Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1804898 times)

Offline TMEubanks

I am not a chemist, but I do know that PAH (Polycyclic Aromatic Hydrocarbons) are common in space, and that these are (or some are) chiral. See
http://www.astrochem.org/sci/Cosmic_Complexity_PAHs.php for more.

I also know that chiral molecules like amino acids have been found in meteorites, but with more or less equal amounts of the two senses of chirality. See
http://en.wikipedia.org/wiki/Murchison_meteorite for an example.

Now, the question you are asking can be converted to are the PAH sorted by chirality in space?, and I do not know the answer to that. Note that magnetic fields in cosmos are typically << 1 Tesla, except near stars, planets, neutron stars, etc. But still, a small motion over a billion years could add up to a lot, and so even a microTelsa field (such as was common in the early solar system) might cause an observable effect.

Offline Prunesquallor

  • Full Member
  • *
  • Posts: 174
  • Currently, TeV Brane Resident
  • Liked: 157
  • Likes Given: 73

You are absolutely correct - there are no other forces accounted for in that plot other than spacecraft thrust acceleration.  As I mentioned in the above quote, I think you need to be clearly above the region where predicted drag deceleration is within an order of magnitude of your expected thrust acceleration.  As time permits, I was going to try to do some calculations and include them as a "keep out zone" in that plot.

I am actually NOT a big fan of a space test, especially CubeSat.  I think the prospect of introducing error sources is much higher than in the lab.  Drag is one, but for example most CubeSats do NOT have active attitude control - they just tumble.  You have no ability, really to shield thruster electronics from spacecraft electronic and vice versa - you just don't have the space or mass available.

Somewhere I saw that the AVERAGE CubeSat deployment altitude was 600 km, which is why I used it in that example, but the analysis was just to answer the question of what kind of altitude change you could expect as a function of predicted thruster performance.  I still maintain that an early CubeSat test at this stage that returns a null results tells us nothing.

That is indeed a worry. Note, however, that reaction wheels (and, for that matter, micro thrusters) are now available for CubeSats. See

http://www.cubesatkit.com/docs/datasheet/DS_CSK_ADACS_634-00412-A.pdf (for example).

However, my thoughts have been moving in a different direction, and I would like to present that here.  I would suggest that a CubeSat test should be 3U, not 1 U, and should have two thrusters,  to enable a spin test, not a thrust test.

The proposed test would have two drives, oriented in opposing direction (i.e., 180 deg relative to each other), to enable the drives to spin the satellite up (i.e., in a "pin-wheel" mode). That way, drag becomes much less important. In Ascii Art mode, the directions of thrust would look (from above the long axis) like

>|
  |
  +
  |
  |<

where + denotes the center of Mass and <,> the directions of thrust.

A 3U CubeSat is 30 cm long (L = 0.3 m) and 10 cm high and wide (h = 0.1 m).  Assume the thrust is 2.5 x 10^-5 N per drive, they are separated by 25 cm, are each 12.5 cm from the center of mass , and the total mass is 3 kg.

The moment of inertia, I,  of a spin perpendicular to the L axis is ~ M * (L^2 + h^2) / 12, or 0.025 kg m^2 (a non-uniform distribution of mass might change that by up to a factor of two, which doesn't matter now but would have to be measured before launch). The torque, T,  from two drives would be 2 x 2.5 x 10^-5 N x 0.125 m = 0.625 x 10^-5 N m =  6.25 x 10^-6 kg m^2 / sec^2. The spin up (or down), d Omega / dt =  T / I =  6.25 x 10^-6  / 0.025 = 2.5 x 10^-4 radians sec^-2

A  d Omega / dt =  2.5 x 10^-4 radians sec^-2 means that in 1 day (86,400 seconds, assuming the orbit was in continual sunlight), the spin rate would be 21.6 radians / second, or a spin period of 0.29 seconds (or 206 rpms).

With two transmitters on either end of the CubeSat, you could easily see the 3 m/sec relative Doppler shift that 21.6 radians represents, and of course gyroscopes, accelerometers and sun-sensors (all for measuring rotation) are available and routinely used on CubeSats.

