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#2520 by Rodal on 12 Feb, 2017 15:34
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You should not use the tensile strength as the limit. You should use the yield stress as the limit.The achieved spring rate is independent of the oil damper if the force is applied long enough to let the torsion pendulum reach its equilibrium position where thrust torque equals the counter torque from the torsion wire. The oil damper will only affect the time it takes to get there.
My calculations say that you should have a constant of 0.686 mN/volt. The attached spreadsheet supports this. The first tab is configured for your set up. The second tab is a check from an independent source on the torsion wire K since I could find no info on how 1178000 is derived. Feel free to check all math! How does this compare to the string method?
With much due respect, I have to disagree with PotomacNeuron. The total weight on the wire is not a major factor preventing calculation; I can find no formulas of relevance that require it. The force pulling on the wire only has a slight effect because it slightly elongates the wire and reduces the wire's diameter. We don't know your actual diameter well enough for this elongation to matter.
Thank you for doing this. I switched from Volts to the proper engineering units, 4-20mA, that the LDS is designed to operate at. The displacement constant I gave you was an approximation. 4-20mA/30mm to 50mm is what is on the spec sheet. That translates into 16mA/20mm (4mA/5mm) as the correct displacement constant. Do you think you could update the spreadsheet with the more accurate engineering units? Then I will compare that to the string method.
#14 piano wire can hold a couple of hundred pounds. My torsional pendulum is in the ~25lbs range. I've not noticed any stretching, and recently eliminated ~8lbs from the rig.
The minimum yield stress for this piano wire I found in the literature is 420 MPa = 60,916 psi
It is not an issue of breaking the wire, it is an issue of making the measurements within the elastic range of the wire, so as to have precise measurements.
That gives you a limit of 52 pound force as the not-to-exceed weight for your #14 music wire.
This is based just on the stretch due to the weight. When you have torsion as well, one has to use a yield rule for combined stresses (for example: von Mises, Tresca, etc.), which will further decrease the allowable elastic limit. However, since the EM Drive force is known to be extremely small ~milliNewton or less, when compared to the weight (or any experimental force artifacts like thermal convection propulsion or Lorentz forces) such effects from the combined torsional stress can be shown to be negligible compared with the force due to the weight.
Therefore at 25 lbs you are safely within 1/2 of the maximum weight allowed to be in the perfectly elastic range.
There has to be stretching, since the modulus of elasticitiy is finite. No material is perfectly rigid. At the yield limit of your wire, which occurs at 52 lbs minimum, and using E=30*10^6 psi as the modulus of elasticity, the stretching strain is 0.2%.
At 25 lbs, it should be about 1/2 of this: 0.1%, so the change in length is 1/1000 of its free length.
Concerning the effect of change in diameter, this is very negligible. At 52 lbs, the change in diameter of the wire is only 0.060 %. At 25 lbs weight it is only 0.030%.
At 52 lbs, the change in cross-sectional are of the wire is only 0.12 %. At 25 lbs weight it is only 0.06%.
It is amenable to an exact solution, which I will post when I have the time (again, since the effect of change in diameter is negligible, as has been noticed by others, no hurry really to show this).
The effect of change in diameter is much smaller than the tolerance dimension of the piano wire (0.0003 inch) by specification (otherwise there are sound problems in pianos).
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#2521 by Rodal on 12 Feb, 2017 16:02
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flux_capacitor78 posted elsewhere the following interesting question:Quote
Can you explain, accounting for the airflow hypothesis due to asymmetric heating of the frustum, how Shawyer measured a thrust force on scales around his demonstrator while it was completely enclosed in a sealed square enclosure?
ANSWER: We know from fluid mechanics solutions of Navier-Stokes equations that such enclosure does not at all eliminate convection, and therefore does not at all eliminate asymmetric heating of the frustum due to thermal convection. Actually, it can increase the asymmetric forces. The way to eliminate this would be: eliminate the air (test in a vacuum).
To help you with an analogy, think of insulated glazing (IG), more commonly known as double glazing (or double-pane, and increasingly triple glazing/pane), which consists of two or three glass window panes separated by a vacuum or gas filled space to reduce heat transfer across a part of the building envelope.
