Gold-pressed Laudenum.
I’ve been picturing something more like a static drop, where the helicopter is carrying a giant, rigid, gold parachute-shaped dome, with somebody hanging from it. Landing might be problematic. You might want to land in water, for instance, and bring some breathing gear.
Stormtroopers for the Gay Agenda!
Gold is soft but is it weak? A tiny nugget of it can be stretched into an extremely long, fine thread without breaking. Wouldn’t the gold parachute just stretch humonguously under high air resistance stress?
Sure, it would stretch, getting larger and thinner until the true stress* exceed the ultimate strength* of the material.
*Definitions:
True stress: True force per unit area
Engineering stress: Force per original unit area
Yield strength: Stress at which material begins to plasticaly deform.
Ultimate (tensile) strength, or UTS: Stress at which material fails.
Off to run the numbers…
Drag coefficient depends on the shape of the thing that’s moving through the air. I don’t think you can get it much higher than 1, though (which is usually about the value you’ll get if you make no particular effort at it).
A flat plate perpendicular to flow is 1.28. A smooth brick is 2.1. The number I found for a regular, circular parachute is 1.5. It doesn’t look like it’s going to work, I’m almost done with the numbers, but any increase in the area is going to be squared with respect to the radius, while the increase in volume (and therefore mass) is going to be cubed.
Oh my. Well done, sir.
%% Gold parachute survival
clear
clc
g = 9.81 %m/s^2 gravity
rhogold = 19300 %kg/m^2 density gold
t = .0001 %m thickness
r1 = 5.642 %m outer diameter
r2 = r1 - t %m inner diameter
V = 4 / 3 * pi * (r1^3 - r2^3)% Volume of chute
m = rhogold * V + 75%kg mass of chute + person
A = pi * r1^2%m^2 cross sectional area
sa = 2 * pi * r1^2%m^2 surface area
rho = 1.2%kg/m^3 density air
Cd = 1.5%drag coeffecient chute
Vt = sqrt(2 * m * g ./ (rho .* A * Cd))%m/s terminal velocity
Fd = 1/2 * (rho * Vt^2 * A * Cd)%N force due to drag
p = Fd/A%N/m^2 pressure in chute
sigmatheta = p * (r1+r2) / (4 * t)%N/m^2 stress due to pressure
g = 9.8100
rhogold = 19300
t = 1.0000e-004
r1 = 5.6420
r2 = 5.6419
V = 0.0400
m = 847.0148
A = 100.0037
sa = 200.0074
rho = 1.2000
Cd = 1.5000
Vt = 9.6084
Fd = 8.3092e+003
p = 83.0891
sigmatheta = 2.3439e+006
So that’s giving a stress of 2.34 MPa, while the ultimate tensile strength of gold is about 120MPa.
Based on these numbers it looks like it would work, but I have a feeling I’m off by a magnitude somewhere. Any help?
Bravo! Bravo!
Without checking the velocity and force #'s, a couple comments: I’m not sure whether you meant 5 m radius or 5m diameter (you wrote diameter, but the formulas are all for radius). Plus the volume is for a whole sphere; divide that in half for a semi-spherical chute.
The other major issue is that I think you need to look at yield strength, not ultimate strength. If the chute ever deforms, it will deform so as to stretch the gold, thinning and weakening it. Since it is highly unlikely that the change in shape would lead to less force on that spot, once it deforms I think it would keep deforming to complete failure.
Finally, in a real chute, there would be much higher stresses at attachment points and probably at the edges of the chute, but I’m fine with assuming miraculous fastening technology and edge reinforcement.
In case anyone was as lost as I:
Gold-pressed latinum - Fictitious precious material in the Star Trek universe, a very rare silver wrapped in gold. They are graded according to quality, slip, strips, bars and bricks.
I was wondering about the diameter myself (mass, too). A standard round parachute, made of nylon, is about 8m diameter, and is about 7kg. Santo Rugger, are you giving a mass for chute + person of 847 kg? If so, and the person is 90 kg, that means the gold canopy is 757 kg! Even with a radius of 5m, I think that thing would be coming out of the sky awfully fast! Or is my thinking flawed? I see you have air density and drag coefficient factored in. I can’t help with the calculations; I can only relate it to my own experience. I know that most skydivers these days are very reluctant to jump a round canopy like I described above (the pussies!). Good luck finding a test jumper; I’ll stay on the ground and watch (with a video camera)!
If, as Quercus stated, the volume is for a whole sphere, that means the mass would actually be about half that for a semi-spherical chute, right? That still means the chute would be about 378 kg, I believe.
Another thing we haven’t considered is how to extract the chute from the container- assuming that Chuck Norris is indeed our test jumper. A standard 1m diameter pilot chute isn’t going to pull that massive chute out. I don’t even think a standard 8m round parachute will do it. Maybe a static line between the chute and the plane will do the trick.
I meant radius, the comments saying diameter are artifacts from the code I modified.
Fixed the volume, I thought I had addressed that when I changed everything from d to r. Guess not.
I can’t find a yield strength for pure gold, I think it’s because it’s too ductile. For steels, IIRC, the yield is usually half the ultimate. My new number for sigma[sub]theta[/sub] is 1.068 MPa, still well below half of 120MPa.
You’re right, Quercus, I was assuming miraculous fasteners made of unobtanium. The parachute is going to weigh much more than the person it’s supporting, hence why I think the assumption is valid (especially since we’re not deploying it, per se).
Bumbershoot, with the volume of the hemisphere halved, I’m getting a new mass of 386 kg (75 kg for the jumper). The terminal velocity of a chute that size is the V[sub]t[/sub] number, which I’m now getting to be 6.5 m/s. So, yes, that’s pretty quick.
Are the jumpers you speak of reluctant to jump because of the mass of the parachute, or because of the lack of performance inherent in them?
Lack of performance is part of it, rate of descent is another factor. Round canopies descend pretty fast; if you don’t do a good PLF (parachute landing fall) you can end up with a sprained ankle or a broken leg or worse. Modern ram-air canopies are extremely maneuverable and, when flared properly, can result in soft tip-toe landings.
ETA: There are other factors as well; ram-air canopies are much safer than rounds. I could go on and on but I don’t want to hijack this thread!
Cool. If you can find the C[sub]d[/sub] numbers and a formula for surface area of the chute and it’s cross sectional area, I can easily re-run the numbers. To try to bring the terminal velocity down.
I’ll see what I can dig up.
I’ve made about 250 jumps, all on square canopies (rectangular really), and you can SO color me in the you first seats
Santo Rugger, I haven’t been able to find anything on drag coefficients for ram-air canopies. I checked the websites for 3 different parachute manufacturers and none of them go into that much detail when listing specs. One problem is that ram-air canopies come in a large variety of sizes and designs. They are much more complex than a round parachute. Here is a photo of a typical ram-air canopy.
Your descent rate of 6.5 m/sec for the gold parachute is probably survivable, but the landing is gonna hurt! A standard round parachute descends at about 5.5 m/sec. A ram-air canopy descends at about 5 m/sec but can be “flared” just before contact with the ground to provide a soft, stand-up landing.