Yes, there's no evidence of mold release, and my recollection of a few 1st hand accounts from factories confirm this. Pity perhaps, because it might have been a naturally compatible lubricant......
Methinks another myth.
LD
Methinks another myth.
LD
Since I'm the Technical Director at the world's top specialty mold release company, I have some small idea of who buys and uses it. 😀 Here's all the record pressing plants on our customer list:
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Some people dream about "mold release agents that have to be removed" while others dream about "...disassembly of Leventhal´s" , is there really difference?
Both assertions can quite easily be evaluated and confirmed or rejected, but that´s obviously bit more tedious than posting such assertions.
People love to believe and generalize things they like to be true. 😉
Where did you get the 3uW dissipation
With an average friction coeff of 0.3 from your posting #272 and a vertical Cart force of 2 gram, horizontal pulling force is 1.96e-2 x 0.3 = 5.9e-3 Kg.m/sec^2.
At an outer track, diam 27cm, travelled distance is 0.47 m/sec.
Energy is Pulling Force times distance/sec: 5.9e-3 x 0.47 = 2.8e-3 kg.m^2/sec^3 or 2.8e-3 Watt or ca. 3mW.
So it seems that with 3uW you are incorrect for a factor 1000.
Hans
Yes for total heat. Memory isn't what it used to be.....😱
On assumption that bulk of heat is conducted away from contact region by the diamond, and ratio of thermal conductivities of diamond and vinyl differ by about three orders of magnitude, or c x1000, conservatively. Thus about 3uW dissipated by vinyl, and, in any event spread over a surface area perhaps x1000000 greater than that of the diamond contact area.
No matter how one assumes heat divides between stylus and vinyl, conclusions don't alter as to thermal effects, they are so small.
LD
LD, take the calculation one step further and consider the relative specific heats. For a given energy transfer, the vinyl will have a temperature change which is a small fraction of that of the diamond.
Opinion presented as fact?
I´d love to read your argument(s) and to consider it/them.
Isn´t that a bit too much simplifying?
Obviously thermal conductivity of the diamant will be much higher than that of vinyl, but what about the thermal conductivity between the tip and the "outer world"? (means tip to stylus and tip to air)
And it is a continuos process; while the vinyl contact zone is moving the tip remains the same during complete replay.
Afair things were a bit complicated further because the contact zone was found to be bigger than expected due to elastic deformation of the vinyl.
Take into account what amount of heat will be generated with a specific heat capacity for Diamond needles and Berillium shaft material around room temp, being resp 0.5 and 1.8 Joule for raising the temp of 1gram with 1 degree Kelvin.
Assume that the whole cantilever plus Diamond tip weights 1mg for a Cart having an effective 0.3mg tip mass, and that cantilever plus tip have a specific heat capacity of 1.8 and that no heat is transferred to the air or to the LP.
With these assumptions, 3mW over a period of 1 seconds accumulates to 3mJoule.
3mJ x 1.8 for 1 mg = 5.4 degree Kelvin temp rise per seconds !!
Somewhere in time, temp rise will stabilise when energy input equals energy output because of radiation and heat transfer to the LP.
I have no idea where this equilibrium will be met, neither whether the 1 mg for the whole cantilever is correct as an order of magnitude.
But with 5.4 degrees Kelvin temp rise per second, temp transfer to LP and to the air has to be substantial not to let the whole cantilever assembly becoming cooking hot.
Regarding the LP however, I think what Hiten said is how this heat issue should be looked at “(remember we can move finger across the candle flame quickly without getting burned)”.
Hans
Far too much speculation here, maybe someone should enter an outline of this problem into a multi-physics package like COMSOLVE. It's clear to me from the few times I did it playing the same test track several times in succession made no difference at all.
COMSOL 🙂
I use it regularly, and while the geometry part wouldn't be too tough, you really do need a good model of the interfacing surfaces. The heat transfer problem would be interesting even if we do bulk down on given numbers. I make absolutely zero promises I'll build up a model anytime soon, though (sincere apologies).
