Wind powered vehicle that goes downwind faster than the wind? - Part 6.

[spork]

Originally Posted by seebs

Whoops!

I just realized that I got my problem spec wrong.

It's considered unusually cruel to get the problem statement wrong on a brainteaser. That's basically a denial of service attack on my brain. It's good you caught it before I lost too many nights of sleep - or worse yet ended up in the funny farm with humber.

I thought I had two really clever solutions, but they both required treating the plane as topologically identical to the sphere (thus the North and East).

D'OH!

Now I must give you my favorite brainteaser:

There is a square table that can spin. There is a coin on each of its corners. There is a cup over each coin so that we can't see if they are heads or tails. There is a light in the middle of the table that turns on if all of the coins are heads, or all are tails. The object is to turn the light on. You must find an algorithm that will do this even if you have the worst luck possible (i.e. you can't say: eventually I will have seen every coin) -or- prove that it can't be done. The rules are as follows: You can pick up any two cups, look at the coins, and place them back in the same corners any way that you want (heads/heads, tails/tails, heads/tails, etc). Once you put them back down (and replace the cups), one of two things happens. Either the light goes on (puzzle solved), or the table spins, and you get to try again (you can't follow the cups while the table spins). Remember you can pick any two
corners after each spin, but you won't know which corners you picked on the previous spin.

[humber]

Originally Posted by seebs

Whoops!

I just realized that I got my problem spec wrong.

In such a way that correcting it will instantly reveal the gimmick.

So, I was wrong -- the southernmost point of A is not north of B. Rather, the point of A on a line extending B to the north is north of B.

You can immediately, then, see how it works; A is a diagonal line segment, and is moving east (or west) fast enough that its intersection with a given vertical line either stays put or moves southwards until you run out of A.

But!

Humber's claim that 101/102 would somehow be different is even more surreally wrong than usual.

That is essentially the same. The length of A is important. If you are driving your car slowly at 1m/s, and a car cuts across you path with its nose 1m ahead of you, and its velocity in the x and y direction is such that you just clear the rear bumper of that car,
what do you think will happen if you add 100m/s to the forward direction only of each car? Bang.

[uncool]

Originally Posted by humber

That is exactly the same. If you are driving your car slowly at 1m/s, and a car cuts across you path with its nose 1m ahead of you, and its velocity in the x and y direction is such that you just clear the rear bumper of that car,
what do you think will happen if you add 100m/s to the forward direction only of each car? Bang.

Entirely wrong. Watch:

In your point of view, it has:
x = v_x t
y = 1.

If you add 100 m/s to both of them, you get:

x_f = v_x t
y_f = 1 + 100 t

y_b = 100 t.

Distance between the two is
in the back car's view
in the observer's view - which is exactly the same.

The frontwards distance is ALWAYS 1.
=Uncool-

[humber]

Originally Posted by uncool

Entirely wrong. Watch:

In your point of view, it has:
x = v_x t
y = 1.

If you add 100 m/s to both of them, you get:

x_f = v_x t
y_f = 1 + 100 t

y_b = 100 t.

Distance between the two is
in the back car's view
in the observer's view - which is exactly the same.

The frontwards distance is ALWAYS 1.
=Uncool-

Do tell, the relative forward velocity remains the same. You fixed the angle at 45degrees?

As in Seebs example, the length if the object is important. I mentioned that, and is why I used cars as an example.
To quote Seebs;
"You can immediately, then, see how it works; A is a diagonal line segment, and is moving east (or west) fast enough that its intersection with a given vertical line either stays put or moves southwards until you run out of A."
Now you have another reason to doubt one dimensional linear transforms

[uncool]

Originally Posted by humber

Do tell, the relative forward velocity remains the same.

As in Seebs example, the length if the object is important. I mentioned that, and is why I used cars as an example.
To quote Seebs;
"You can immediately, then, see how it works; A is a diagonal line segment, and is moving east (or west) fast enough that its intersection with a given vertical line either stays put or moves southwards until you run out of A."
Now you have another reason to doubt one dimensional linear transforms

But the length stays the same here. The car still acts as a line straight in the same direction - so no bang.
ETA: What angle are you talking about? I only see the straight forward movement in the examples...
ETA2: Let's do this for any case at all.

In any case, we have (from the back car's view)
x_f = v_dx t
y_f = 1 + v_dy t
x_b = 0
y_b = 0

where d denotes the difference between the front and back cars.

and
x_f = v_dx t + v_x t
y_f = 1 + v_dy t + v_y t
x_b = v_x t
y_b = v_y t

No matter what, the y-ward distance is always 1 at time 0. That is, there is no "bang".
=Uncool-

[Requiem for your dreams.]

Originally Posted by recursive prophet

I’ve always thought of it like this: the air blown by the prop was what provided that extra push. The tailwind acts as a buffer for the air pushed against it by the prop, right? Again language fails as I realize this is an over-simplification, but might it be another way of looking at it?

The Cart is not just a passive passenger like some paper bag in the wind.
When the wind blows The Cart, The Cart blows back.