It is not uncommon for the vacuum to be lost in defective window panes, and then it is obvious to the consumer that even this very small distance between panes loses its insulation properties due to flow of air which manifests itself as humidity between the panes.
==> Such enclosures (as for example the transparent enclosure in Monomorphic's excellent build) are meant to minimize the effect of external air drafts produced by objects (for example humans) moving in the vicinity. Such enclosures do not eliminate the effect of asymmetric thermal convection propulsion forces, unless the enclosure has a vacuum inside. -
#2522 by flux_capacitor on 12 Feb, 2017 16:13
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Thank you for your answer, I was about to post my question here also, you are quick
When you speak of the ambient airflow issue due to asymmetric heating of the frustum, do you mean:
- a rising flow of hot air, more or less in the z vertical axis (the "more or less" is important, as it can impress a direction in the x/y horizontal plane when rising unevenly around the body),
- or a faster hot air jet, which can have any direction?We know from fluid mechanics solutions of Navier-Stokes equations that such enclosure does not at all eliminate convection, and therefore does not at all eliminate asymmetric heating of the frustum due to thermal convection.
In that sentence you seem to imply the asymmetric heating of the frustum is due to thermal convection, but the thermal convection of air is the consequence of the asymmetric heating of the frustum, which itself is due to eddy currents in the copper produced by electromagnetic induction of the internal EM energy. -
#2523 by tleach on 12 Feb, 2017 16:22
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In that spirit, Monomorphic`s with cone.
Interesting.... Can we just expand the diameter of that torus and put a spaceship inside?
I suppose that would be to easy...
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#2524 by Rodal on 12 Feb, 2017 16:23
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Thank you for your answer, I was about to post my question here also, you are quick
I mean that when the end plate is at a higher temperature than the adjacent air, the adjacent air will decrease in density, and there will be natural convection flow within the enclosure.
When you speak of the ambient airflow issue due to asymmetric heating of the frustum, do you mean:
- a rising flow of hot air, more or less in the z vertical axis (the "more or less" is important, as it can impress a direction in the x/y horizontal plane when rising unevenly around the body),
- or a faster hot air jet, which can have any direction?We know from fluid mechanics solutions of Navier-Stokes equations that such enclosure does not at all eliminate convection, and therefore does not at all eliminate asymmetric heating of the frustum due to thermal convection.
In that sentence you seem to imply the asymmetric heating of the frustum is due to thermal convection, but the thermal convection of air is the consequence of the asymmetric heating of the frustum, which itself is due to eddy currents in the copper produced by electromagnetic induction of the internal EM energy.
Natural convection is the flow that occurs in a fluid due to difference of density within the fluid.
I don't understand whether this is a question:QuoteIn that sentence you seem to imply the asymmetric heating of the frustum is due to thermal convection, but the thermal convection of air is the consequence of the asymmetric heating of the frustum, which itself is due to eddy currents in the copper produced by electromagnetic induction of the internal EM energy.
Yes, the asymmetric heating of the air is due to asymmetric induction heating.
This explains that all 3 theories (McCulloch, Shawyer and Notsosureofit) that correlate with experiments have a force dependence based on Power*Q, since the induction heating itself is proportional to Power*Q.
This may address:
1) Shawyer has never conducted tests in vacuum, even though he has been at this longer than anybody else
2) Shawyer's preference for TE013 mode, as most effective way to asymmetrically induction heat the small end
3) Shawyer's insistence on his cut-off rule (operating beyond cut-off results in less induction heating of the small end)
4) person that reported Boeing measuring "nothing" (at Estes conference) if they tested the Flight Thruster in a vacuum chamber
5) Why Tajmar's and NASA's tests in partial vacuum result in much smaller forces than the Shawyer tests (recall however, in great contrast, that Tajmar's test in air resulted in close results to what Shawyer predicted in air)
6) NASA's test in air not as much affected by asymmetric thermal convection because NASA used epoxy layer on the outside of the small and big ends, which insulates the copper from the outside air.
