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About randomic

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  1. On topic: Nice try, telescope thief! But seriously I have displaced a wardrobe of my fiancée's clothes and it is now my astro gear cupboard.
  2. The whole glass being a very slow flowing liquid thing is a myth. You don't have to worry about your lenses. https://www.cmog.org/article/does-glass-flow
  3. The EdgeHD 800 is amazing for planets. On a good night I can make out details on Mars and can see the disk and colour of Uranus, it's awesome. I use the Baader Zoom 8-24mm eyepiece. For imaging planets I have a ZWO ASI 462MC but I've not been able to take it for a spin yet due to weather. You technically can do it with a DSLR but what you really want is high framerate video capture. For galaxies, the EdgeHD 800 is again pretty great for visual. I use the 40mm eyepiece which comes with the OTA and can make out the core of Andromeda even from very light polluted skies. Your experiences here will depend mostly on light pollution. The darker the skies the more you'll see. When it comes to photographing large objects like galaxies the EdgeHD 800 will struggle. You say you have a full frame camera which is good but even with that your field of view is narrow enough that you'd have to mosaic to get a complete picture of, say, Andromeda. If you're photographing smaller galaxies or nebulae then it's absolutely fine. Just be aware that the mount is the biggest factor here, even small imperfections in tracking can easily ruin long exposures when you're at 2000mm focal length.
  4. I almost bought that combo but the general consensus is you can get better mounts for the same money. So, I bought the EdgeHD 8" OTA and HEQ5 Pro separately. CEM25p is a good alternative. An 8" SCT OTA is on the limit of what these mounts can perform well with for long exposures so you might want to stretch for a Celestron CGX or CEM40 mount. Alternatively you might get an autoguiding setup to help the mount perform at its best. I've really enjoyed my EdgeHD 800 so far but it might be just as good getting a C8 XLT, unless you know for sure that you want the internal flattener (which comes with its own set of pitfalls).
  5. If I had to spend that much I'd probably go with iOptron CEM40 and a Celestron EdgeHD 925 but it depends a LOT on what kind of targets you want to shoot. BUT If you're just starting out I don't think you need to spend that much. A SkyWatcher HEQ5 Pro or iOptron CEM25p paired with an 80mm apo like SkyWatcher Evostar 80ED will net you some amazing shots of wide field targets.
  6. Just seen this thread, maybe it's the same for you?
  7. Yes, however tides in gasses tend to be dominated by heating effects so you don't notice the gravitational atmospheric tide so much. It's still a contributor though.
  8. I made a diagram which might help visualise. These points are not interacting so in the right-hand case they would drift apart indefinitely. Obviously, the oceans are gravitationally bound to the Earth so they don't just leave The red ellipses are not supposed to be an accurate representation of the tidal bulge shape, rather just a visual aid. P.S. The end of the caption on the right hand diagram should say "the outer two move away from it."
  9. Hypothetically, couldn't everything happen on the gpu? Since as long as the frames end up in the right order it doesn't matter in what order they're processed (assuming you don't run out of vram).
  10. Hot damn, I'd better make sure I crop the frame!
  11. Out of boredom I built oacapture from source this afternoon and you'll be pleased to know that the ASI462MC is now recognised correctly. It seems to work fine although only manages to record at 6 fps. On a semi-related note, is a 67 frame SER supposed to be ~500MB?
  12. Yep, the bulge is caused by the non-uniform gravitational field across the Earth. We see the bulge in the oceans because the Earth is (relatively) rigid and water is not. Check my posts in this thread for more information.
  13. It shouldn't be too hard to run the numbers. We'll keep things Newtonian for simplicity. So a = GM/r^2 G is Gravitational constant 6.67e-11 Radius of Earth, re, is 6,371km Mass of Sun is 2e30 kg Distance to Sun from Earth centre, rs, is 149,785,000km (going off the above diagram) Mass of Moon is 7.3e22 kg Distance to Moon from Earth centre rm, is 384,835km a at Earth centre due to Sun = 6.67e-11 x 2e30 / (1.5e11 + 0)^2 = 0.0059289 m/s^2 a at Earth close edge due to Sun = 6.67e-11 x 2e30 / (1.5e11 - 6.371e6)^2 = 0.0059294 m/s^2 The difference between these two is the tidal force due to the Sun = 0.0000005 m/s^2 a at Earth centre due to Moon = 6.67e-11 x 7.3e22 / (3.8e8 + 0)^2 = 0.000033719 m/s^2 a at Earth close edge due to Moon = 6.67e-11 x 7.3e22 / (3.8e8 - 6.371e6)^2 = 0.000034879 m/s^2 The difference between these two is the tidal force due to the Moon = 0.00000116 m/s^2 So the Sun imparts a tidal force around 43% of the Moon's tidal force. It's a bit different from the 46% in the diagram above because I did a lot of rounding. To me, it's astonishing that the Sun imparts such a significant tidal force given how far away it is. It really helps appreciate just how ridiculously massive the Sun is compared to everything else in the solar system.
  14. As a side note: counterintuitively, this means that you not only weigh less when the moon is directly overhead but also when the moon is directly underneath.
  15. The key part which is being missed is that it's the non-uniformity in the field which is important. If we imagine the Earth as a rigid sphere then: the water closest to the Moon "feels" a stronger acceleration towards the Moon than the Earth does the Earth "feels" a stronger acceleration towards the Moon than the water furthest from the Moon does the net result is that you get a tidal bulge at each side. It's got nothing to do with the springiness of water or pressure or anything like that. You see tidal forces in action in merging galaxies, where stars are (effectively) only interacting gravitationally. You still see two bulges. However, there is a whole other interesting topic which is due to the "springiness" of the system: tidal locking. It's no coincidence that one side of the moon always faces us. If you're in a uniform gravitational field there is no tidal bulging in the direction of the field. This graphic from the wikipedia may help visualise: Graphic of tidal forces. The top picture shows the gravity field of a body to the right, the lower shows their residual once the field at the centre of the sphere is subtracted; this is the tidal force.
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