Physics; Help

[Matos de Matos]

Hi everybody:

My acquaintance with numbers and formulas are very bad, (we do not like each other ), and this is bugging me , please, I need your help to solve this.
I can’t stop seen the unbalanced lever as a pendulum, and can´t figure out the velocity with the counter weight.
We may be able to use the use the leverage advantage to pull it up, but probably is another wrong perception, and the lack of mathematical skills; doesn´t let me move on.


Thank you
David

[Matos de Matos]

Hi Guys:

I guess that my question is too confusing.

I have a feeling that when we pump a swing, the force that we add to the centripetal force to gain height, is less than if we have to lift our weight the distance gained in the swing.
Can someone resolve this for me?
Thank you
David

[Harvey]

Hi Dave,

I'll try and get back to this as soon as I can -

At first glance your last post, Frame A1; V should equal 9.899 m/s I think that wrong number got carried into the other calcs.

[Harvey]

Looking at Frame C we find two images that both have V = ?; the upper image has a weight named (1) and the lower image has two weights named (3). We will only refer to the rightmost (3) in this case:

Q1: What is V?
The velocity for (1) and (3) is the same if their weight is the same because the angle and the rod length is the same. This velocity is already accurately calculated in Frame A as 14 m/s if there were no counter weight extending out the other side. In accordance with my last post, there will be a negative velocity of 9.899 m/s on the counter weight. So we have an effective 4.101 m/s for that set. What is unclear here, is whether or not the rod and weights (2) are connected in some way to (1) and (3). If the hub is constructed in such a way that they are all kept at 90° to each other at all times then we have to recalculate. Especially if (2) is sliding while the weights are moving around the common axle as this creates a complex arc and gets rather involved. So let us work on the supposition that (2) is elevated prior to the axial motion and in this case the velocity for (2) is also 4.101 m/s so the system is happy with that.

Q2: What is the Force F1?:
Now in this case you have a vector drawn with no angles in reference (green arrow) which seems to imply that you want to add some force in that direction to lift (2) 5 meters. That would imply that you are looking to have (2) move while the assembly is in axial rotation and as I have mentioned before, this is a complex arc and gets into rather involved calculations. So again, we will suppose the worst case here; lifting (2) vertically. Here we have a direct relationship between gravity and the weight of the assembly being lifted. In this case two 10kg weights plus the connecting rod. Up until now, I believe we have been treating the connecting rod as a very light weight rigid rod, so we will suppose that its weight is negligible and only concern ourselves with the 20kg that needs to be lifted 5 meters. So a valid question arises as to how much time we want to take to lift that 20kg. Supposing the system is stopped at this point and we have as much time as we need, we could choose an optimum time imposed by gravity and inertia. Let us suppose that we would like to accelerate the 20kg to a velocity such that a 2.5 meters it's own inertia will carry it on to the final 2.5 meters. To accomplish this we need a midpoint velocity of 7 m/s. The equation d = x + V0t + 1/2at² helps in this regard where d is distance in meters, x is a starting position, Vo is the starting velocity, t is the time involved and a is the acceleration involved. So let us put in some values:

x = 2.5m (midpoint in the travel)
V0 = 7m/s
t = 0.7142857 seconds
a = -9.8m/s²

So we get d = 2.5 + (7 * 0.7142857) + (0.5* -9.8 * 0.51020406122449)
Or d = 2.5 + 5 + -2.5
So our final distance (d) would be the 5m.

This means that we are looking for an initial acceleration that will get us to 7 m/s at the midpoint and optimum time to do that is 0.7142857 seconds. Using the same formula again we have some new values to put in:

x = 0
V0 = 0
t = 0.7142857 seconds
a = -9.8m/s²

d = 0 + (0 * 0.7142857) + (0.5 * 9.8 * 0.51020406122449) = 2.5m

And our velocity at that point is √ V0²+2ad = √ 0² + 2 * 9.8 * 2.5 = 7m/s

So we need a positive acceleration of 9.8m/s over and above that needed to offset gravity and therefore need 19.6 m/s². So what is the force? F = ma so 20kg * 19.6 m/s² = 392N. Now we apply this force for 2.5m and therefore end up with 980 Newton Meters which is the same as 980 Joules. And that will provide a smooth transition from a stopped position, to 7m/s and back to a stopped position at the end of the 5m travel.
So that is the input energy needed to reset the system.

Now how much energy does the system provide in that 90° span? I won't go into the details of integrating Torque, but I will state that W = τ · θ where Work = Torque times the angle in Radians that the torque is applied. We could then take that value for W and equate that to energy. But in this case for horizontal weight (1), the torque is a gradient that ranges from 9.8 x 10(rod length) Newton Meters at 90° to zero Newton Meters at 0° and that gradient is not linear but is sinusoidal (τ= rFsinθ). Not to mention there are four weights and thus four individual torques involved, one for each weight where two are positive torque and two are negative torque. Things get a bit complex in that regard.

Fortunately, gravity is typically a conservative field and the motion in the X axis in non-consequential with regards to exchanges in potential energy. This is helpful because we can easily derive the 4 values of potential energy using the familiar 'mgh' equation. Then all we need to do is evaluate the differences in potential energy to arrive at how much has been converted to kinetic energy for extraction. So let us suppose our axle is 15 meters off the ground and let us label our weights as X1,-X1, Y2,-Y2 where the positives represent top-right and the negatives represent bottom-left. All the weights are 10kg. The starting height for each is as follows:x1 = 15m, -x1 = 15m, Y2 = 25m and -y2 = 10m. So the total starting potential energy at the beginning is 1470 + 1470 + 2450 + 980 = 6370J. At the end of the travel we have x1 = 5m, -x1 = 20m, Y2 = 15m and -y2 = 15m for the ending heights, and again multiplying ('mgh') 10 * 9.8 * height we get 490 + 1960 + 1470 + 1470 = 5390J. We then can expect that the difference has been converted to kinetic energy (or lost in friction and drag) and this value is 6370 - 5390 = 980J.

       Start    End    Difference
 X1    1470     490    980
-X1    1470    1960   -490
 Y2    2450    1470    980
-Y2     980    1470   -490
=============================
       6370    5390    980J

You will notice that this is the same as the difference of dropping 20kg 5m ('mgh'): (20 · 9.8 · 10) - (20 · 9.8 · 5) = 980J.

So barring all other losses, we end up with a balanced system.

[Matos de Matos]

So barring all other losses, we end up with a balanced system.

Hi Harvey:

Thank you very much for the math analysis.
Here you can see how I am bad with numbers. It is like a handicap. I couldn´t transcript 9,899, and everything was wrong.
You explain everything so simple, that soon I can´t have excuses doing my own math.
I believe that now you will understand my point insisting in the parametric pumping using sliding weights.
Please correct me, but I see the box 3 as a gain, because gravity will work pushing the leverage to the right and down until it reaches 90º.
The movement from 90º to bottom will be balanced with the weights going up.

What am I missing?

Thank you
David

[Harvey]

Zoom Image

In the beginning we have a balanced system.

Then we add Leverage Potential Energy 'A' by moving 20kg to the right 5 meters - note that the Gravitational Potential remains unchanged here as 'mgh' remains constant but energy is used to slide the 20kg 5m and this is manifested as a leverage force. (does the force = the work done? )

Then we add Gravitational Potential Energy 'B' by lifting 20kg 5m - This is Gravitational Potential Energy because we have changed the 'h' in 'mgh'.

We now allow the Potentials to be applied and they are converted to Kinetic Energy. The kinetic energy after 90° exactly equals B

At this point we can either apply that energy as C to reset the system, or we can allow the full measure to continue another 45° and our total Kinetic Energy will equal A and B combined. However, if both A and B are extracted, then the new rest positions are shown in the Full Rest diagram.


So as long as we leave A in the system, we have B and C = 980J and that is all we can extract for 90°.

[Matos de Matos]

In the beginning we have a balanced system.

So as long as we leave A in the system, we have B and C = 980J and that is all we can extract for 90°.

Hi Harvey

Thank you, the perception is that, not static, but pushing in tuned with the swing, we could have a gain, but the force applied is for the pumping, balanced, no gain.

Thanks
David

[Harvey]

Hi Harvey

Thank you, the perception is that, not static, but pushing in tuned with the swing, we could have a gain, but the force applied is for the pumping, balanced, no gain.

Thanks
David

I was playing with a pendulum applet where I could change the length of the arm at different places in the swing. It behaved very odd

If I lengthen the arm at the very bottom of the swing, and then shorten it at the top, the applet would gain energy. It is a parametric pump but in essence I was allowing the centrifugal force to extend the arm at precisely BDC and then I used Gravity to retract the arm at 1° BTDC. I don't see how the applet gained energy, but somehow the algorithms showed it as an increase.

It would be very interesting to make a machine that did that and see how it worked.

[Matos de Matos]

I was playing with a pendulum applet where I could change the length of the arm at different places in the swing. It behaved very odd

If I lengthen the arm at the very bottom of the swing, and then shorten it at the top, the applet would gain energy. It is a parametric pump but in essence I was allowing the centrifugal force to extend the arm at precisely BDC and then I used Gravity to retract the arm at 1° BTDC. I don't see how the applet gained energy, but somehow the algorithms showed it as an increase.

It would be very interesting to make a machine that did that and see how it worked.

Hi Harvey

That is my feeling, I have a device that is similar to that, and just don´t go the full rotation. Acts like a pendulum and swings back doing the effect, too.
Please see:
http://www.energeticforum.com/renewa...um-hammer.html
Regarding the double weights sliding, I think that the top weight when goes up, we do not have to use a force, the force to push the bottom up, will reduce the centripetal of the top one, and its goes up by centrifugal force.

That is my feeling, that when you did the math analysis you did not count for the centrifugal force of the top going up on the same direction.

Thank you

David

[Harvey]

You are correct, I did not introduce centripetal or centrifugal forces in the calculations.

But if you consider it, you will see that something must be given up to make that happen.

1. The centrifugal force works against you until you pass the midpoint so any gains by centrifugal force is first spent getting from the bottom to the midpoint.

2. If we use a stop and go approach, where we only raise to the midpoint and then allow the horizontal arm to accelerate it and fling it out, then we lose the potential energy of lowering it's height before it gets to the outside edge and this reduces the kinetic energy we would otherwise have. The reason, is that we used that energy to extend the arm instead by allowing the wheel to turn before the arm was fully extended.

3. A system that has a variable arm length could allow us to lift only one weight so that it is positioned over the hub and therefore has no centripetal or centrifugal forces on it - this way there is no negative centrifugal force on that weight to pull back on the other side and all the force can be used to extend the other one. So instead of lifting 20kg 5m we are lifting 10kg 7m and allowing centrifugal force to extend the other one at a payment of angular displacement and a lower height at the point of full extension.

4. Another approach can be to delay vertical alignment of the upper arm by incorporating flexible spring arms and allowing the lower arm to complete to BDC before releasing that spring. In this way, the snap of the spring can fling the weight out to the end as it reaches TDC. But I have a feeling that winding up that spring will match the energy we have available. But there is something to that rapid motion that will fling a weight outward that you can't get from a slow motion. You could bend it a full 90° if necessary, but there is probably some optimum value and tension to get the desired result.

5. I think I've seen attempts at using fluid dynamics to do this stuff as a means to store the energy. I have toilet tank that uses water pressure and air for the flush mechanism. It is quite powerful, much more so than just water and gravity. It works on the concept of building up pressure over time and the compression of air. Then when the air is released into the atmosphere pressure the action is almost violent. Perhaps something like that could be used to force the lower weight up to the center then use technique 4. to centrifugally move the upper weight to the outside.

However it is done, in order to get at the energy in the field, the conservative field needs to be separated into non conservative parts so that the energy can be extracted midpoint. A spring against gravity could provide a way to pit two conservative forces into an asymmetrical relationship the converts gravity into kinetic energy where it can be taken while letting the system refill from the field in some way.

[boguslaw]

For many physical effects we can apply the same method : take one parameter and make it changing in time . You can always choose so interesting time function that other parameter will be rising like energy for example.
Famous Tesla show ball example from his autobiography.

[Matos de Matos]

But if you consider it, you will see that something must be given up to make that happen.


2. If we use a stop and go approach, where we only raise to the midpoint and then allow the horizontal arm to accelerate it and fling it out, then we lose the potential energy of lowering it's height before it gets to the outside edge and this reduces the kinetic energy we would otherwise have. The reason, is that we used that energy to extend the arm instead by allowing the wheel to turn before the arm was fully extended.


However it is done, in order to get at the energy in the field, the conservative field needs to be separated into non conservative parts so that the energy can be extracted midpoint. A spring against gravity could provide a way to pit two conservative forces into an asymmetrical relationship the converts gravity into kinetic energy where it can be taken while letting the system refill from the field in some way.

Hi Harvey:

I think that the effect that I use on the double pendulum hammer is different from stop and go.
I have a ratchet that will prevent the first arm to move back when the extending arm reaches and passes the alignment line of the first arm.
My opinion is that will transfer the momentum from the first arm to the swing of the hammer making the hammer go higher than suppose to go if free swing.
The velocity down doesn’t seem to slow, and the first arm stops when they align and goes back after alignment. The ratchet prevents that “go back”. Well probably you are right and will slow down, and again, balance.

I never seem that toilet, but is a great idea. The water in the tank will match the city water pressure, and will flush pressurized by the air trapped on the top on the tank, instead of going down by gravity.
Great water jet clean up.

I will think about a set up as you proposed delaying the top arm to use centrifugal, after I finish the Buzzsaw graphics, with my idea. It seems that everything is matching and that I will have 4 weights coming down on the out side and two up in the inside and with double velocity.

The sketch attached shows the idea.
Please can you tell me if there is any gain doubling the velocity of the short arm going up?

Thank you
David

Sketch 14.JPG

[Harvey]

I have some things stealing my time at the moment - and I know I have neglected a few posts here at EF - but I will try and get back to them as time allows.

Looking at your pics gives me a perception that an elliptic hub of a proper size and shape could force the arm to extend and retract causing a pumping effect.

Naturally, if the contraction was sufficient to pull the weight all the way into a centric spot over the hub prior to the upswing, then total conservation of momentum occur. And conversely, if the weight was then allowed to extend out at TDC gravity could pump the system.

I think the secret there would be to make sure the elliptic only operates orthogonal to the rotation. Different sized pulleys and a belt or gears and chain or a cam lobe are all methods of getting an elliptic that could effectively change the arm lengths.

Just remember that the orthogonal pressure increases proportional to the centrifugal force during the arm contractions and that pressure often results in serious losses do to increased friction. Perhaps magnetic bearings could absorb the pressure while have zero friction

As long as you have a mechanical way to reduce the arm length orthogonally, there will be no reduction in torque from that mechanical action. But there will be an exchange between velocity and momentum.

So I think that when the arm extends, the velocity decreases and when the arm contracts in increases. This can lead to a problem if the opposing arms are connected to the same chain because the chain will transfer the velocity from one to the other unless something allows the weights themselves to move at different speeds.

We end up back at our arms being made of rope and we have that rope wrap around the tight elliptic and then release the hold pin to allow it to extend straight up like a sling shot. But then we need to make the rope stiff at the top and hold it like that throughout the down swing . . .

Nitinol Wire may work:

Nitinol wire, tubing, and sheet from NDC | Materials Sourcing

Get in touch with Nitinol Development Corp in Fremont, CA.

YouTube - Nitinol Memory Wire

YouTube - Nitinol Memory Wire

But then we have to provide a thermal reaction somehow, like perhaps solar energy to heat it. But if the energy it takes to stiffen the wire is less than the energy extracted from the gravitational fall then we have a generator.

It is all about separating the field energy into non conservative sections and stealing the energy from one of those conversions.

[Harvey]

Hi Dave,

Are you familiar with Phun?

Phun - 2D physics sandbox - Home

Phun - Wikipedia, the free encyclopedia

Some of the concepts you have could be modeled there with gravity to see how they respond.

If something looks promising, then you could try and build a 3D version to see how close it matches the design.

[Matos de Matos]

Hi Dave,

Are you familiar with Phun?

If something looks promising, then you could try and build a 3D version to see how close it matches the design.

Hi Harvey:

Living in Africa is like living in another world.
It is too bad that my parents did not choose US to live and make me in a developed country. I guess that is the way had to be, and Internet is given me all those wonderful discoveries like the Nitinol wire. Great material.
Thank you for sharing, I am learning a lot with you.

I just manage download Algodoo, and didn’t had the time to study it in detail.
It is my next task.
Juan did the double pendulum hammer, but could not put the latch, looks nice, but not really what I would like to demonstrate.
YouTube - double pendulum 2

I sketch in ACAD without dimensions and it is easy to come up with the 3D graphics that I am posting. I am going to follow your advice, and after this one, I will learn Phun.

Thank you
David