simple logic tells me gravity is not conservative!

[arthur]

ok, I have to say this.
Gravity is not the "conservative" force that we have been led to believe.
Gravity can share it's power!!

I hope one of you folks can understand what I am getting at here.....

gravity can enhance the force of a moving weight,
gravity can also detract from the force of a moving weight.

the force of a weight moving horizontally is not affected by gravity.
the force of a weight moving (vertically) up is weakened by gravity.
the force of a weight moving (vertically) down is strengthened by gravity.

people say that: gravity is a "conservative" force.
people is claiming that: a falling weight can attain no more energy than is used to lift it.

Is that always true?
NO!

I will give you a scenario:

A weight is attached to a wheel at 9:00.
We will use force (ENERGY INPUT) - to make this weight travel clockwise from 9:00 to 3:00 where it will strike with force downwards like a hammer - (ENERGY OUTPUT).

The point I will try to make here is that:
It is possible for the weight to strike 3:00 with more force than was used to move this weight from 9:00 to 3:00.
This means that ENERGY OUTPUT can be greater than ENERGY INPUT.

So, let me explain 2 scenarios:
we will calculate ENERGY INPUT vs. ENERGY OUTPUT regarding a weight's movement between positions 9:00 and 3:00 on the wheel.

scenario #1)
ENERGY INPUT: We will use force (x) to lift/rotate a weight clockwise from 9:00 to 12:00.
We will now let this weight fall from 12:00 to 3:00.
ENERGY OUTPUT: It will land at 3:00 with force (x)
ENERGY INPUT = ENERGY OUTPUT
----no gain there.

However, scenario 2 is different.

scenario #2)
Okay lets do the same thing as scenario #1 ... but this time ....
.....we will give the weight a good clockwise push with force (y) at 12:00.

ENERGY INPUT: We have just used forces (x) +plus (y) to move the weight from 9:00 to 3:00.
ENERGY OUTPUT: Now this weight will land at 3:00 with the force of (x) +plus (y)

ENERGY OUTPUT is greater.
WHY?, look more closely at the force (y):
(force (y) = push given to the wheel at 12:00)

---ENERGY INPUT: force (y) was applied horizontally at 12:00, therefore force (y) is not affected by gravity
---ENERGY OUTPUT: force (y) is now applied vertically (down) at 3:00, therefore force (y) is now strengthened by gravity.

So a falling weight can indeed attain more force than was used to lift it.

.....the force of a weight (w) with a downward velocity (v) is greater than the force of weight (w) with horizontal velocity (v).

[LustInBlack]

Yes But... The energy needed to go from 3:00 to 12:00 so to reset your mechanism is higher than your total remaining energy...

Your mech will not reset, unless you have a clever way to overcome this and reset the mech. ?

[arthur]

this scenario I describe is not the full picture of bessler's wheel.

it's not a closed path.
therefore not a complete wheel.

However I am just using this logic to prove that gravity can do work,

and that it is clearly possible for a weight to lift and fall,
....falling with more force than it took to lift the weight.

= free energy from gravity

[LustInBlack]

I see ..

But closing the loop is what you should seek.. ;]

I am thinking about your idea inside a rotating reference frame, looks good..

Imagine a big wheel with your mech inside.. The big wheel provide the 1/4 of turn needed for your setup to reset . The bigger wheel must be driven by your inner wheel excess energy..

Maybe what I think is not possible and still will keel however.

[ovyyus]

people say that: gravity is a "conservative" force.
people is claiming that: a falling weight can attain no more energy than is used to lift it.

The established view that gravity is a conservative force is based upon centuries of careful observation and measurement - it is demonstrated.

On the other hand you are saying that gravity is not conservative based solely upon your "simple logic" - you have not demonstrated it.

Again, how can you disagree with the established scientific view that gravity is a conservative force when you clearly don't understand what it means?

[arthur]

ovvyus,

I don't feel that my logic depends on a physical confirmation for validity.
It's like wondering what would happen if I dropped a rock on my foot.
I feel certain that I already know.

After all, my whole argument simplifies into one idea:

the force of a weight (w) with a downward velocity (v)
is greater than the force of weight (w) with horizontal velocity (v).

I know that if a rock was dropping on my foot at 5 mph,
that it would hurt more than that same rock hitting my foot from the side at 5 mph.

the force of the rock is obviously greater when velocity is aimed downwards.

this simple concept is the basis for my claims, because fact is that in both cases the rock is traveling the at the same speed, therefore the initial input of energy required to obtain this speed is the same in both cases. energy IS gained from gravity.

If centuries of careful observation and measurement really shows that I am wrong, would you care to fill me in on exactly why and where my logic is flawed?

[Fletcher]

Arthur .. you are talking in terms of Force. This is only part of the picture. Work is what is required to be done & known in this context imo.

Wk = force x distance

Here's a site you might like to spend some time at to see the difference ... start at Mechanics

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

[LustInBlack]

Err .. Two objects traveling at the same speed and assuming they are exactly the same (mass form etc) they will have the same inertia and the same relative mass.

If the impact point is in the 2 cases 5 mph, then you will feel the same pain in the two cases.. Gravity acts on the object not on your foot, it accelerated the object to 5 mph, something else accelerated the second object..

The rate of acceleration have no outcome on the total energy of those rocks (IF I am not wrong, I believe I am not) ..

If that is so, then your analogy is not logic.

You may feel less pain on certain area on your foot, and that's your physiology, it is NOT physics.. 8]

[arthur]

Arthur .. you are talking in terms of Force. This is only part of the picture. Work is what is required to be done & known in this context imo.

Wk = force x distance

yes distance is a factor,
thats why I chose to eliminate vertical distance by starting at 9:00 and ending at 3:00.

the only distance that remains is horizontal distance which is negligent considering gravity does not affect horizontal distance.

therefore in my model, work is the same as force.

[ovyyus]

I don't feel that my logic depends on a physical confirmation for validity...I feel certain that I already know.

Feelings of certainty are certainly no proof that something will work the way you think it might. If it were that simple I'd have dozens of working wheels by now :D

[arthur]

If the impact point is in the 2 cases 5 mph, then you will feel the same pain in the two cases.. Gravity acts on the object not on your foot, it accelerated the object to 5 mph, something else accelerated the second object..

ok I was waiting for this.

you are saying that since the speed is the same, force is the same, direction does not matter
and there would be no greater pain on my foot.

I beg to differ.

Even though speed is exactly the same and the downward rock has no room to accelerate past 5 mph,
force would indeed be greater, and therefore pain would be greater.

for the same reason that I can feel the weight of a rock on my foot when there is no velocity. no weight on my foot with the sideswiping rock.

gravity is not conservative because downward rock will always inflict MORE pain.

[LustInBlack]

Are we talking forces or work there?!

Speed is work not forces.


Pain is not a Mathematical unit.

[arthur]

when I say speed I mean velocity.

Like I've said since distance is not a factor in this model, force and work represent the same thing.

pain may not be a mathematical unit but "centuries of observations have concluded" that it is directly proportional to the force of impact!!!!

[LustInBlack]

How can you calculate speed if there is no distance involved?!

Centuries of Observations prove you are wrong too..

Unless I see a video of you throwing similar rocks at your foot at the same spot on your foot from horizontal to vertical and the intensity of your scream and the color and look on your face is not similar by a factor of 10 I will say what you say is unproved and probably false.

Next thing, show me how you calculate speed or velocity if you want, when no distance in time is observed....

[arthur]

of course distance is involved, in this case it is just not relevant to the definition of work.