I use it regularly, and while the geometry part wouldn't be too tough, you really do need a good model of the interfacing surfaces. The heat transfer problem would be interesting even if we do bulk down on given numbers. I make absolutely zero promises I'll build up a model anytime soon, though (sincere apologies).
No, because essentially the situation is one of a heat pipe, such is the high thermal conductivity of both cantilever and diamond.
For a 6mm long Al hollow tube cantilever, outer diameter 0.6mm, wall thickness 0.15mm, thermal conduction is c 5mW per deg C difference in temp between one end of the cantilever and the other. Diamond has such high thermal conductivity and stylus is so small, its contribution is negligible.
As long as the thermal junction between stylus and cantilever is good, there will be a temperature difference of c 1 degC between stylus and the suspension end of the cantilever, assuming 3mW heat flow. So temperature of the stylus/cantilever is mostly determined by thermal conductivity of the cartridge suspension, and/or losses from surface of the cantilever, with a 3mW heat flow. Nothing to get excited about then, methinks.
Obviously, if one deprives a heatpipe of its outlet, it will act as a body and get hot in a hurry, Hans. That is what you have calculated, but it isn't the case because in this case there is always an outlet via thermal junction with the cartridge suspension.
Measure the temperature of the cantilever, know the temperature of the stylus and contact region. I tried this a few years ago, by attaching a thermocouple to the cantilever at the stylus shank, and obtained a negligible rise in temperature above ambient - however it was not a good experiment as the thermal conductivity of the thermocouple was significant.
Anyone wishing to double check, I used:
236 Wm-1K-1 for thermal conductivity of Al
Cross sectional area 1.24E-7 m2
Length 6E-3m
It's all about losses and dissipation, esp into the cartridge suspension. Yes, modelling software is all well and good, but needs the right model, Scott, and I agree: don't think there is anything worthy of the effort here.
LD
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Scaling the physics is a tough problem, remember the old grade school biology argument of why there can't be an 18' man and why an ant can carry several times its own weight with ease.
If the vinyl were melted and frozen, the surface "shine" would change after the first play. It should be visible to the naked eye from a distance, as a change of reflection.
Wrong. I was very clear in stating that the heat build up of 5.4 degree Kelvin per second was only in case no heat could be taken away.
I regarded the stylus cantilever assembly as one homogeneous body with excellent heat conductivity having the same temp all over.
So to say that 3mW would be nothing to worry about is pointless.
Yes, that is exactly what I did to get some feeling what 3mW can possibly do to a cantilever with no cooling.
And of course this is not the case because the LP and the air are consuming heat from the stylus/cantilever.
We all know that the cantilever is not becoming cooking hot, not even close, so heat is quite efficiently absorbed by its surrounding.
An other way of looking at the matter is to caculate how much energy is needed to rise the temp of a 180 gram LP by 1 degree.
Assuming a specific heat for PVC of 1 Joule to raise 1 gram with 1 degree, it follows that 180 Joule is needed to raise the whole 180 gram by 1 degree.
One side of an LP takes maximally 30 minutes generating because of friction 3mW x 1800 sec = 5,4 Joule, way below the 180 Joule needed for 1 degree temp rise.
And yes I know PVC does not have the heat conductivity of copper, but nevertheles it also gives some feeling for orders of magnitude, making it very unlikely that PVC could melt because of heat transfer.
I'm a bit surprised that you mention the cantilever suspension as being a (substantial) heat conductor.
In general elastic polymers are not known for their good heat transfer.
Hans
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Yes. But though cantilever is small, it's not that small in the scheme of things, and stuff like bending models seem to work.
On the premise that 'normal' convection and radiation constants apply with such dimensions, just for cantilever convection and radiation losses I calculate 3mW heat flow can be bought for about 10 deg C rise in cantilever temperature, bulk of which is from convection.
Shout if anyone's interested in detail of calcs for that.
LD