==> Why don't builders testing in air try gluing thermal insulation on both end plates outside surfaces (big and small ends) and see what difference it makes to their force measurements ?
Or.. at the very least use the same FR-4, as done by NASA...
Otherwise measurements are suspect as due to asymmetric thermal convection... -
#2525 by flux_capacitor on 12 Feb, 2017 16:40
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The square symmetric metallic box (or symmetric spherical plastic as I proposed a few weeks ago) does not of course eliminate the asymmetric heating of the frustum within, nor does it eliminate the thermal convection of the air inside the box. Still it would mitigate and delay the issue of asymmetric airflow outside the box for ambient air tests of limited duration, and I think all experimentalists should try to mitigate air heating asymmetries for non-vacuum tests.
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#2526 by SeeShells on 12 Feb, 2017 17:00
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I would like to re-visit this idea of yours in a couple months after I finish off my new lab. I can see where your ideas would make sense on pure copper endplates where the microwave induction from the patterns created by the mode of operation could create a pressure differential on the plates.Thank you for your answer, I was about to post my question here also, you are quick
I mean that when the end plate is at a higher temperature than the adjacent air, the adjacent air will decrease in density, and there will be natural convection flow within the enclosure.
When you speak of the ambient airflow issue due to asymmetric heating of the frustum, do you mean:
- a rising flow of hot air, more or less in the z vertical axis (the "more or less" is important, as it can impress a direction in the x/y horizontal plane when rising unevenly around the body),
- or a faster hot air jet, which can have any direction?We know from fluid mechanics solutions of Navier-Stokes equations that such enclosure does not at all eliminate convection, and therefore does not at all eliminate asymmetric heating of the frustum due to thermal convection.
In that sentence you seem to imply the asymmetric heating of the frustum is due to thermal convection, but the thermal convection of air is the consequence of the asymmetric heating of the frustum, which itself is due to eddy currents in the copper produced by electromagnetic induction of the internal EM energy.
Natural convection is the flow that occurs in a fluid due to difference of density within the fluid.
I don't understand whether this is a question:QuoteIn that sentence you seem to imply the asymmetric heating of the frustum is due to thermal convection, but the thermal convection of air is the consequence of the asymmetric heating of the frustum, which itself is due to eddy currents in the copper produced by electromagnetic induction of the internal EM energy.
Yes, the asymmetric heating of the air is due to asymmetric induction heating.
This explains that all 3 theories (McCulloch, Shawyer and Notsosureofit) that correlate with experiments have a force dependence based on Power*Q, since the induction heating itself is proportional to Power*Q.
This may address:
1) Shawyer has never conducted tests in vacuum, even though he has been at this longer than anybody else
2) Shawyer's preference for TE013 mode, as most effective way to asymmetrically induction heat the small end
3) Shawyer's insistence on his cut-off rule (operating beyond cut-off results in less induction heating of the small end)
4) person that reported Boeing measuring "nothing" (at Estes conference) if they tested the Flight Thruster in a vacuum chamber
5) Why Tajmar's and NASA's tests in partial vacuum result in much smaller forces than the Shawyer tests (recall however, in great contrast, that Tajmar's test in air resulted in close results to what Shawyer predicted in air)
6) NASA's test in air not as much affected by asymmetric thermal convection because NASA used epoxy layer on the outside of the small and big ends, which insulates the copper from the outside air.
==> Why don't builders testing in air try gluing thermal insulation on both outside surfaces (big and small ends) and see what difference it makes on their measurements ?
Whereas the frustum I built had ceramic endplates 3/8" and 1/4" thick with copper 101 O2 free copper .032 "bonded onto them.
http://global.kyocera.com/prdct/fc/list/material/alumina/alumina.html
(BTW, I understand Kyocera is still using my machines to process their ceramics)
Powering on my tests from a cold start I would see an immediate thrust anomaly well before any thermal conductivity through the endplates could take place.
I'll be looking at this again as in this case observed observations don't agree with theory. As always I respect your breadth and depth of knowalage Dr. Rodal.
My Best,
Shell
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#2527 by Rodal on 12 Feb, 2017 17:09
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...