Even in 8 hours of Sun, the spin would be 68 RPM or 0.87 seconds.

I, personally, think that spinning a test satellite up to 200 rpms or even 68 RPMS in a day would be pretty conclusive. (Of course, you would need a "null mode," with power but no expected thrust, as a control.) And, note, this could be done at ISS altitude (or on a "next available" orbit, which is cheapest on commercial flights) - drag at those altitudes does not spin satellites up like that. If these thrusts are at all realistic, then the test would succeed or fail in a very short period of time (2 or 3 days), which is also a plus.

Also, the first such test you run, if successful (a big if in my opinion) would establish a market for EM drives for attitude control.

I think that would be a much better test.  Another you could do is to thrust out-of plane during half an orbit (e.g., the sunlit half) and measure orbital inclination change/node shift - that should be independent of drag also.  You could initially spin up (by conventional means) the CubeSat so it pointed in the direction you wanted (along the thrust axis) and then not worry about attitude control after that.
Retired, yet... not

Offline TMEubanks

About that spinning idea. T = I dw/dt is the rotational equivalent of F = m a. T = F r and I is the moment of inertia of the thing being accelerated. So for a given F, what r maximises dw/dt?

dw/dt = F r / I, so naively maximum r results in maximum dw/dt
But I itself depends on r; I = m r2, the result of some integral and averaging, so
dw/dt = F / (m r), so here minimum r results in maximum dw/dt.

Which is it?

Suppose you had a uniform (very thin) rod, of (linear!) density rho, so that M = rho L and I = M L^2 / 12 = rho L^3 / 12

dw/dt = F L / I = 12 F L / rho L^3 = 12 F / (rho L^2) so you are correct, a large lever arm will have a smaller  d Omega / dt. However, if your structure's size is fixed, then you want the separation of the thrusters to be as large as possible, as then I is fixed (to first order) and T depends linearly on the separation.

A 1U test would be better in terms of a smaller I, maybe, but I don't think all of this would fit in 1U, and I worry about doing a rotation test on a small cube (the thrusters would not be symmetrically placed with respect to the face of the cube they were pointing at). (I would have the same response to a rotational test _about_ the long axis of a 3-U CubeSat.)

But, these are all implementation details, and FWIW I could certainly be convinced differently if there was a good reason for it.

Offline Prunesquallor

  • Full Member
  • *
  • Posts: 174
  • Currently, TeV Brane Resident
  • Liked: 157
  • Likes Given: 73
About that spinning idea. T = I dw/dt is the rotational equivalent of F = m a. T = F r and I is the moment of inertia of the thing being accelerated. So for a given F, what r maximises dw/dt?

dw/dt = F r / I, so naively maximum r results in maximum dw/dt
But I itself depends on r; I = m r2, the result of some integral and averaging, so
dw/dt = F / (m r), so here minimum r results in maximum dw/dt.

Which is it?

Suppose you had a uniform (very thin) rod, of (linear!) density rho, so that M = rho L and I = M L^2 / 12 = rho L^3 / 12

dw/dt = F L / I = 12 F L / rho L^3 = 12 F / (rho L^2) so you are correct, a large lever arm will have a smaller  d Omega / dt. However, if your structure's size is fixed, then you want the separation of the thrusters to be as large as possible, as then I is fixed (to first order) and T depends linearly on the separation.

A 1U test would be better in terms of a smaller I, maybe, but I don't think all of this would fit in 1U, and I worry about doing a rotation test on a small cube (the thrusters would not be symmetrically placed with respect to the face of the cube they were pointing at). (I would have the same response to a rotational test _about_ the long axis of a 3-U CubeSat.)

But, these are all implementation details, and FWIW I could certainly be convinced differently if there was a good reason for it.