This is how I see it:
Drag is proportionate to the effective area, and is equal in either direction for a stationary simple prop. As the wind blows the power delivered to some surface is proportionate to the relative velocity of the wind multiplied by the effective surface-area of the surface.

When the cart is moving, the wheels are turning the prop and so the effective area of the prop (its effective 'size' is its ability to displace air molecules coming the other way) increases rearwards.

Therefore the acceleration experienced by the cart - beyond the initial stages of mechanical resistance as the propeller 'seeks' a less stressful path - is greater than for the wind alone, because the wind energy is converted into effective surface-area by the equivalent velocity of the prop. Hence the action-reaction is greater than for a static sail, the cart pushes back and so gets a bigger push back and so gives a bigger push in return.

The acceleration from the difference in wind velocity versus the ground alone dies off at wind speed, but since the cart develops thrust when the air is still in relation to the cart, there remains some acceleration potential in the cart or looked at another way...

...when the cart is going faster than the wind some of that forward momentum pushes the prop from the other side, i.e. the prop doesn't resist it but instead transfers it to the wheels, effectively reducing the forward drag of the cart compared to a stationary prop. The faster the cart goes the faster the the prop spins, the less the effective surface area of the prop from the front. IOW the cart exchanges (edt, duh) kinetic energy with the air (now going the other way relative to the cart) as a turbine, until friction and other frontal drag forces push it back into the clutches of the big bad wind.

Put it all together and you got feedback baby.

[humber]

Originally Posted by uncool

But the length stays the same here. The car still acts as a line straight in the same direction - so no bang.
ETA: What angle are you talking about? I only see the straight forward movement in the examples...
ETA2: Let's do this for any case at all.

In any case, we have (from the back car's view)
x_f = v_dx t
y_f = 1 + v_dy t
x_b = 0
y_b = 0

where d denotes the difference between the front and back cars.

and
x_f = v_dx t + v_x t
y_f = 1 + v_dy t + v_y t
x_b = v_x t
y_b = v_y t

No matter what, the y-ward distance is always 1 at time 0. That is, there is no "bang".
=Uncool-

The North going object, with the East-West motion.

[uncool]

Originally Posted by humber

The North going object, with the East-West motion.

East-West motion doesn't affect the Northward distance, which is what we're looking at - so still, no "bang".
=Uncool-

[humber]

Originally Posted by uncool

But the length stays the same here. The car still acts as a line straight in the same direction - so no bang.
ETA: What angle are you talking about? I only see the straight forward movement in the examples...
ETA2: Let's do this for any case at all.

In any case, we have (from the back car's view)
x_f = v_dx t
y_f = 1 + v_dy t
x_b = 0
y_b = 0

where d denotes the difference between the front and back cars.

and
x_f = v_dx t + v_x t
y_f = 1 + v_dy t + v_y t
x_b = v_x t
y_b = v_y t

No matter what, the y-ward distance is always 1 at time 0. That is, there is no "bang".
=Uncool-

You are missing the point?
You say that the cars keep the same 1m, and same y component. The car still takes the same time to move the same distance in the x direction, even though the y_speed is much higher. At 100m/s, the south car will cover that 1m in 10mS. The north car will not move far enough in the x direction in that time to avoid a collision.

[uncool]

Originally Posted by humber

You are missing the point?
You say that the cars keep the same 1m and same y component. The car still takes the same time to move in the the same distance in the x direction, even though the y is much higher. At 100m/s, the south car will cover that 1m in 10mS. The north car will not move far enough in the x direction in that time to avoid a collision.

But the north car will have moved far forward enough in the meantime not to get hit. After 10 mS, it will look like the same situation as without the 100 m/s change, only translated by a meter. They keep the same relative y-component - which is the only thing that's important.
=Uncool-

[seebs]

You already posted that one here, spork.

Without loss of generality, refer to the cup closest to you as 0, and the other three as 1, 2, 3 moving clockwise around the table. Each step presumes that the previous step did not result in all-heads or all-tails.

Step one: Turn 0 and 1 to heads. Either you're done, or two coins are definitely heads, and at least one of the other two is tails.

Step two: Turn 0 and 2 to heads. Either you're done, or three coins are heads and one is tails.

Step three: Reveal 0 and 1. If either is tails, flip it. You're done. Otherwise, flip one coin to tails. Now either you're done, or you have HHTT or HTHT.

Step four: Reveal 0 and 2. If they're both tails, flip them and you're done. If they're both heads, flip them and you're done. So, if you're not done, you had HHTT. Change nothing.

Step five: Reveal 0 and 1. If they're both heads, or both tails, flip them and you're done. Otherwise, flip them anyway; this turns HHTT into either HTHT or THTH.

Step six: Reveal 0 and 2. Flip them both. You're done.

Maybe this can be done in one step fewer.

Sorry about getting the qualifier wrong. It's obvious, if you drop that, that the slower object can hit the faster object from behind.

[seebs]

You know, I should have noticed right away that I must have misstated the problem when Humber claimed that an equivalent problem had a solution.