I would like to re-visit this idea of yours in a couple months after I finish off my new lab. I can see where your ideas would make sense on pure copper endplates where the microwave induction from the patterns created by the mode of operation could create a pressure differential on the plates.
Whereas the frustum I built had ceramic endplates 3/8" and 1/4" thick with copper 101 O2 free copper .032 "bonded onto them.
http://global.kyocera.com/prdct/fc/list/material/alumina/alumina.html
(BTW, I understand Kyocera is still using my machines to process their ceramics)
Powering on my tests from a cold start I would see an immediate thrust anomaly well before any thermal conductivity through the endplates could take place.
I'll be looking at this again as in this case observed observations don't agree with theory. As always I respect your breadth and depth of knowalage Dr. Rodal.
My Best,
Shell
Hi Shell,
Nice to have you back.
I don't know what specific grade of alumina you used, but as a quick comparison, alumina has a thermal conductivity of 12-34 W/(m·K) which is much higher than NASA's FR-4 0.29 W/(m·K), alumina ceramic is almost 100 times more thermally conductive than NASA's FR-4 though its thickness
The table http://global.kyocera.com/prdct/fc/list/material/alumina/alumina.html#mat03 shows a range between 12 and 34 W/(m·K) depending on Kyocera number (alumina content ).
Beyond this, I can't comment on your results with ceramic endplates because I don't recall seeing Measured Force vs. Time plots for such tests. I don't even know how your measuredForce/InputPower, and Q compare with Shawyer, Tajmar and NASA.
Can you post your results again?
Would like to compare your ceramic results with NASA's FR-4 ends and Tajmar and Shawyer's copper ends.
Thanks
JR -
#2528 by SeeShells on 12 Feb, 2017 17:41
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Dr. Rodal,...
I would like to re-visit this idea of yours in a couple months after I finish off my new lab. I can see where your ideas would make sense on pure copper endplates where the microwave induction from the patterns created by the mode of operation could create a pressure differential on the plates.
Whereas the frustum I built had ceramic endplates 3/8" and 1/4" thick with copper 101 O2 free copper .032 "bonded onto them.
http://global.kyocera.com/prdct/fc/list/material/alumina/alumina.html
(BTW, I understand Kyocera is still using my machines to process their ceramics)
Powering on my tests from a cold start I would see an immediate thrust anomaly well before any thermal conductivity through the endplates could take place.
I'll be looking at this again as in this case observed observations don't agree with theory. As always I respect your breadth and depth of knowalage Dr. Rodal.
My Best,
Shell
Hi Shell,
Nice to have you back.
I don't know what ceramic you used, but as a quick comparison, alumina has a thermal conductivity of 18-36 W/(m·K) which is much higher than NASA's FR-4 0.29 W/(m·K), alumina ceramic is almost 100 times more thermally conductive than NASA's FR-4 though its thickness
Beyond this, I can't comment on your results with ceramic endplates because I don't recall seeing Measured Force vs. Time plots for such tests. I don't even know how your measuredForce/InputPower, and Q compare with Shawyer, Tajmar and NASA.
Can you post your results again?
Would like to compare your ceramic results with NASA's FR-4 ends and Tajmar and Shawyer's copper ends.
Thanks
JR
I didn't publish my test results and don't expect any until I finish this new lab rebuild. Until I'm happy with mitigating all the errors I can.
Shell
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#2529 by Rodal on 12 Feb, 2017 17:50
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...
Looking forward to this summer and great testing at your Lab
Dr. Rodal,
I didn't publish my test results and don't expect any until I finish this new lab rebuild. Until I'm happy with mitigating all the errors I can.
Shell
The story is not over: your new Lab, Paul March, Monomorphic, Peter Lauwer, WarpTech?
and...
Gilo & Shawyer ?
Chinese testing Space?
Cannae testing in Space? -
#2530 by otlski on 12 Feb, 2017 18:29
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The achieved spring rate is independent of the oil damper if the force is applied long enough to let the torsion pendulum reach its equilibrium position where thrust torque equals the counter torque from the torsion wire. The oil damper will only affect the time it takes to get there.