And it still begs the question, why not just do it in the lab on a low friction turntable or magnetic suspension?
Retired, yet... not

Offline TMEubanks

About that spinning idea. T = I dw/dt is the rotational equivalent of F = m a. T = F r and I is the moment of inertia of the thing being accelerated. So for a given F, what r maximises dw/dt?

dw/dt = F r / I, so naively maximum r results in maximum dw/dt
But I itself depends on r; I = m r2, the result of some integral and averaging, so
dw/dt = F / (m r), so here minimum r results in maximum dw/dt.

Which is it?

Suppose you had a uniform (very thin) rod, of (linear!) density rho, so that M = rho L and I = M L^2 / 12 = rho L^3 / 12

dw/dt = F L / I = 12 F L / rho L^3 = 12 F / (rho L^2) so you are correct, a large lever arm will have a smaller  d Omega / dt. However, if your structure's size is fixed, then you want the separation of the thrusters to be as large as possible, as then I is fixed (to first order) and T depends linearly on the separation.

A 1U test would be better in terms of a smaller I, maybe, but I don't think all of this would fit in 1U, and I worry about doing a rotation test on a small cube (the thrusters would not be symmetrically placed with respect to the face of the cube they were pointing at). (I would have the same response to a rotational test _about_ the long axis of a 3-U CubeSat.)

But, these are all implementation details, and FWIW I could certainly be convinced differently if there was a good reason for it.

And it still begs the question, why not just do it in the lab on a low friction turntable or magnetic suspension?

Well, that should be done first, ideally in vacuum. Any idea what the torque drag is on a good low friction turntable? I was thinking of a torsion pendulum for ground tests, which I think can be made very low drag,

Even on the ground, I would think rotation tests would have the great advantage that static perturbations (of the "the drive is attracted to that hunk of metal over there" type) will average out in a rotational test.  And, if this was ever flown, the actual flight hardware should be tested this way too.

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5925
  • Likes Given: 5269
...

Well, that should be done first, ideally in vacuum. Any idea what the torque drag is on a good low friction turntable? I was thinking of a torsion pendulum for ground tests, which I think can be made very low drag,

Even on the ground, I would think rotation tests would have the great advantage that static perturbations (of the "the drive is attracted to that hunk of metal over there" type) will average out in a rotational test.  And, if this was ever flown, the actual flight hardware should be tested this way too.
NASA has been using a low torsional stiffness torque pendulum for their measurements.  The forces they are reporting are extremely small (from a few microNewtons to 100 microNewtons).  They have an inclination problem, discussed up by @frobnicat in very detailed posts. 

The main issue has been that they have measured in vacuum chamber a 5*10^-6 Torr partial vacuum, forces of 55 microNewtons with 50 watts power input, but when they turn it around by 180 degrees pointing in the opposite direction, they only measured 9.9 microNewtons under an input power of 35 watts.

See:

http://emdrive.wiki/Experimental_Results

NASA Glenn has offered to do measurements but they have a minimum threshold of 100 microNewtons for their measuring equipment, so they need NASA Johnson (Eagleworks) to up their thrust to over 100 microNewtons to repliate the experiments.
« Last Edit: 06/10/2015 06:15 PM by Rodal »

Offline wallofwolfstreet

  • Full Member
  • *
  • Posts: 165
  • Liked: 169
  • Likes Given: 436
Wallofwolfstreet

Would you care to tell us about Your theory on the em-drive...

Sure @arc.  Unfortunately, it's pretty mundane and a little/lot pessimistic.  I wonder what that says about me as a person ;)

First things first, I feel it is very unlikely the emdrive is producing a thrust that would be useable for space propulsion.  I want this to work as much as anyone, but it's a scientific long shot.  I am fully supportive of all the DIY experimenters here who are trying to get real data and validation on this thing, but I don't hold out much hope.

Now, there is definitely measured thrust in the experiments we have seen.  I basically listed out all of the experimental error possibilities that could have led to this thrust in an earlier post, but (embarrassingly) I can't figure out how to quote your own posts without having to just search the thread for it.