[humber]

Originally Posted by uncool

But the north car will have moved far forward enough in the meantime not to get hit. After 10 mS, it will look like the same situation as without the 100 m/s change, only translated by a meter. They keep the same relative y-component - which is the only thing that's important.
=Uncool-

I am talking about the original question posed by seebs where the South car is slower the the North carm but 1m behind.
When the cars are moving slowly, the North cars motion is the superposition if the x any y motions. We say that results in no collision.

By adding 100m/s to the y component, the differential speed of both cars can remain at 1m/s, ( butboth are traveling at nearly 100m/s) while changing the ratio of the y to the x. If the distance to collision is still set at 1m, collision is inevitable. As both cars move, car B will pass across the path of Car B. Dependent upon the length of car A, car B can travel much slower than Car B.
=

[humber]

Originally Posted by seebs

You know, I should have noticed right away that I must have misstated the problem when Humber claimed that an equivalent problem had a solution.

But I made your misstatement work, Seebs.

[humber]

Anyway, it is quite clear that simple translations lead to errors. If you are standing by a river, and it changes speed from 1m/s to 30m/s in a minute, do you think that if you started with a constant 1m/s river, and cycled to 29m/s in that same time, that the river may not look a bit odd for a river said to be flowing at 30m/s?

[spork]

Originally Posted by seebs

You already posted that one here, spork.

D'OH! My favorite brainteaser - all used up. Here's another (although I probably posted this one already as well):

You mark a 1 meter stick in two places at random. After cutting the stick at these positions, what's the probability that the pieces could be used to form a triangle?

[semper]

Originally Posted by humber

Anyway, it is quite clear that Humber translations lead to errors. If you are standing by a river, and it changes speed from 1m/s to 30m/s in a minute, do you think that if you started with a constant 1m/s river, and cycled to 29m/s in that same time, and thought this meant that the river wrt to the river bank had change velocity, that the river may not look a bit odd for a river said to be flowing at 30m/s?

There ya go Humber.

[spork]

Imagine there is a culture that prefers boys to girls. So they live by the following rule: all couples have babies until their first born male, and then stop. Assuming that each birth is just as likely to be male as female, what is the percentage of males to females in this culture?

[spork]

Make the following figure of 5 squares with toothpicks. Then convert it to 4 squares (each of the same size as the original 5) by moving two of the toothpicks. All toothpicks must be used, none can overlap, and none are hanging out in the breeze.

[humber]

Originally Posted by semper

There ya go Humber.

Do you think that because you are riding a bike, the river is imbued with all the force and energy of a 30m/s river? Or cart in still air takes on all the force of windspeed travel?

[humber]

Originally Posted by spork

Imagine there is a culture that prefers boys to girls. So they live by the following rule: all couples have babies until their first born male, and then stop. Assuming that each birth is just as likely to be male as female, what is the percentage of males to females in this culture?

M
FM
FFM
FFFM
FFFFM
FFFFFM
So just short of 2/3 are male

Imagine that you are an internet bullshitter, trying to avoid talking about how his ideas of Galilean transforms are entirely wrong.

[spork]

Originally Posted by humber

M
FM
FFM
FFFM
FFFFM
FFFFFM
So just short of 2/3 are male

Very nice. I didn't think you could maintain your spectacular level of wrongitude. But you've outdone yourself here. Congratulations.

Imagine that you are an internet bullshitter, trying to avoid talking about how his ideas of Galilean transforms are entirely wrong.

Indeed. I have no trouble fooling all sort of people with advanced degrees in aero and/or physics, but I just can't pull one over on you.

[Requiem for your dreams.]

Originally Posted by humber

M
FM
FFM
FFFM
FFFFM
FFFFFM
So just short of 2/3 are male

P(X=j)=(1-p)^j-1p, j=1,2,3,....
Where p = 1/2 then the mean number of pregnancies required to obtain a boy is 1/p = 2. So the population should stabilise at 50:50.

[humber]

Originally Posted by spork

Very nice. I didn't think you could maintain your spectacular level of wrongitude. But you've outdone yourself here. Congratulations.

I don't really care. First guess.

Originally Posted by spork

Indeed. I have no trouble fooling all sort of people with advanced degrees in aero and/or physics, but I just can't pull one over on you.

You can fool them, but that is all. Perhaps you fail to tell them of all obvious errors they assume you have not made. You big cart will bomb.

[semper]

Originally Posted by humber

Do you think that because you are riding a bike, the river is imbued with all the force and energy of a 30m/s river? Or cart in still air takes on all the force of windspeed travel?

I think if I am riding my bike at 5m/s wrt to the ground and I fall into the middle of the river which is flowing at 30m/s in the opposite direction wrt ground I will hit the water at 35m/s wrt the water.

On the other hand, if I am riding at 30m/s wrt ground and fall into the middle of the river which is flowing at 5m/s the opposite direction wrt ground, I will hit the water at 35m/s wrt the water.

In each case I will feel the same splash forces (I happened to be riding at windspeed in both cases), but you will maintain it must have been different. Therefore I would first attach you to a large concrete bike before throwing you in the river, then you would sound the same as usual, but only for a short time.