Bzzzt, wrong answer Mangetout, the OP says max speed 
All it takes is a bit of creativity with the definition of ‘speed’ …
Creativity with the definition of speed, or of maximum? 
Anything you like, even creativity in the definition of ‘What would be’ (although that will cost a little more).
Snipers uses bullets which are typical for rifles, a shape called a spitzer boatttail. A pointy but curved front with a truncated taper at the back. In the US it is typically the Sierra MatchKing HPBT, hollowpoint boat tail. Scroll down a bit for the .308 caliber, 168 grain MatchKing though I have heard that heavier versions of the same bullet are now being used for sniper use. You can click to the right of each photo for actual ballistic data. The hollowpoint is used so the jacket can be drawn with a copper base which can be made perfectly smooth rather than leaving exposed lead as with a normal full metal jacket military bullet as the unevenness of the lead can cause it to be less stable.
Not the best shape for the OP as the most stable flight is ass end first which puts the center of mass infront of the center of pressure. When fired from a rifle the gyriscopic forces of the spin, usually one turn every eleven inches or thereabouts, keeps it flying pointy end first. Certainly not enough density being constructed of copper and lead.
Cold Steel makes a throwing spike shaped like a tiny but somewhat bhubby javelin, maybe 1-1/2" in diameter and 12-14" long. That shape made from depleted uranium with a section of one end made of a lighter material like tungsten would have a stable fall without needing any spin and the shape would possibly allow hypersonic speeds.
Not to mention that rain drops are not raindrop-shaped; they’re spheres, or in the case of larger ones flattened spheres (flattened on the bottom because of wind resistance.) There’s nothing aerodynamically efficient about a rain drop, and why should there be?
Raindrops behave something like air bubbles in water, small ones are spherical, as they grow they begin to flatten, if they´re too big eventually the drop breaks up in smaller droplets due to their own aerodynamic turbulence tearing them appart.
Using a fall of 10,000 ft and G = 32 ft/sec/sec and ignoring air friction entirely, I come up with a time of fall of 25 seconds and a velocity at impact of 800 ft/sec or 545.45… mph or Mach 0.71 at STP. So that’s the absolute fastest possible terminal velocity for the OP’s conditions if we made something that magically had zero aerodynamic drag.
In the real world, the aerodynamically-optimal shape depends on the top speed. For speeds well below Mach 1, minimizing wetted surface and flow separation at the trailing end is most important. For speeds above about Mach .90 and up into the hypersonic range, managing shock drag is more important. Hence 747s and such have fairly blunt noses, whereas Concordes & fighters have fairly sharp noses. Mach 0.71 is a very slow jet airliner.
So the needle-noses are out. I’m going with a hemisherical or parabolic nose.
How long a body after the nose? Just a wag, but I’m gonna say a length/diameter ratio of about 8-10. As it gets longer you get more flow separation effects along the length wherein eventually a marginal slice of length adds more drag than it does weight.
The trailing end wants to be tapered, but more tapered than the nose was. Managing flow separation is key.
You’ll want some sort of large lightweight fins at the back to keep the thing aligned with the airstream. Relying on just a lighter material at the back to move CG ahead of CP won’t have enough control power & the flight will be wobbly. Any time the airframe isn’t exactly alighted with the airstream the drag skyrockets.
Funny enough, we just built an airliner fuselage. You don’t suppose those shapes were designed by the pros with an eye to minimum drag over a speed range of zero to, say, just for example, Mach .71 or so? Nah. I didn’t think so either.
Finally once we have the shape picked out, how big overall should it be? Drag is driven almost entirely by wetted surface area & nosecone cross-sectional area, whereas mass (for a uniform cylinder which we approximately have) is proportional to cross-section. As the diameter grows, the surface area of drag grows more slowly than the cross-sectional area of mass. So bigger is better. There’s an upper limit as the size grows truly huge, but that gets beyond my limited expertise.
So, we get a rod of tungsten (~10% heavier than U) about 100 ft in diameter and 1000 ft long. Round off one end, bolt some fins on the back and let 'er go. 25-30 seconds later … WHUMP!!
p.s. Hoisting this thing to the drop point is gonna be a trick too!
LSLGuy, you put more into this than anyone else so I’ll build on what you did.
We’d need to look for an airfoil with the lowest possible ratio of drag from frontal area and surface to section area since mass will be the driving force. My WAG is that it will look like the airfoil of a P-51 which already works well at those mach numbers or the horizontal section of a trout though we need to put a few of them through wind tunnel tests. I don’t think it will look like a modern airliner fuselage as the long cylindrical section is an aerodynamic compromise made for flexibility of interior space.
I still don’t think we’ll need fins. A rounded nose with a tail tapered to a point will already have inherent stability from surface area. If not there are ways around that without adding large, high drag fins. A small set of fins mounted on a long thin boom can leverage a tiny bit of parasite drag to give high stabilization force. Small fins could also be used to give spin stabilization. It wouldn’t need as high a rate as a rifle bullet which itself may add drag as we’re already starting with a shape that has marginal stability.
Because all those things need to do something more than fall. Cars need wheels. The space shuttle and bombs need control surfaces. The teardrop shape has a very good drag coefficient relative to other shapes especially in laminar flow, but it’s hard to do anything but fall straight down unless you tack some other gear on it.
As pointed out by others, no they don’t. They’re spherical when small because surface tension dominates. At larger sizes they flatten and break up due to drag forces. I only reiterate what others have said to point out that there are lots of links available with pretty pictures.
I can’t do better than LSLGuy did on the OP. The fluid mechanics is pretty basic, but optimization takes a little time.
I’m sorry, but what does density have to do with this? Terminal velocity is all about drag. It doesn’t matter if your super-efficient shape is made out of plastic or tungsten, if its the same shape and has the same coef, its going to have the same terminal velocity. Acceleration due to gravity on at this level is independent of the object’s mass. Compared to the Earth’s mass, there is nothing you are going to come up with dense enough to make gravitational attraction anywhere remotely relevant.
Now if you want to talk about which object will make the biggest crater when it hits the ground, then you’ll want the densest materiel you can get your hands on.
Insipired by mangetout’s rod… are we losing falling space as we increase length?? If we’re allowed to drop our object from a height of 10,000 feet… is that the top of it is 10,000 feet above the ground, or the bottom of it? (If the object is 1000 feet, it makes a difference.)
If 10,000 feet is the top of the object, then there has to be a tradeoff point somewhere between lost falling space, and increased mass in the height to overcome drag. Heck if I know how to figure THAT one out though.
“I have no worthwhile skills or knowledge in this area - I compensate by ‘raising issues’” (From a dilbert cartoon.)
I’m not sure I believe this… let’s take our shape as being a one-foot diameter sphere, for simplicity.
Drop one sphere made out of ultra-lightweight plastic and one made out of tungsten out of a high window at the same moment… are you sure that they’re going to hit the ground at the very same instant??
They are accelerated at the same rate by gravity. They have identical drag forces acting against them (at similar speeds.) But acceleration rate and force are not identical – the missing factor in that equation is mass.
The tungsten sphere is being pulled by gravity with much greater force, which is necessary to accelerate its higher mass at the same rate. But that higher force can more easily overcome the constant forces of drag. To put it another way; the force of drag is the same (more or less) but it cannot push a heavier item as well as a lighter item.
Did that make sense, or am I totally out to lunch??
I think we have at least something here to guide discussions.
http://www.lancastermuseum.ca/s,tallboy.html
Apparently, imparting a spin by careful design of the tail fins not only improved accuracy, but also increased the rate of fall, beyond the speed in the calculations of LSL_guy
http://www.bismarck-class.dk/tirpitz/miscellaneous/tallboy/tallboy.html
http://www.bismarck-class.dk/tirpitz/miscellaneous/tallboy/tallboy.html
Nope. You are entirely correct. Acceleration due to gravity remains the same, but force due to gravity is proportional to mass. At the extreme end, a lightweight balloon filled with helium will be so light it won’t fall at all, and the same size ballon filled with tungsten or DU will fall with enormously force which will affect its terminal velocity in a non-linear manner. As the object gets heavier, velocity increases but in significantly less than proprotion to its weight.
One other thing to take into account is that, even if the ball is falling as subsonic velocity, you can create shock wavefronts in front of it at high subsonic speeds through prefered geometry (very sharp leading edges/points) and rotation. Think, for instance, of a golf ball, which is dimpled to create and maintain a turbulent layer around it. The motion of the ball (spinning) in flight combined with the discontinuous surface prevents laminar flow and reduces drag and “air friction”. There is an energy drain associated with creating the shock wave, but a benefit in the way that the fluid (air, in this case) behaves when “seperating” (at the rear) of the object. This is, in part, why rain is “teardrop” shaped; the pointy rear end creates a smaller void and less wake. Think of the difference in wake between a square-transomed dinghy and that of a torpedo-shaped canoe or kayak.
The OP doesn’t state whether you can use some kind of active mechanism to create turbulence, so I’ll assume that’s not available, but in response to the questions:
[ol]
[li]A sharp-pointed, narrow projectile with broad, thin fins, something like the Picatinny finned KE sabot round.[/li][li]Tungsten or depleated uranium would be the most dense, readily available material.[/li][li]Based on LSL Guy’s no-drag calc (which I quickly confirmed…whew, didn’t want to make any foolish assumptions) I’d guess somewhere in the area of .60-.65 Mach, assuming no loss of stability or deformation due to heating.[/li][/ol]
Stranger
I thought about the helium angle, but didn’t want to complicate things with it. Seems to me that there is a ‘buoyancy effect’, in which the air is not just slowing down a falling object, but actively trying to push it up so that the air can ‘fall’ into the same place as the object is taking up. That probably won’t be significant once we get to an object that’s, oh, say three times as dense as the medium through which it is falling, or more.
I still think thats wrong. Lets simplify that scenario even more, drop those two spheres in a vacuum. If your reasoning is correct, the tungsten shere will hit the ground even quicker than the the plastic one, and the heads of physics students everywhere will explode.
Er, you may gain a higher terminal velocity via decreasing energy loss in boundary seperation, but you aren’t going to get any more speed than v=√(2ad) from an unpowered “free-fall”. 800fps from 10,000 feet is the physical maximum in a 32 ft/s[sup]2[/sup] field.
Stranger
Why??
when they’re in a vacuum, the friction and drag forces disappear, and the spheres hit at the same time. they’re both being accelerated by forces of different strength, perfectly in proportion to their mass. Acceleration becomes equal.
The point I was making was that drag forces, being equal, would affect objects of unequal mass unequally. In the absence of drag forces that point does not apply.
Nope. In a vacuum, you are dispensing with drag, which is a significant consideration in the OP. They are accelerated at the same rate, but the forces on them are different in proportion to their masses.
Don’t believe me? Drop an empty two-liter plastic bottle on your foot. Then fill it up with water, and repeat. Just make certain you are wearing steel-toed boots for the latter action.
Stranger