I used to favour this explanation, but not so sure anymore:

The emdrive actually produces a net thrust that has been correctly measured, but there is also an equal and opposite net force that appears on the magnetron/radio frequency source.  As far as I know, the only setup so far where the source and the emdrive were on the same measuring platform is Shawyer's dynamic rig.  Personally, I have no faith in the validity of that setup, so I wouldn't say it invalidates this hypothesis.  Exactly how the force is felt on the magnetron(and how the force was transmitted electromagnetically through the feed antenna/feed lines) is hard to say, but doesn't seem impossible.

Like I said, it's not a very good theory, and I would love to find myself needing to change it.  It falls on the backs of all the experiments coming online in the next few months to do just that!

Also, I feel like I just need to get this out there.  If this thing works, it will have been discovered completely by accident (it won't have been the first time in science).  I have read everything graciously supplied by @TheTraveller on Shawyer's work, and I have to say it really is nonsensical.  It contradicts itself constantly, ex: The net radiation pressure is greater on the big end, which causes thrust, but then it actually moves in the direction of the little end because of Newton's 2nd law???  Either he doesn't know how F=ma works, or he is hiding something.  What's more, in 2001 his company was given a UK SMART grant of 45,000 pounds to build and test a drive.  15 years later, and we still can't say if this works or not with any certainty! 

Iulian was able to build a model in what, a few weeks, and with a maybe a few hundred dollars?  A bit more time and money and he might have quite a nice setup.  How can Shawyer justify 15 years and exponentially more money to produce a worse result?  All we have is a few confused papers and a 30 second youtube video!  Strange, very strange. (By now you should realize I'm a cynical person, so I acknowledge that those Shaywer criticisms might be unfounded)

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
15 years? I had no idea that Shawyer had been fiddling about for so long with this.

Offline TMEubanks

...

Well, that should be done first, ideally in vacuum. Any idea what the torque drag is on a good low friction turntable? I was thinking of a torsion pendulum for ground tests, which I think can be made very low drag,

Even on the ground, I would think rotation tests would have the great advantage that static perturbations (of the "the drive is attracted to that hunk of metal over there" type) will average out in a rotational test.  And, if this was ever flown, the actual flight hardware should be tested this way too.
NASA has been using a low torsional stiffness torque pendulum for their measurements.  The forces they are reporting are extremely small (from a few microNewtons to 100 microNewtons).  They have an inclination problem, discussed up by @frobnicat in very detailed posts. 

The main issue has been that they have measured in vacuum chamber a 5*10^-6 Torr partial vacuum, forces of 55 microNewtons with 50 watts power input, but when they turn it around by 180 degrees pointing in the opposite direction, they only measured 9.9 microNewtons under an input power of 35 watts.

See:

http://emdrive.wiki/Experimental_Results

NASA Glenn has offered to do measurements but they have a minimum threshold of 100 microNewtons for their measuring equipment, so they need NASA Johnson (Eagleworks) to up their thrust to over 100 microNewtons to repliate the experiments.

Note the difference between a torsional pendulum test and a true rotational test. A torsional pendulum is basically a null test (motions are small, and any thrust is opposed by the torsion of the pendulum). A torsional test will _not_ cancel out external environmental effects on the test equipment. A rotational test (think a turntable) will, if there is thrust, start rotating, and _will_ cancel out (some) external effects and also can build up to a decent rotation period over time. Rotating at (say) 1 RPM would be a lot further on the road to being convincing IMHO than moving a torsion pendulum by some fraction of a mm.

FWIW, my take on this is the same as @wallofwolfstreet. I assume these results are due to errors of some sort, and is highly unlikely to be real, much less useful in spacecraft propulsion. However, that needs to be shown, not just asserted.

Offline SeeShells

  • Senior Member
  • *****
  • Posts: 2326
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2956
  • Likes Given: 2589

By placing one of my tests in water, any and all abnormalities can show up and I'm looking to address any of them. Things you'll not see on a bench strapped down to a air bearing or simply pushing on a force gauge or hanging in free air with a wire. Being submerged with the device water cooled I can reduce the issues with the thermal expansion coefficients of the case. See any unknown effects and log them, measuring in real time the EmDrive to move freely in XYZ & T.  It's all about the data and getting data in 3D is better than in 2D.