My calculations say that you should have a constant of 0.686 mN/volt. The attached spreadsheet supports this. The first tab is configured for your set up. The second tab is a check from an independent source on the torsion wire K since I could find no info on how 1178000 is derived. Feel free to check all math! How does this compare to the string method?
With much due respect, I have to disagree with PotomacNeuron. The total weight on the wire is not a major factor preventing calculation; I can find no formulas of relevance that require it. The force pulling on the wire only has a slight effect because it slightly elongates the wire and reduces the wire's diameter. We don't know your actual diameter well enough for this elongation to matter.
Thank you for doing this. I switched from Volts to the proper engineering units, 4-20mA, that the LDS is designed to operate at. The displacement constant I gave you was an approximation. 4-20mA/30mm to 50mm is what is on the spec sheet. That translates into 16mA/20mm (4mA/5mm) as the correct displacement constant. Do you think you could update the spreadsheet with the more accurate engineering units? Then I will compare that to the string method.
#14 piano wire can hold a couple of hundred pounds. My torsional pendulum is in the ~25lbs range. I've not noticed any stretching, and recently eliminated ~8lbs from the rig.
Happy to help. New spreadsheet posted with updated units.
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#2531 by SeeShells on 12 Feb, 2017 18:44
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It looks like the rest of the year is going to be a very productive one for all. Even though there remains many unanswered questions, many have "stepped up to the plate" to build better tests, labs and devices....
Looking forward to this summer and great testing at your Lab
Dr. Rodal,
I didn't publish my test results and don't expect any until I finish this new lab rebuild. Until I'm happy with mitigating all the errors I can.
Shell
The story is not over: your new Lab, Paul March, Monomorphic, Peter Lauwer, WarpTech?
and...
Gilo & Shawyer ?
Chinese testing Space?
Cannae testing in Space?
This group has been my "go to" group and some of the finest minds to pick.
My Best,
Shell -
#2532 by Monomorphic on 13 Feb, 2017 00:19
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It was a long day in the lab. I found some $2 raspberry pi heat sinks at micro-center that worked perfectly for the pre-amps and HD monitor. I used a piece of aluminum angle as a heatsink for the mini-computer. Thermal compound was used.
I also worked on the RF system. I now have 34dBm (2.5W).
I would be a lot further along if the USB spectrum analyser had better documentation, but I won't go there now.
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#2533 by qraal on 13 Feb, 2017 02:01
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This might be relevant:
Experimental demonstration of a fifth force due to chameleon field via cold atoms
Hai-Chao Zhang
(Submitted on 10 Feb 2017)
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#2534 by rq3 on 13 Feb, 2017 02:31
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You should not use the tensile strength as the limit. You should use the yield stress as the limit.The achieved spring rate is independent of the oil damper if the force is applied long enough to let the torsion pendulum reach its equilibrium position where thrust torque equals the counter torque from the torsion wire. The oil damper will only affect the time it takes to get there.
My calculations say that you should have a constant of 0.686 mN/volt. The attached spreadsheet supports this. The first tab is configured for your set up. The second tab is a check from an independent source on the torsion wire K since I could find no info on how 1178000 is derived. Feel free to check all math! How does this compare to the string method?
With much due respect, I have to disagree with PotomacNeuron. The total weight on the wire is not a major factor preventing calculation; I can find no formulas of relevance that require it. The force pulling on the wire only has a slight effect because it slightly elongates the wire and reduces the wire's diameter. We don't know your actual diameter well enough for this elongation to matter.
Thank you for doing this. I switched from Volts to the proper engineering units, 4-20mA, that the LDS is designed to operate at. The displacement constant I gave you was an approximation. 4-20mA/30mm to 50mm is what is on the spec sheet. That translates into 16mA/20mm (4mA/5mm) as the correct displacement constant. Do you think you could update the spreadsheet with the more accurate engineering units? Then I will compare that to the string method.
#14 piano wire can hold a couple of hundred pounds. My torsional pendulum is in the ~25lbs range. I've not noticed any stretching, and recently eliminated ~8lbs from the rig.
