier, sturdier and thinner requires better materials, and my folks are just getting the hang of smelting iron. [and only because more oxygen means hotter fires]

I should really think about how a denser atmosphere [Somewhere around 1.2atm] and a lower gravity ['round 0.80m/s^2] affect weapons development and stuff

Your planet's gravity is 1/12 that of Earth's? Or did you mean 0.8x Earth's gravity?

But yeah; Mebhara would be similar in that respect, as its atmospheric pressure is 1.43 atm and its gravity is 94% Earth's.

xD lol... I meant 8,0m/s^2
that's... what, 0,815G ?

[Curan Roshac]

According to Google:

gravity on earth = 9.80665 m / s2

[su_liam]

Daistallia. Looking at the power supply on your tank, I'm somewhat confused. It states that your tank has fuel cells generating a total of 1440 kWs. The M-1 tank has a 1500hp(1119kw) gas turbine. Those fuel cells would generate about 1.3 seconds of power for that. Also fuel cells aren't batteries. The lithium polymer batteries would do a bit better, giving a total of one hour and 12 minutes of power.

Further down you have 12 fuel canisters with a total of 24 kg of hydrogen, giving 40 hours of fuel supply. Okay, so now I can assume that your fuel cells are actually supplying 1440 kW of power, not 1440 kJ of energy. This is good, though normally we might write kilowatts out fully with pluralization, adding the s to kW causes confusion with the somewhat unusual kilowatt-second(I've seen defibrillators with watt-second units ). Now the M-1 tank carries 1920 liters of gasoline to supply a less powerful engine. That's about 1440kg. I don't think the M-1 can even manage 40 hours on that. The fuel cells will be more thermally efficient than the gas turbine, but I don't think, even if hydrogen has a significantly higher specific energy than gasoline and the conversion of electricity to motive power is essentially perfectly efficient, that you could get a factor of 120 improvement in mass to energy output.

[Lyhoko Leaci]

24kg of hydrogen would be roughly 12,000 moles of hydrogen gas. Completely burning this with 6000 moles of oxygen gas results in 12,000 moles of water vapor. Oxygen gas and hydrogen gas both have a free energy of formation of 0 kJ/mol. Water vapor, meanwhile has a free energy of formation of -228.61 kJ/mol. To find out the total energy released, you multiply the number of moles of each of the reactants (the hydrogen and the oxygen) by their free energy of formation. Because both of their free energy of formations are 0, this results in 0kJ. Then multiply the number of moles of the product (the water vapor) by it's free energy of formation, and subtract that result from the first result of 0kJ, to get 2,743,320kJ of energy output. (If this value was negative, it would require that much energy to cause the reaction to happen: For example, splitting water into hydrogen and oxygen)

If the energy is being generated at a rate of 1440kW, the hydrogen would be used up after 2,743,320kJ/1440kW = 1905 seconds, or about 31 minutes and 45 seconds. And that is assuming 100% efficiency... You'd get a little less than 34 minutes of power if the fuel cell emitted liquid water instead of water vapor, again assuming 100% efficiency.

Unless there's a fusion reactor in there, or you drastically lower power output, you're not going to get 40 hours out of 24kg of hydrogen.

[Daistallia]

Daistallia. Looking at the power supply on your tank, I'm somewhat confused. It states that your tank has fuel cells generating a total of 1440 kWs. The M-1 tank has a 1500hp(1119kw) gas turbine. Those fuel cells would generate about 1.3 seconds of power for that. Also fuel cells aren't batteries. The lithium polymer batteries would do a bit better, giving a total of one hour and 12 minutes of power.

Further down you have 12 fuel canisters with a total of 24 kg of hydrogen, giving 40 hours of fuel supply. Okay, so now I can assume that your fuel cells are actually supplying 1440 kW of power, not 1440 kJ of energy. This is good, though normally we might write kilowatts out fully with pluralization, adding the s to kW causes confusion with the somewhat unusual kilowatt-second(I've seen defibrillators with watt-second units ). Now the M-1 tank carries 1920 liters of gasoline to supply a less powerful engine. That's about 1440kg. I don't think the M-1 can even manage 40 hours on that. The fuel cells will be more thermally efficient than the gas turbine, but I don't think, even if hydrogen has a significantly higher specific energy than gasoline and the conversion of electricity to motive power is essentially perfectly efficient, that you could get a factor of 120 improvement in mass to energy output.

Heh - that's several years old and hasn't really been used in years - I'll have to go have a look at it again before commenting... (It's probably full of flaws.)

[Daistallia]

Heh - having had a look at it again, and some of the source material where I got it, yes. Given the time lapsed, I'm not sure of exactly where the calculations went awry - probably multiple places.