Shell

It will be very interesting.  My prediction is that you will see a random walk


I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind.  :P

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5925
  • Likes Given: 5269
...
Iulian was able to build a model in what, a few weeks, and with a maybe a few hundred dollars?  A bit more time and money and he might have quite a nice setup.  How can Shawyer justify 15 years and exponentially more money to produce a worse result?  All we have is a few confused papers and a 30 second youtube video!  Strange, very strange. (By now you should realize I'm a cynical person, so I acknowledge that those Shaywer criticisms might be unfounded)
I share your skepticism.

What compounds the issue is that Prof. Yang and her team at a Chinese University publishes in a Chinese peer-reviewed journal her results that exceed Shawyer's claims, achieving 1 N/Kw.

Her Force/InputPower results are thousands of times greater than the one reported by NASA Eagleworks.



And her papers claim low errors between the tests (look at the curves in the figure above).  Very difficult to reconcile all of this disparate data...

See:

http://emdrive.wiki/Experimental_Results#cite_note-design_factor-3
« Last Edit: 06/10/2015 06:41 PM by Rodal »

Offline SeeShells

  • Senior Member
  • *****
  • Posts: 2326
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2956
  • Likes Given: 2589
...
Iulian was able to build a model in what, a few weeks, and with a maybe a few hundred dollars?  A bit more time and money and he might have quite a nice setup.  How can Shawyer justify 15 years and exponentially more money to produce a worse result?  All we have is a few confused papers and a 30 second youtube video!  Strange, very strange. (By now you should realize I'm a cynical person, so I acknowledge that those Shaywer criticisms might be unfounded)
I share your skepticism.

What compounds the issue is that Prof. Yang and her team at a Chinese University publishes in a Chinese peer-reviewed journal her results that exceed Shawyer's claims, achieving 1 N/Kw.

Her Force/InputPower results are thousands of times greater than the one reported by NASA Eagleworks.



And her papers claim low errors between the tests (look at the curves in the figure above).  Very difficult to reconcile all of this disparate data...

See:

http://emdrive.wiki/Experimental_Results#cite_note-design_factor-3
It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.
Sorry to ask but I've tried to dig it out on my own and nadda.

Shell

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5925
  • Likes Given: 5269
...
It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.
Sorry to ask but I've tried to dig it out on my own and nadda.

Shell
Because (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.

The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.
« Last Edit: 06/10/2015 06:56 PM by Rodal »

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5925
  • Likes Given: 5269
...
I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind.  :P

md2z/dt2 + ρgAz = 0

if you use an encasing cylinder, it will have a constant cross-section A, and it will act like a spring with stiffness ρgA

http://www.codecogs.com/library/engineering/fluid_mechanics/floating_bodies/the-oscillation-of-floating-bodies.php

http://physics.stackexchange.com/questions/64154/shm-of-floating-objects

http://www.okphysics.com/1-oscillations-of-a-body-floating-in-a-liquid/
« Last Edit: 06/10/2015 07:08 PM by Rodal »

Offline SeeShells

  • Senior Member
  • *****
  • Posts: 2326
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2956
  • Likes Given: 2589
...
I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind.  :P

md2z/dt2 + ρgAz = 0

if you use an encasing cylinder, it will have a constant cross-section A, and it will act like a spring with stiffness ρgA

http://www.codecogs.com/library/engineering/fluid_mechanics/floating_bodies/the-oscillation-of-floating-bodies.php

http://physics.stackexchange.com/questions/64154/shm-of-floating-objects

http://www.okphysics.com/1-oscillations-of-a-body-floating-in-a-liquid/

Perfect! Thnaks!!!!  ;D

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5925
  • Likes Given: 5269
...
I'm looking for a vertical moment that means a displacement of any water, not so much a over zealous partier on St Patrick's Day. But it is something to keep in mind.  :P