The minimum yield stress for this piano wire I found in the literature is 420 MPa = 60,916 psi
It is not an issue of breaking the wire, it is an issue of making the measurements within the elastic range of the wire, so as to have precise measurements.
That gives you a limit of 52 pound force as the not-to-exceed weight for your #14 music wire.
This is based just on the stretch due to the weight. When you have torsion as well, one has to use a yield rule for combined stresses (for example: von Mises, Tresca, etc.), which will further decrease the allowable elastic limit. However, since the EM Drive force is known to be extremely small ~milliNewton or less, when compared to the weight (or any experimental force artifacts like thermal convection propulsion or Lorentz forces) such effects from the combined torsional stress can be shown to be negligible compared with the force due to the weight.
Therefore at 25 lbs you are safely within 1/2 of the maximum weight allowed to be in the perfectly elastic range.
There has to be stretching, since the modulus of elasticitiy is finite. No material is perfectly rigid. At the yield limit of your wire, which occurs at 52 lbs minimum, and using E=30*10^6 psi as the modulus of elasticity, the stretching strain is 0.2%.
At 25 lbs, it should be about 1/2 of this: 0.1%, so the change in length is 1/1000 of its free length.
Concerning the effect of change in diameter, this is very negligible. At 52 lbs, the change in diameter of the wire is only 0.030 %. At 25 lbs weight it is only 0.014%.
At 52 lbs, the change in cross-sectional are of the wire is only 0.060 %. At 25 lbs weight it is only 0.029%.
It is amenable to an exact solution, which I will post when I have the time (again, since the effect of change in diameter is negligible, as has been noticed by others, no hurry really to show this).
The effect of change in diameter is much smaller than the tolerance dimension of the piano wire (0.0003 inch) by specification (otherwise there are sound problems in pianos).
In some degree, I must respectively disagree with this analysis. In the real, nuts and bolts world, mechanical assemblies are retained with inclined planes (nuts and bolts, or screws and nuts). In critical situations, such as the assembly of an aircraft engine, the lubrication of the nut and bolt, and it's mating surface, are critical to the life of the final assembly.
In the case of a torsion pendulum, the diameter of the wire, and the yield strength of such wire, are not only the critical parameters, but indeed the only parameters of importance. In this case, lubrication does not enter in to the equation, but yield and ulimate tensile strength certainly do.
As a real world example, the bolts that retain the connecting rods in an aircraft piston engine are not torqued to spec, they are STRETCHED to spec. In other words, most aircraft bolts are torqued to a specified inch-pound number. But these highly stressed bolts are tightened until the bolt STRETCHES until it reaches a pre-determined length over it's resting state. This is because the bolt stress can be determined based on the worst case engine operating conditions, and the bolt can be "pre-stressed" so that it can never relax below, or be stressed above, the conditions that the bearing it is responsible for retaining will see, plus a fairly large safety factor.
How does this apply to torsion pendulums? In the case of a torsion pendulum, you want the wire to be of the highest possible tensile strength (piano wire, carbon fiber, quartz), with the minimum possible diameter to support the load without
exceeding yield strength (Poissons ratio). The thinner the wire, the more sensitive the pendulum, until it breaks.
Be aware also, that most spring steel "music wire" is hardened to maximum, and then minimally tempered to, at most, 400F to maintain it's spring temper. It is VERY sensitive to Poisson ratio changes with temperature, not just in the short term, but in the long term with exposure to temperatures approaching it's fabricated temper quench temperature (100-400 degrees Fahrenheit).
Enough bloviating. The end result is that, for a maximum sensitivity torsion pendulum, it should be constructed of the thinnest wire capable of supporting the expected load. The wire should be gently warmed with a heat gun or hair drier, repetitively, allowed to cool between heat cycles, until the pendulum attains minimum reaction to external thermal conditions. -
#2535 by Bob Woods on 13 Feb, 2017 05:33
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This might be relevant:
Thank you for the message.
Experimental demonstration of a fifth force due to chameleon field via cold atoms
Hai-Chao Zhang
(Submitted on 10 Feb 2017)
谢谢你的消息。 -
#2536 by Tcarey on 13 Feb, 2017 06:02
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Hi Shell,
Nice to have you back.
