## Poll

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**5 members have voted**

2nd hint, you are probably going to need some kind of vector analysis of fluid flow. But maybe you have some tricks to come up with an estimate without it? Regardless the answer will definitely be an estimate, it's just who's estimate will be most accurate or highest σ.

I probably didn't manage to understand the exact details but I'm guessing the puzzle is about a sphere of unit radius falling vertically. It currently has its centre at (0,0,z) (z=1: so just touching the surface; z=0: has dropped further?) and is travelling at 10 m/s (since there's reference to .1 second, terrminal velocity and g is about 10m/s^2) and then asking how a flat body of liquid/water (formerly z<=0) would be displaced.

Puzzles are supposed to be fun and the really good ones have an ahh! moment or use logic rather than mathematics. Personally I tend to ignore the ones where you need to do non-trivial integration or the like. So I will be passing on this one.

Quote:charliepatrickI'm sure there are people, such as gas pipe analysts, who rely on fluid dynamics, but suspect the typical WoV reader doesn't use it.

At work, we use an $80,000 software program to simulate fluid dynamics. And by we I don’t mean me, I mean people smarter than me. Even with the expensive software it’s an extremely complicated process.

Quote:charliepatrickI suspect this is fluid dynamics and is what we covered in second year degree, but can't remember any details. - btw if Dr Goddard ever reads this, thanks for your help. I'm sure there are people, such as gas pipe analysts, who rely on fluid dynamics, but suspect the typical WoV reader doesn't use it.

I probably didn't manage to understand the exact details but I'm guessing the puzzle is about a sphere of unit radius falling vertically. It currently has its centre at (0,0,z) (z=1: so just touching the surface; z=0: has dropped further?) and is travelling at 10 m/s (since there's reference to .1 second, terrminal velocity and g is about 10m/s^2) and then asking how a flat body of liquid/water (formerly z<=0) would be displaced.

Puzzles are supposed to be fun and the really good ones have an ahh! moment or use logic rather than mathematics. Personally I tend to ignore the ones where you need to do non-trivial integration or the like. So I will be passing on this one.

Actually, from the problem statement, the sphere's x,y position is not 0,0 but x,y! And, I'm not sure about your interpretation of the reference to terminal velocity; its not clear to me that the sphere isn't moving downward at the start of the problem. Also, falling for 0.1 sec the sphere's velocity would be about 1 m/s, not 10 m/s.

Other than that, I do agree with everything you are saying. If I want fluid dynamics problems, I can look in textbooks on my shelf. But . . .actually . . . I don't want fluid dynamics problems, lol.

This is a gambling forum, and, IMO, puzzles involving number theory, combinatorial math, geometry and logic are particularly suitable.

Also, anyone posing a "math puzzle" to the forum should put sufficient work into the puzzle statement such that it is clear, sufficient in its information and reduces the chance of misinterpretation to a minimum.

I calculate that the sphere weight, W, = 0.154 newtons (approx.), assuming a sphere radius = 0.1 meter

Terminal velocity, v = sqrt( 2mg/ C ρ

_{air}A)

v in m/s

mg = weight of ball in newtons = W

C = drag coefficient of a sphere ~ 0.4

ρ

_{air}=air density = 1.29 kg/m

^{3}

A = frontal area of sphere =πr

^{2}; but radius r is not defined in problem statement

v ~ sqrt( (2/(0.4*π*1.29 )*W/ r

^{2})

v ~ sqrt( W/ r

^{2})

For convenience, let's assume r= 10 cm = 0.1 meter so that r

^{2}= 0.01

so v ~ sqrt(100 W)

The problem statement says "the equation of the spheres location x² + y² + (z-1)² = 1 at approximately 0.1 second of impact." I assume this is an equation for the center of the sphere.

at 0.1s, z= 0.1v =0.1*(sqrt( 100*W) ≈ sqrt(W)

so, x² + y² = 1-(sqrt(W) - 1)

^{2}

x² + y² ≈ 1- (W-sqrt(W)+1)

x² + y² ≈ sqrt(W) - W

the water displacement is stated to be ≈ to sin(x² + y²)

So, displacement ≈ sin(sqrt(W) - W)

Now, the initial displacement should be related to the impact force which is F (newtons) = mv

^{2}

we know that v

^{2}= 100 W and m = W/g where g ~10

so the impact force, mv

^{2}is 10 W

^{2}(newtons).

Thus, I take 10*W

^{2}= sin(sqrt(W) - W)

Solving this numerically I get that the weight of the sphere, W ~ 0.154 newtons (assuming r =0.1 m)

I have no conviction that this is correct, because so much of the problem statement is mysterious to me and because I did not use any of the hints.

Quote:gordonm888This is for the original problem.

I calculate that the sphere weight, W, = 0.154 newtons (approx.), assuming a sphere radius = 0.1 meter

Terminal velocity, v = sqrt( 2mg/ C ρ_{air}A)

v in m/s

mg = weight of ball in newtons = W

C = drag coefficient of a sphere ~ 0.4

ρ_{air}=air density = 1.29 kg/m^{3}

A = frontal area of sphere =πr^{2}; but radius r is not defined in problem statement

v ~ sqrt( (2/(0.4*π*1.29 )*W/ r^{2})

v ~ sqrt( W/ r^{2})

For convenience, let's assume r= 10 cm = 0.1 meter so that r^{2}= 0.01

so v ~ sqrt(100 W)

The problem statement says "the equation of the spheres location x² + y² + (z-1)² = 1 at approximately 0.1 second of impact." I assume this is an equation for the center of the sphere.

at 0.1s, z= 0.1v =0.1*(sqrt( 100*W) ≈ sqrt(W)

so, x² + y² = 1-(sqrt(W) - 1)^{2}

x² + y² ≈ 1- (W-sqrt(W)+1)

x² + y² ≈ sqrt(W) - W

the water displacement is stated to be ≈ to sin(x² + y²)

So, displacement ≈ sin(sqrt(W) - W)

Now, the initial displacement should be related to the impact force which is F (newtons) = mv^{2}

we know that v^{2}= 100 W and m = W/g where g ~10

so the impact force, mv^{2}is 10 W^{2}(newtons).

Thus, I take 10*W^{2}= sin(sqrt(W) - W)

Solving this numerically I get that the weight of the sphere, W ~ 0.154 newtons (assuming r =0.1 m)

I have no conviction that this is correct, because so much of the problem statement is mysterious to me and because I did not use any of the hints.

I have a lot I can say about this question but I like your solution and it is surprisingly within the range of answers I got. But there should definitely be a range of answers since there are multiple fluid functions that could represent the water displacement. But without a vector flow, I don't see how you could get a range of possible answers. Newton gets you a close enough answer but who here wants to use microphysics to get a more precise answer?

Quote:WizardIt isn't often that I say this, but this one is over my head. I don't even understand what is being asked. I got lost with the "diameter" of the cube. That is a term that I thought applied only to circles and spheres.

Sorry, I meant length of the side of the cube but wrote diameter since my mind was thinking of the base of water as a palabera expression.