Surprised nobody jumped on 12*2=100 before:

12 alkali-modified fullerene nanotube lattice hydrogen canisters, containing 2kg of hydrogen each (100 kg)

[su_liam]

24kg of hydrogen would be roughly 12,000 moles of hydrogen gas. Completely burning this with 6000 moles of oxygen gas results in 12,000 moles of water vapor. Oxygen gas and hydrogen gas both have a free energy of formation of 0 kJ/mol. Water vapor, meanwhile has a free energy of formation of -228.61 kJ/mol. To find out the total energy released, you multiply the number of moles of each of the reactants (the hydrogen and the oxygen) by their free energy of formation. Because both of their free energy of formations are 0, this results in 0kJ. Then multiply the number of moles of the product (the water vapor) by it's free energy of formation, and subtract that result from the first result of 0kJ, to get 2,743,320kJ of energy output. (If this value was negative, it would require that much energy to cause the reaction to happen: For example, splitting water into hydrogen and oxygen)

If the energy is being generated at a rate of 1440kW, the hydrogen would be used up after 2,743,320kJ/1440kW = 1905 seconds, or about 31 minutes and 45 seconds. And that is assuming 100% efficiency... You'd get a little less than 34 minutes of power if the fuel cell emitted liquid water instead of water vapor, again assuming 100% efficiency.

Unless there's a fusion reactor in there, or you drastically lower power output, you're not going to get 40 hours out of 24kg of hydrogen.

Thanks for putting numbers up for that. I'm recovering from a flu and still half dizzy and two-thirds stupid. Wasn't really up to rummaging through my old thermodynamics textbooks and trying to remember how to do all that. Muhh...

Heh - that's several years old and hasn't really been used in years - I'll have to go have a look at it again before commenting... (It's probably full of flaws.)

I didn't mean to be critical. Well, at least not in an unconstructive way . Your tank may still work essentially as is. If we assume it is otherwise similar to a slightly larger M-1, we could say it has a capacity of 2000 liters of liquid hydrogen. At a density of 70kg per cubic meter and understanding that there are 1000 liters to a cubic meter, this gives us a capacity of 140kg. Using Ebba Leaci's numbers, that gives you 11 hours and about seven minutes of power. Taking various inefficiencies into account, I figure that amounts to 9 or 10 hours in practice. That, at least seems more reasonable.

I'm not sure what the cubic capacities of the various hydrates are. Is it possible that 2 cubic meters of some sort of hydrate storage could contain more than 140 kg of hydrogen? I may look into that in the future if some smart person doesn't beat me to it .

Heh - having had a look at it again, and some of the source material where I got it, yes. Given the time lapsed, I'm not sure of exactly where the calculations went awry - probably multiple places.

Surprised nobody jumped on 12*2=100 before:

12 alkali-modified fullerene nanotube lattice hydrogen canisters, containing 2kg of hydrogen each (100 kg)

I just assumed the 100kg was the total weight of the loaded cannisters .

[Daistallia]

I didn't mean to be critical. Well, at least not in an unconstructive way .

Heh - no worries. Interestingly enough, I put it up for critique on two different NationStates boards, one of which specializes in critiquing militaria, and it wasn't caught either time.

(And Ebba Leaci, your contribution did not go unnoticed. )

[blank stare II]

My conpeople are on about the technological level of the American Indians before the settlers.
The only weapon of theirs that I'm for sure of is the sling. I've even gone to the trouble of learning to sling. I figure it's the closest thing I'll get to physically interacting with them. Learn to sling at http://www.slinging.org

[devzoo]

The Klingon bat'leth is a neat con-weapon, with a well-developed history and mythology in the Star Trek universe.

It reminds me of an ulu, which is an Inuit knife used for many purposes (but not as a weapon). An ulu has a curved blade, which is used in a rocking motion to cut through meat and bone.

I'm considering a con-weapon for a con-world. I wonder if you could have a weapon that is a cross between a short sword and an ulu. The handle would be parallel to the blade, like in this photo of an ulu...



...but the blade would extend a foot or so on one side. Here is a mock-up:

ulusword.jpg (6.3 KiB) Viewed 3846 times

(This mock-up is pretty close to what I'm envisioning, except the bottom of the blade should curve as well, and act as a pommel.)

The key point is that the handle is behind the blade instead of under it. It seems to me that this construction would naturally protect the wielder's hand (as the guard does on a traditional sword). I'm just not sure it would be possible to keep your grip on the thing in battle.

[Salmoneus]

I just can't imagine people stupid enough to not realise they could put the hilt at the end of the blade and so get a stronger and more wieldable weapon that also allowed more powerful strikes and was lighter to boot. The only advantage of moving the hilt there is that the hand is protected from one direction - but that can be just as easily achieved with normal hand-guards.

[Torco]

Such a sword loses the advantage of the lever. It looks like the weapons on anime: impractical but so cool it doesn't matter. a longer blade has advantages and disadvantages; longer means more level and therefore stronger blow, but it also means heavier and slower and more expensive and tiring to wield: putting the hilt behind the blade gets the worse of both worlds.

[Přemysl]

The Mohuana tradtionally used spears, bows, and the huero. Huero means stinger in Majiusgaru and it is also sometimes called a joy meaning paddle. Since the Mohuana no longer get involved in war the huero has become largely ceremonial weapon and a cultural skill at best (there has to be a joke in there about a huero being unused). Occasionally there are honor battles fought with the huero. The huero is a stocky wooden paddle with shark teeth tied into groves along the edge. It is wielding like an e-tool if it had a serrated edge. Any firearms owned personally are for hunting. Mohuana in the modern military use the M16.