I don't know what specific grade of alumina you used, but as a quick comparison, alumina has a thermal conductivity of 12-34 W/(m·K) which is much higher than NASA's FR-4 0.29 W/(m·K), alumina ceramic is almost 100 times more thermally conductive than NASA's FR-4 though its thickness
The table http://global.kyocera.com/prdct/fc/list/material/alumina/alumina.html#mat03 shows a range between 12 and 34 W/(m·K) depending on Kyocera number (alumina content ).
Beyond this, I can't comment on your results with ceramic endplates because I don't recall seeing Measured Force vs. Time plots for such tests. I don't even know how your measuredForce/InputPower, and Q compare with Shawyer, Tajmar and NASA.
Can you post your results again?
Would like to compare your ceramic results with NASA's FR-4 ends and Tajmar and Shawyer's copper ends.
Thanks
JR
Dr. Rodal,
Wouldn't it be relatively easy to make the question of asymmetric heating thrust from the end plates close to moot by insulating the end plates with something light and easy to apply such as fiberglass batting? I would think that such insulation would delay the temperature rise of the insulation air interface sufficiently to make any thrust generated by the heating visible in the data by the time delay from onset of power.
I suppose the whole frustum could be insulated in that same manner if the input power was not too high. -
#2537 by SeeShells on 13 Feb, 2017 06:35
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Great summary! I love it.
You should not use the tensile strength as the limit. You should use the yield stress as the limit.The achieved spring rate is independent of the oil damper if the force is applied long enough to let the torsion pendulum reach its equilibrium position where thrust torque equals the counter torque from the torsion wire. The oil damper will only affect the time it takes to get there.
My calculations say that you should have a constant of 0.686 mN/volt. The attached spreadsheet supports this. The first tab is configured for your set up. The second tab is a check from an independent source on the torsion wire K since I could find no info on how 1178000 is derived. Feel free to check all math! How does this compare to the string method?
With much due respect, I have to disagree with PotomacNeuron. The total weight on the wire is not a major factor preventing calculation; I can find no formulas of relevance that require it. The force pulling on the wire only has a slight effect because it slightly elongates the wire and reduces the wire's diameter. We don't know your actual diameter well enough for this elongation to matter.
Thank you for doing this. I switched from Volts to the proper engineering units, 4-20mA, that the LDS is designed to operate at. The displacement constant I gave you was an approximation. 4-20mA/30mm to 50mm is what is on the spec sheet. That translates into 16mA/20mm (4mA/5mm) as the correct displacement constant. Do you think you could update the spreadsheet with the more accurate engineering units? Then I will compare that to the string method.
#14 piano wire can hold a couple of hundred pounds. My torsional pendulum is in the ~25lbs range. I've not noticed any stretching, and recently eliminated ~8lbs from the rig.
The minimum yield stress for this piano wire I found in the literature is 420 MPa = 60,916 psi
It is not an issue of breaking the wire, it is an issue of making the measurements within the elastic range of the wire, so as to have precise measurements.
That gives you a limit of 52 pound force as the not-to-exceed weight for your #14 music wire.
This is based just on the stretch due to the weight. When you have torsion as well, one has to use a yield rule for combined stresses (for example: von Mises, Tresca, etc.), which will further decrease the allowable elastic limit. However, since the EM Drive force is known to be extremely small ~milliNewton or less, when compared to the weight (or any experimental force artifacts like thermal convection propulsion or Lorentz forces) such effects from the combined torsional stress can be shown to be negligible compared with the force due to the weight.
Therefore at 25 lbs you are safely within 1/2 of the maximum weight allowed to be in the perfectly elastic range.
There has to be stretching, since the modulus of elasticitiy is finite. No material is perfectly rigid. At the yield limit of your wire, which occurs at 52 lbs minimum, and using E=30*10^6 psi as the modulus of elasticity, the stretching strain is 0.2%.
At 25 lbs, it should be about 1/2 of this: 0.1%, so the change in length is 1/1000 of its free length.
Concerning the effect of change in diameter, this is very negligible. At 52 lbs, the change in diameter of the wire is only 0.030 %. At 25 lbs weight it is only 0.014%.