md2z/dt2 + ρgAz = 0

if you use an encasing cylinder, it will have a constant cross-section A, and it will act like a spring with stiffness ρgA

http://www.codecogs.com/library/engineering/fluid_mechanics/floating_bodies/the-oscillation-of-floating-bodies.php

http://physics.stackexchange.com/questions/64154/shm-of-floating-objects

http://www.okphysics.com/1-oscillations-of-a-body-floating-in-a-liquid/

Perfect! Thnaks!!!!  ;D

md2z/dt2 + ρgAz = F

md2z/dt2 + k z = F

k = ρgA

F= force from the EM Drive

ρg is a constant (equal to the weight density of the liquid, water in this case ρg= 9.81kN/m3) you can't do much about (unless you change the density of the liquid).  All you have as a variable is A and it is determined by the size of your EM Drive, which for Shawyer's., Yang and Eagleworks is fairly large.

The steady state displacement will be

z = F / k


You can calculate the effective stiffness k = ρgA (of your "water buoyancy pendulum") due to bouyancy in water and compare it with the stiffness you would get in other set-ups, for example in a Cavendish pendulum, or compare it with Iulian's set-up

Plug in F ~ 50 microNewtons to a few hundred milliNewtons, divide by k = ρgA and that's your expected steady-state displacement

for NASA Eagleworks A= 0.0613 m2

k=  9.81kN/m3 * 0.0613 m2 = 601 N/m

z = F /  (601 N/m) = 1.664*10^(-3) micrometer/microNewtons

for 50 microNewtons , z = 0.08 micrometers

for 601 microNewtons z = 1 micrometers

for 300 milliNewtons z= 500 micrometers
« Last Edit: 06/10/2015 08:08 PM by Rodal »

Offline SeeShells

  • Senior Member
  • *****
  • Posts: 2326
  • Every action there's a reaction we try to grasp.
  • United States
  • Liked: 2956
  • Likes Given: 2589
...
It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.
Sorry to ask but I've tried to dig it out on my own and nadda.

Shell
Because (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.

The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.
And that's it? Thanks. I thought it might be it but I could never know if I missed something.

Offline Rodal

  • Senior Member
  • *****
  • Posts: 5845
  • USA
  • Liked: 5925
  • Likes Given: 5269
...
It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.
Sorry to ask but I've tried to dig it out on my own and nadda.

Shell
Because (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.

The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.
And that's it? Thanks. I thought it might be it but I could never know if I missed something.
This may be one of the very few things with the EM Drive that has a simple answer  ;)

Offline wallofwolfstreet

  • Full Member
  • *
  • Posts: 165
  • Liked: 169
  • Likes Given: 436
...
It may seem like a question that has been asked before and tried to find it using the search function (right... good luck) but why 2.45 GHZ? Considering you can make a cavity in any size with a new set of harmonics at a different input frequency.
Sorry to ask but I've tried to dig it out on my own and nadda.

Shell
Because (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.

The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.
And that's it? Thanks. I thought it might be it but I could never know if I missed something.

Minor thing to remember is that when this cylinder is being submerged, your z-coordinate, if measured from the water level, is not a constant.  When the cylinder goes down some incremental step dz, it will actually raise the water level in your container by:

dwaterlevel= (A*dz)/AreaGarbageCan)

The changing of the water level will mean more of your cylinder is submerged that just a simple calculation would indicate, which is something you need to include into the math.  Easy to do the iterative calculation.  There is also a closed form solution if you are into that sort of thing :P.       

You can see just from the above equation that one way to eliminate this issue is to choose a large area container, so the change in water height becomes negligible.

Offline deltaMass

  • Full Member
  • ****
  • Posts: 955
  • A Brit in California
  • Liked: 671
  • Likes Given: 275
Because (2.45 GHz) that's the standard frequency for Magnetrons used in home microwave cooking ovens.
The same reason why it was easy for Iulian to get a magnetron at that frequency at a reasonable price.
You can get microwaves from Freecycle for nothing.

Tags: