4 ms·

Holy crap. 125,000RPM for a 5mm disc means the edge is spinning at nearly 9,000mph. On top of that, the g-force would be a = v^2 / r v = 2 * pi * 5mm

by tbabb 10y ago

Holy crap. 125,000RPM for a 5mm disc means the edge is spinning at nearly 9,000mph. On top of that, the g-force would be a = v^2 / r v = 2 * pi * 5mm * 125000/60 s^-1` r = 5mm ...giving 87,000 gees?! I start to wonder if general relativity effects (frame-dragging?) start to become noticeable at that acceleration.

jessriedel 10y ago
This is an impressively high spin rate. The sub-$1k centrifuges on Amazon look like they about 10,000 RPM, although at least one gets 100,000 RPM. https://www.amazon.com/s/ref=s9_dnav_bw_ct02_s?node=317985011&pf_rd_m=ATVPDKIKX0DER&pf_rd_s=merchandised-search-2&pf_rd_r=2MEEMSYNQJR3WY5B5YXN&pf_rd_t=101&pf_rd_p=47218708-b9f8-48c8-b89e-556a5cc1a394&pf_rd_i=317983011 https://www.amazon.com/s/ref=s9_dnav_bw_ct02_s?node=31798501...
hguant 10y ago
Quick math and google says no: 9,000 MPH = 0.00001342 C Threshold for relativistic effects is .01 C according to this paper: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/rellim.html http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/rellim.ht... EDIT: Rereading that paper, they're focused at the particle level, which may or may not make this irrelevant. I know anecdotally that GPS satellites have to take relativity into account. Geostationary satellites move at 1.9 miles per seconds, which is 6840 mph - quite a bit lower than our centrifuge. That being said, the precision required for GPS means that very small changes due to relative effects have a rather large impact. In short, where there is motion, there is relativity. Is it enough to measure here? Possibly - sticking a small microcontroller and having it report the time would be interesting. Is it enough to matter? The answer is relative.
- wosos 10y ago
  Yes, putting MC and check it's time after a device-lifetime of use would be quite an interesting experiment.
- andrepd 10y ago
  I believe in GPS satellites relativistic effects are due in greater part to the difference in gravity between the surface of the earth and the satellite, more so than the special relativity effects.
  - sobani 10y ago
    According to a reference[1] from Wikipedia[2] the effect would be 45 μs for general relativity 7 μs for special relativity, combining into a 38 μs drift. So although you are technically correct (45 > 7) they are both significant. [1] http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.... [2] https://en.wikipedia.org/wiki/Global_Positioning_System#History https://en.wikipedia.org/wiki/Global_Positioning_System#Hist...
- dogma1138 10y ago
  Relativity has almost no effect on GPS at it's worse it would have an accuracy degradation of less than 1cm, there are considerably bigger sources of errors in GPS than relativity and various corrections to those. Shadows, echoes, atmospheric effects and the doppler effect have considerable effect on GPS accuracy, relativity will not.
  - function_seven 10y ago
    Relativity has a massive effect on GPS. If unaccounted for, the error would increase by 11km per day. EDIT: I might be wrong on the “11km” part. See reply http://physics.stackexchange.com/questions/1061/why-does-gps-depend-on-relativity http://physics.stackexchange.com/questions/1061/why-does-gps...
    - dogma1138 10y ago
      No it doesn't http://www.physicsmyths.org.uk/gps.htm http://www.physicsmyths.org.uk/gps.htm This would only be an issue if you had substantial drift between different sattlites the 38 microsecond delay is irrelevant since all satellites in the constellation have a similar delay. It's also less relevant when solving using the more common method which is time-tagging at the receiver, relativity might be a bigger pain if you are doing time-tagging at the transmitter but since the clocks of different satellites are not synchronized and can differ by as much as 1ms that delta would be more of an issue than relativity. If you are doing normal GPS accuracy which is 15m you don't need to use relativistic correction in the receiver, is you are solving for almanac data you do not need to do relativistic correction in the receiver, you only really do relativistic corrections when you are using centimeter level accuracy GPS and even then you have considerably bigger corrections to do. Yes relativity was accounted for when GPS was designed, but it's by no means would send you to China, not even to the wrong town. What GPS needs to work is for the receivers to be able to calculate a proper geometric range delay or at least be constantly and consistently wrong when calculating the delay for different satellites in the cluster. http://www.navipedia.net/index.php/GNSS_Basic_Observables http://www.navipedia.net/index.php/GNSS_Basic_Observables Discounting relativity would reduce accuracy, but it would be consistent and constant and so would not present that much of an issue, it's also by far not the biggest offender when it comes to corrections.
pvg 10y ago
No it isn't, you must have mistyped something. The diameter, not radius is 5mm, not that it makes a big difference. The top speed is a little over 30 meters per second or 70 mph. (Edit: possibly because you turned seconds into 1/s for some reason? Either way, the edge of the disk is not going hypersonic or anywhere close to it)
- Nition 10y ago
  Actually the radius is 5mm in the paper, it was nneonneo that got that part wrong.
  - pvg 10y ago
    The wrong part doesn't come from the 5mm, it comes from being orders of magnitude wrong by some other method.
    - Nition 10y ago
      Oh, yeah I know. Didn't mean to imply your main point was incorrect.
- nneonneo 10y ago
  It's my fault - I misread R_d as diameter and not radius. The disc is indeed 5mm in radius (10mm in diameter). With this value, the speed of the disc edge is 65m/s or about 145mph.
Yetanfou 10y ago
Where do you get those 9000 mph from? as far as I can see the edge velocity will be 65.45 m/s or 236 km/h: Fill in the blanks, m/s: v (in m/s) = pi * D * rpm / 60 (D in m) 3.141593 * 0.01 * 125,000 / 60 = 65.45 (m/s) or, in km/h v (in km/h) = pi * D * rpm * 60 / 1000 (D in m) 3.141593 * 0.01 * 125,000 * 60 / 1000 = 235.62 (km/h) These values are valid for a disc with a 5 mm radius. If that '5 mm' indicates the diameter the values would be halved to 32.72 m/s or 118 km/h.
yosyp 10y ago
You're wrong. 125,000 RPM with a 5 mm diameter is 235.58 km/hour or 146.4 mph, giving you 1,712,957 m/s^2 or 174,613 Gs (weird!)
taneq 10y ago
You used 5m instead of 5mm, putting you out by a factor of 1000: 2 * Pi * 0.005m * 125000/60 s^-1 = 65.4m/s = 235km/h.

by tbabb 10y ago

jessriedel 10y ago

This is an impressively high spin rate. The sub-$1k centrifuges on Amazon look like they about 10,000 RPM, although at least one gets 100,000 RPM. https://www.amazon.com/s/ref=s9_dnav_bw_ct02_s?node=317985011&pf_rd_m=ATVPDKIKX0DER&pf_rd_s=merchandised-search-2&pf_rd_r=2MEEMSYNQJR3WY5B5YXN&pf_rd_t=101&pf_rd_p=47218708-b9f8-48c8-b89e-556a5cc1a394&pf_rd_i=317983011 https://www.amazon.com/s/ref=s9_dnav_bw_ct02_s?node=31798501...

hguant 10y ago

Quick math and google says no: 9,000 MPH = 0.00001342 C Threshold for relativistic effects is .01 C according to this paper: http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/rellim.html http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/rellim.ht... EDIT: Rereading that paper, they're focused at the particle level, which may or may not make this irrelevant. I know anecdotally that GPS satellites have to take relativity into account. Geostationary satellites move at 1.9 miles per seconds, which is 6840 mph - quite a bit lower than our centrifuge. That being said, the precision required for GPS means that very small changes due to relative effects have a rather large impact. In short, where there is motion, there is relativity. Is it enough to measure here? Possibly - sticking a small microcontroller and having it report the time would be interesting. Is it enough to matter? The answer is relative.

wosos 10y ago
Yes, putting MC and check it's time after a device-lifetime of use would be quite an interesting experiment.
andrepd 10y ago
I believe in GPS satellites relativistic effects are due in greater part to the difference in gravity between the surface of the earth and the satellite, more so than the special relativity effects.
- sobani 10y ago
  According to a reference[1] from Wikipedia[2] the effect would be 45 μs for general relativity 7 μs for special relativity, combining into a 38 μs drift. So although you are technically correct (45 > 7) they are both significant. [1] http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.... [2] https://en.wikipedia.org/wiki/Global_Positioning_System#History https://en.wikipedia.org/wiki/Global_Positioning_System#Hist...
dogma1138 10y ago
Relativity has almost no effect on GPS at it's worse it would have an accuracy degradation of less than 1cm, there are considerably bigger sources of errors in GPS than relativity and various corrections to those. Shadows, echoes, atmospheric effects and the doppler effect have considerable effect on GPS accuracy, relativity will not.
- function_seven 10y ago
  Relativity has a massive effect on GPS. If unaccounted for, the error would increase by 11km per day. EDIT: I might be wrong on the “11km” part. See reply http://physics.stackexchange.com/questions/1061/why-does-gps-depend-on-relativity http://physics.stackexchange.com/questions/1061/why-does-gps...
  - dogma1138 10y ago
    No it doesn't http://www.physicsmyths.org.uk/gps.htm http://www.physicsmyths.org.uk/gps.htm This would only be an issue if you had substantial drift between different sattlites the 38 microsecond delay is irrelevant since all satellites in the constellation have a similar delay. It's also less relevant when solving using the more common method which is time-tagging at the receiver, relativity might be a bigger pain if you are doing time-tagging at the transmitter but since the clocks of different satellites are not synchronized and can differ by as much as 1ms that delta would be more of an issue than relativity. If you are doing normal GPS accuracy which is 15m you don't need to use relativistic correction in the receiver, is you are solving for almanac data you do not need to do relativistic correction in the receiver, you only really do relativistic corrections when you are using centimeter level accuracy GPS and even then you have considerably bigger corrections to do. Yes relativity was accounted for when GPS was designed, but it's by no means would send you to China, not even to the wrong town. What GPS needs to work is for the receivers to be able to calculate a proper geometric range delay or at least be constantly and consistently wrong when calculating the delay for different satellites in the cluster. http://www.navipedia.net/index.php/GNSS_Basic_Observables http://www.navipedia.net/index.php/GNSS_Basic_Observables Discounting relativity would reduce accuracy, but it would be consistent and constant and so would not present that much of an issue, it's also by far not the biggest offender when it comes to corrections.

pvg 10y ago

No it isn't, you must have mistyped something. The diameter, not radius is 5mm, not that it makes a big difference. The top speed is a little over 30 meters per second or 70 mph. (Edit: possibly because you turned seconds into 1/s for some reason? Either way, the edge of the disk is not going hypersonic or anywhere close to it)

Nition 10y ago
Actually the radius is 5mm in the paper, it was nneonneo that got that part wrong.
- pvg 10y ago
  The wrong part doesn't come from the 5mm, it comes from being orders of magnitude wrong by some other method.
  - Nition 10y ago
    Oh, yeah I know. Didn't mean to imply your main point was incorrect.
nneonneo 10y ago
It's my fault - I misread R_d as diameter and not radius. The disc is indeed 5mm in radius (10mm in diameter). With this value, the speed of the disc edge is 65m/s or about 145mph.

Yetanfou 10y ago

Where do you get those 9000 mph from? as far as I can see the edge velocity will be 65.45 m/s or 236 km/h: Fill in the blanks, m/s: v (in m/s) = pi * D * rpm / 60 (D in m) 3.141593 * 0.01 * 125,000 / 60 = 65.45 (m/s) or, in km/h v (in km/h) = pi * D * rpm * 60 / 1000 (D in m) 3.141593 * 0.01 * 125,000 * 60 / 1000 = 235.62 (km/h) These values are valid for a disc with a 5 mm radius. If that '5 mm' indicates the diameter the values would be halved to 32.72 m/s or 118 km/h.

yosyp 10y ago

You're wrong. 125,000 RPM with a 5 mm diameter is 235.58 km/hour or 146.4 mph, giving you 1,712,957 m/s^2 or 174,613 Gs (weird!)

taneq 10y ago

You used 5m instead of 5mm, putting you out by a factor of 1000: 2 * Pi * 0.005m * 125000/60 s^-1 = 65.4m/s = 235km/h.