Can someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?

Quote from: ClaytonBirchenough on 07/31/2015 04:10 PMCan someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:t=P*r/sigmaSigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-AluminumP=1000 psir=48/2=24"t=1000psi*24"/84000=0.286"Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"

Quote from: strangequark on 07/31/2015 07:03 PMQuote from: ClaytonBirchenough on 07/31/2015 04:10 PMCan someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:t=P*r/sigmaSigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-AluminumP=1000 psir=48/2=24"t=1000psi*24"/84000=0.286"Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"You did a pressure calculation in ACU? It is important to note that that only takes into consideration the pressure stress and not other kinds of stresses (but the higher the pressure the better it handles anything else). And that the top and bottoms need special calculations later depending on design. BTW, the usual definition of "thin walled" pressure vessel is thickness<1/10 Diameter. So The example that ClaytonBirchenough used was exactly on the limit (which must have some margin in itself).

Quote from: baldusi on 07/31/2015 09:26 PMQuote from: strangequark on 07/31/2015 07:03 PMQuote from: ClaytonBirchenough on 07/31/2015 04:10 PMCan someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:t=P*r/sigmaSigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-AluminumP=1000 psir=48/2=24"t=1000psi*24"/84000=0.286"Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"You did a pressure calculation in ACU? It is important to note that that only takes into consideration the pressure stress and not other kinds of stresses (but the higher the pressure the better it handles anything else). And that the top and bottoms need special calculations later depending on design. BTW, the usual definition of "thin walled" pressure vessel is thickness<1/10 Diameter. So The example that ClaytonBirchenough used was exactly on the limit (which must have some margin in itself).Thank you both for the replies/comments!I arbitrarily chose the example above, but I did have "thin" walled tanks in mind. Thanks also for the helpful math strangequark!With regards to the "other stresses" you mention baldusi, what do you mean? I was starting to imagine that there would be a complex interaction of different stresses like tensile strength and compressive strength. Is there anyway to estimate this interaction and its resulting "felt" stresses? For example, if we use the above 48" diameter aluminum tube/tank used for discussion/calculation purposes, what would the resulting "felt" stress on the tank be if it was accelerating at 4g with a 1000 kg payload on top of it? It's ok if we don't get to the forward closure of the tank, but just assume that the load is dispersed equally around the radius of the cylindrical tank. I also know the math might get a little crazy, but was just wondering if there was some theory/explanation for the "compounding" effects of such forces? Thank you both again!

What you said makes sense to me (I think.. heh), but from what I can tell, it doesn't coincide with what's stated in the wikipedia entry (please tell me the wiki is wrong!).

From what I make of your explanation and the diagram from Sutton, the vertical component consists of the sum of mg and D sin θ.

In other words, if we're thrusting perpendicular to g, we can say there's no gravity drag, but we're still accelerating perpendicular to the direction of thrust at the rate of g, right?

What was the largest mass launched into low earth orbit in a single launch?

At the other end of the scale of manned (personned to be strictly pc ) spacecraft is Liberty Bell 7 at 1286.4 kilos.

The heaviest spacecraft from the Apollo era is generally reckoned to be Apollo 15 at 52819.5 kilos.

Quote from: Hoonte on 08/29/2015 09:33 PMWhat was the largest mass launched into low earth orbit in a single launch?The SA-513/Skylab 1 Flight Evaluation Report lists the mass actually inserted into orbit at 147,363 kg, including 88,474 kg for Skylab 1 itself.The similar report for SA-512/Apollo 17 lists 140,893 kg as the mass inserted into the initial parking orbit. That included 90,257 kg for the partially used S-IVB stage and 2,027 kg for the Instrument Unit. - Ed Kyle