At 52 lbs, the change in cross-sectional are of the wire is only 0.060 %. At 25 lbs weight it is only 0.029%.
It is amenable to an exact solution, which I will post when I have the time (again, since the effect of change in diameter is negligible, as has been noticed by others, no hurry really to show this).
The effect of change in diameter is much smaller than the tolerance dimension of the piano wire (0.0003 inch) by specification (otherwise there are sound problems in pianos).
In some degree, I must respectively disagree with this analysis. In the real, nuts and bolts world, mechanical assemblies are retained with inclined planes (nuts and bolts, or screws and nuts). In critical situations, such as the assembly of an aircraft engine, the lubrication of the nut and bolt, and it's mating surface, are critical to the life of the final assembly.
In the case of a torsion pendulum, the diameter of the wire, and the yield strength of such wire, are not only the critical parameters, but indeed the only parameters of importance. In this case, lubrication does not enter in to the equation, but yield and ulimate tensile strength certainly do.
As a real world example, the bolts that retain the connecting rods in an aircraft piston engine are not torqued to spec, they are STRETCHED to spec. In other words, most aircraft bolts are torqued to a specified inch-pound number. But these highly stressed bolts are tightened until the bolt STRETCHES until it reaches a pre-determined length over it's resting state. This is because the bolt stress can be determined based on the worst case engine operating conditions, and the bolt can be "pre-stressed" so that it can never relax below, or be stressed above, the conditions that the bearing it is responsible for retaining will see, plus a fairly large safety factor.
How does this apply to torsion pendulums? In the case of a torsion pendulum, you want the wire to be of the highest possible tensile strength (piano wire, carbon fiber, quartz), with the minimum possible diameter to support the load without
exceeding yield strength (Poissons ratio). The thinner the wire, the more sensitive the pendulum, until it breaks.
Be aware also, that most spring steel "music wire" is hardened to maximum, and then minimally tempered to, at most, 400F to maintain it's spring temper. It is VERY sensitive to Poisson ratio changes with temperature, not just in the short term, but in the long term with exposure to temperatures approaching it's fabricated temper quench temperature (100-400 degrees Fahrenheit).
Enough bloviating. The end result is that, for a maximum sensitivity torsion pendulum, it should be constructed of the thinnest wire capable of supporting the expected load. The wire should be gently warmed with a heat gun or hair drier, repetitively, allowed to cool between heat cycles, until the pendulum attains minimum reaction to external thermal conditions.
On my last build I captured both ends of the SS piano wire and used a guitar tuner to set the tension. I knew it would lesson the oscillations of the attached hardware to it and also by pre-streching it to a set tone. Once that was done it held that same 10k "note". I even measured it with my O-scope and a mic.
http://imgur.com/a/LSwQN
FYI...
My Best,
Shell -
#2538 by Mulletron on 13 Feb, 2017 07:00
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A new perspective on how to think about gravity.
https://medium.com/starts-with-a-bang/ask-ethan-if-gravity-attracts-how-does-the-dipole-repeller-push-the-milky-way-45ea32f6cef1#.ss2bsekuyQuoteGravity is never repulsive, but a less attractive force in one direction than all the others behaves indistinguishably from a repulsion.
Edit:
I see this was discussed a few pages back.
http://forum.nasaspaceflight.com/index.php?topic=41732.msg1639919#msg1639919
What I find interesting is that this phenomena does away with the need for strange things like negative mass* for instance, and in the case of gravitational induction** I can see how this could work.
* https://en.wikipedia.org/wiki/Breakthrough_Propulsion_Physics_Program#Diametrical
** https://en.wikiversity.org/wiki/Physics/Essays/Fedosin/Gravitational_induction#Theory_of_phenomenon
** https://en.wikipedia.org/wiki/Gravitoelectromagnetism#Equations -
#2539 by Peter Lauwer on 13 Feb, 2017 09:15
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This might be relevant:
Experimental demonstration of a fifth force due to chameleon field via cold atoms
Hai-Chao Zhang
(Submitted on 10 Feb 2017)
Very interesting.
