How are you accessing the internet? :dubious:
I guess what I was getting at with the ocean currents is the affect it will have on objects with different terminal velocities.
If I’m understanding the replies, the BB, because of the low terminal velocity will land nowhere near the 1 ton sphere if there is any current. Will a strong lateral current change the rate of descent for the different sized objects or does that remain constant?
Much easier than getting the 1 ton sphear over the railing.
Thank you for confirming our intuitive guesses, that terminal velocity is affected by surface area, so that the large items will have a higher terminal velocity (and fall faster) than the small items.
On the other hand, terminal velocity is also affected by the pressure of the fluid which our object is falling through. For example, an object’s terminal velocity through the atmosphere will be much higher 50 miles above sea level than at sea level. So too, an object’s terminal velocity through the atmosphere will be much higher at sea level than at 5 miles below sea level.
My point is that while the sphere will fall much faster than the BB, I suspect that your calculated Time For Descent is mistaken. Or, maybe you did take this into account. Can you give us just a little more info about your calculations? Thanks!
There is no direct effect of pressure–only indirect, through the effect on density.
Terminal velocity is where the aerodynamic drag force is equal to the weight of the object. This is written:
(1/2)[symbol]r[/symbol]C[sub]d[/sub]AV[sup]2[/sup] = mg
where [symbol]r[/symbol] is the density of the fluid, C[sub]d[/sub] is the drag coefficient (about 0.7 for a sphere), A is the frontal area, V is the velocity, m is the mass, and g is the gravitational acceleration. Solving,
V = SQRT[(2mg)/([symbol]r[/symbol]C[sub]d[/sub]A)]
which matches Wikipedia’s page that Keeve linked to. The terminal velocity is a function of density (but not pressure). That’s important when the density changes appreciably–as it does when you’re talking about dropping the object through the atmosphere, where the air 50 miles above the Earth’s surface is substantially less dense than it is on the surface.
When you’re talking about water, the density change is slight (4% or so, per above), and the effect on terminal velocity is the square root of that. Not much, in other words.
In any case, the terminal velocity is approached quite quickly for all four spheres, so the time of descent, without too much inaccuracy, is just the total depth divided by the terminal velocity. To calculate V, I used a drag coefficient of 0.7, water density of 1.94 slugs/ft[sup]3[/sup], and gravitational acceleration of 32 ft/sec[sup]2[/sup]. The BB I assumed was about 2mm in diameter (just a guess), and the density of steel is about 7.8 g/cm[sup]3[/sup]; that’s enough to get mass and frontal area. Bastard units, yes, but enough to do the calc. (And now that I look at it, I think I miscalculated the volume, but not enough to substantially change the answer.)
You didn’t know some ships had internet access via satellite?
It’s a standard feature on most cruise ships. Doing a quick search at Carnival’s website shows that most of their ships have internet cafes.
This is not quite true. Pressure increases with depth, so the pressure on the bottom of the ball will be greater than on the top of the ball. In fact, it is precisely this difference in pressure which results in buoyancy. Of course, the weight of a steel ball (of any size) is significantly greater than its buoyancy, so the balls will all still reach the bottom, and the effect of buoyancy will be the same on all of them.
Very true - by a factor of something like 60,000. Your point is quite correct: in the atmosphere there is a substantial density gradient with altitude, whereas in the ocean it’s small.
A standard size is 4.5mm.
I had no idea that such a small change in density could correspond to such a huge increase in pressure. I stand corrected. Thank you.
Don’t fight the hypothetical.
About an hour after your steel sphere bonks it on the head, the Kraken will surface and begin to angrily dismantle your ship.
OK, a little more accurate calculation. I looked up the standard size of a BB (4.5mm-thanks Xema) and a bowling ball (8.5 in), fixed my miscalcualtion of area, then calculated the weight of the object the terminal velocity, and the distance it would take to get to 99% of terminal velocity starting with 0 velocity. Then I calculated time of descent, assuming terminal velocity the whole way: 35,900 ft (the real time would be slightly higher, of course, but the inaccuracies from this assumption are probably less than the inaccuracies of assuming 0.7000 as the drag coefficient). Results:
Terminal Distance to Time of
Object Diameter Weight Velocity 99% of Term V Descent
BB 0.177 inch 8.2E-4 lb 2.65 ft/sec 0.43 ft 226 min
1" sphere 1 inch 0.147 lb 6.31 ft/sec 2.3 ft 94.8 min
Bowl. ball 8.5 inch 90 lb 18.4 ft/sec 20.5 ft 32.5 min
1 ton sph 23.8 inch 2000 lb 30.8 ft/sec 57.2 ft 19.4 min
Not substantially different that above. Still not guaranteed accurate, etc etc.
I’m still wondering how long a one ton steel ball would take to reach the bottom.
And of course, once we get that answer, there’s no need to repeat. Just once will do.
but we do need to specify in which year this is being done…
(well, after 35 posts, somebody had to say it…)
What if the Marianas Trench was on a giant treadmill?
What if Opal were dancing to Rio by Deathray?
You may be over the Mariana(s) trench, but the SDMB sure isn’t.
From a practical standpoint, we usually say liquids don’t compress. What happens at those pressures? Where does that 4% come from? Do the molecules actually compress, or do they align themselves into a more space efficiant pattern?