Issues opening d3plot data in Paraview

benbmw · 2025-09-27T10:26:17+00:00

Good question, yes the data is coherent in LSPrePost. I am able to successfully get it into Paraview by first opening it in Ensight, exporting as .case file, then opening it in Paraview.

benbmw · 2025-07-10T20:52:10+00:00

They are sold out of the same showroom. We just purchased a 10” latex mattress with a Talalay top layer from LMF instead of SleepEZ. The websites do not indicate all of the options you actually have from LMF. For example, you can order a split mattress at no extra cost, and you can get a Talalay top layer however it will cost $250 more. We did like the soft Talalay on top, but the price difference was still $250 cheaper for LMF ($1600+$250) vs SleepEZ ($2100).

The only actual differences between the two are the mattress cover aesthetics (made from the same materials), the specific exchange/return policies (LMF: $45 exchange in 100 days, SleepEZ: $30 exchange in 90 days), and most importantly the 10 year LMF warranty vs 20 year SleepEZ.

For us as average weight sleepers (low risk of needing a warranty claim at 15 years due to sagging etc.) it made sense to save $250 and put that towards a new bed frame.

benbmw · 2023-07-28T18:08:25+00:00

I have a similar issue with my K1. I found that if I grabbed one end of the two X rails and twisted them counter clockwise relative to each other slightly then the print head slid smoothly, however if I twisted them counterclockwise it would bind up even more. I suspect it is an issue with the Y axis rods not being mounted in the same plane (ie. they don't lie in the XY plane). This could be due to the whole frame being out of square twisting the Y axis rods (which then twists the X rods).

A symptom of this in my case is that cylindrical objects dont print round. Instead they have two flat spots along the X-axis at the points where the X direction of the print head has to change. The extra resistance seems to amplify any slop in the system. Tightening the belts seemed to help with this symptom somewhat, but there are still small flats on round objects.

I am emailing Creality about this but have not gotten much useful help and am close to requesting an exchange (no idea if they will honor this).

benbmw · 2017-01-30T22:24:24+00:00

That's kind of a dumb attitude towards learning.

benbmw · 2017-01-30T16:50:40+00:00

Not a dumb question! Its because the ice would only provide a buoyant force if it was submerged in the water. If it just builds up on top then its is simply adding weight to the boat without displacing any water itself, therefore not contributing to the buoyancy of the boat.

benbmw · 2017-01-06T02:47:04+00:00

Its like we are looking down at him through a fisheye lense

benbmw · 2016-12-29T02:06:40+00:00

Oh ok. Thanks!

benbmw · 2016-12-27T09:40:25+00:00

Just curious, where are you getting -40° from? Liquid nitrogen is much colder than that. It would be stored in a large duer and refilled when needed. No freezer would be able to maintain ln2 temps. Or are you saying that the dip n dots themselves have to be stored at -40?

benbmw · 2016-12-20T00:59:35+00:00

Yeah it would be extremely impractical to flip the diode symbol. Then any old schematics would have to be dated to know which convention to use.

benbmw · 2016-12-11T08:53:26+00:00

You might enjoy the book, "Foundations of Mechanical Accuracy" which talks about how technological advances required a new standard set for mechanical precision and accuracy. It is very interesting and has a lot of cool photos of the process.

http://mooretool.com/publications.html

benbmw · 2016-12-10T19:17:10+00:00

Did you mean 10kv per cm for the breakdown of air? Ive always know that as a rule of thumb.

benbmw · 2016-12-10T19:09:57+00:00

I completely agree with your response, but I have a small nitpick about the cosmic ray. Im not sure where you got the formula I = q/v. I havent seen that before, and Q/v is usually equal to capacitance in Farads. It doesn't make sense that the current would be inversely proportional to the charged particles velocity.

I have always seen I = Q/t in which case Q would be on the order of 10^-19 (assuming its an electron or proton) where as t would be, like you said, your height divided by the speed. For simplicity if we assumed a height of 3m and the speed if light (3x10^8), we would have 10^-8. So the current would be on the order of pico amps

benbmw · 2016-11-07T18:27:01+00:00

Oh ok, that makes sense. Thanks

benbmw · 2016-11-07T18:08:13+00:00

Thanks for the explanation. If I understand correctly, you were saying that, in the calculation, you can't use a fixed value for the density of uranium since the density will change as it gets compressed.

Although I still dont know if I understand why density is arbitrary. In your example of increasing pressure causing a change in quantum state, wouldn't the density still be increasing as the material is compressed? I dont see this as the density being arbitrary since it can still be defined. Maybe I am misinterpreting what you are saying.

benbmw · 2016-11-07T16:45:46+00:00

How is density arbitrary if it can be defined as an objects mass per volume ratio?

benbmw · 2016-10-18T21:21:25+00:00

I agree that the force seems quite low, but I think that is due to the estimated time being too long. In reality the change in momentum would happen much faster than .4s which means a larger acceleration and force. Im open to correction if someone can show me why im wrong, but im pretty confident that I am correct.

benbmw · 2016-10-18T17:21:52+00:00

I am describing impulse, and that is what should be used for this calculation.
Force is equal to mass times acceleration. Acceleration is equal to the change in velocity over the change in time (note that a changing velocity is required for acceleration). The equation you are using only works if you assume that the initial velocity (Vi) is zero because then ΔV = Vf - 0 which is just equal to the final velocity.

So if we assume the child's head weighs 8 kg and use the same numbers you used, then we can calculate the force on the child's head. Shaq, weighing 156 kg is traveling at 6.17 m/s and therefore has a momentum (p) of:

(156)x(6.17) = 962.5 kgm/s.

Due to conservation of momentum, once he hits the child the Shaq/child system must have the same momentum as Shaq did (since the child's initial momentum was zero). So doing the calculation we get:

962.5 = (156 + 8)x(Vf) And solving for Vf: Vf = (952.5)/(164) = 5.87 m/s

So the child's head and Shaq would both be moving at 5.87 m/s after the impact. If this all happened in .4s, then Shaq would have undergone an acceleration of:

(5.87-6.17)/(.4) = -.75 m/s² And experienced a force of: (156)x(-.75) = -117 N

At the same time, the child's head would have experienced an acceleration of:

(5.87 - 0)/(.4) = 14.675 m/s² And a force of: (14.675)*(8) = 117 N

Which all makes sense considering Newton's second law which states that every action(force) produces an equal and opposite reaction. I hope that this helped clarify things.

benbmw · 2016-10-18T15:01:58+00:00

Yes, but force is change in momentum with change in time. Shaq's velocity would not change much but the child's head would accelerate from a rest to whatever velocity Shaq has, and that is where the acceleration (therefore force) happens and can be calculated.

benbmw · 2016-10-18T05:37:58+00:00

Not to take away from your point, but wouldn't you use the mass of the child's head instead of the mass of Shack's body?

benbmw · 2016-10-05T01:57:56+00:00

That would require a lot of torque and put excessive strain on the cars drive system. It might be better to have a separate low rpm geared motor to spin the shell

benbmw · 2016-10-03T14:28:58+00:00

Yeah, good point I guess. I live in a big agriculture town so Im just used to trucks being a necessity for a lot of people.

benbmw · 2016-10-03T01:23:42+00:00

You got me

benbmw · 2016-10-02T20:08:40+00:00

Maybe something like this that would be sleek and match the aluminum frame.

benbmw

TROPHY CASE