5 ms·

My girlfriend's mom has a teaching oven (used in home economics classes) from the 1950s. The 'teaching' part mostly means it has a more accurate and detailed te

My girlfriend's mom has a teaching oven (used in home economics classes) from the 1950s. The 'teaching' part mostly means it has a more accurate and detailed temperature gauge. This gas oven is so heavy that light in the kitchen gravitationally lenses around it, but it's still going strong and the best oven I ever used.

PartiallyTyped 3y ago
On a serious note, how heavy does an object have to be for gravitational lensing or bending to be noticeable with a naked eye? s/visible/noticeable s/noticeable/noticeable with a naked eye s/lensing/gravitational lensing
- deleted 3y ago
  [deleted]
- leidenfrost 3y ago
  Depends on how you define "noticeable". If you can measure a small enough distance, you can see the lensing effect of any object.
  - idlewords 3y ago
    Comments like yours are what keep this site great.
  - PartiallyTyped 3y ago
    Noticeable with the naked eye.
  - LorenPechtel 3y ago
    Only in a Newtonian world--but there wouldn't be lensing in a Newtonian world. Once you consider Heisenberg and quantum mechanics you find your signal swamped by noise for smaller objects. Now, figuring out this limit is left as an exercise for the reader as it's way beyond my abilities.
    - leidenfrost 3y ago
      Haha you're right. I was mistaken.
- ben_w 3y ago
  That you'd be able to see it with the naked eye? α = 4GM/((c^2)b), where b is the impact parameter[0]. Apparently human visual acuity is 0.3 milli-radians, so if b = 1 meter, that's approximately "the moon" (in at most a 1 meter radius volume)… …assuming I didn't mix up my units in this formula I never used before, though it feels about right given the Schwarzschild radius of the Earth is ~ centimetres. [0] never heard of this before just now; I think it's the shortest distance between the central point and the path the light would have taken if it hadn't been deflected?
  - adtac 3y ago
    The Schwarzschild radius [0], which defines the radius at which the escape velocity equals the speed of light, is given by 2GM/c^2. I don't know what the impact parameter is either, but given these two expressions, it sounds like b is dimensionless. I don't know how fast the radius of curvature drops off as a function of the Schwarzschild radius, but I'd imagine it's at least R^-1. So assuming a spherical cow^H^H^H oven with gravitational lensing dying out at ~100x the event horizon, we need a Schwarzschild radius the size of a tennis ball in order to still see the curvature a few metres out. The oven needs to weigh ~4x the Earth's mass for that. [0] https://en.wikipedia.org/wiki/Schwarzschild_radius https://en.wikipedia.org/wiki/Schwarzschild_radius
  - treeman79 3y ago
    There is also Superman’s key. Made of neutronium. https://dcmovies.fandom.com/wiki/Fortress_Key_(All-Star_Superman) https://dcmovies.fandom.com/wiki/Fortress_Key_(All-Star_Supe... Last I tried to calculate (poorly), if you were to touch it. You liquify and be sucked into it just before contact.
    - ben_w 3y ago
      (Superhero physics is vague and unrealistic at the best of times). Assuming "dwarf star material" means "neutronium" (which is unstable at any level less than a neutron star) and the "half million tons" mass quote for the key… That's about the energy content of all the world's fossil fuel reserves being released in a 10 minute half-life.
- adolph 3y ago
  > On a serious note, how heavy does an object have to be for lensing or bending to be noticeable with a naked eye? 22 grams. That is how heavy my glasses are.
- dav_Oz 3y ago
  A quick search would tell you: >The angle of deflection (theta) is: theta = (4GM)/(cr^2) toward the mass M at a distance r from the affected radiation, where G is the universal constant of gravitation and c is the speed of light in vacuum.[0] The best resolution our eyes can offer is about one arcminute (1/21600 of a turn). Depending on your distance from the object, just plug in some numbers. Say at the earth-moon distance 384400 km the object must be about 24x the mass of the sun to bend the incoming light at one arcminute (~0,0002909rad). The sun actually bends light at about 2 arcseconds as seen from Earth; the focal point would be about 542x the distance Sun-Earth. [1] Alternatively the object of say 1m^3 volume at a distance of 10 meters will bend light by 1 arcminute if it weighs 3.27x10^16 kg, the density of about 1/10th of a neutron star. To conclude: one will be instantly overwhelmed by the gravitational forces before being able to see an object bend light with one own eyes. That's why this kind of extreme bending/lending is reserved for galaxy clusters. [0]https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_spacetime_curvature https://en.wikipedia.org/wiki/Gravitational_lens#Explanation... [1]https://en.m.wikipedia.org/wiki/Solar_gravitational_lens https://en.m.wikipedia.org/wiki/Solar_gravitational_lens
ethbr0 3y ago
I have a Sunbeam Radiant Control toaster that still works flawlessly. https://m.youtube.com/watch?v=1OfxlSG6q5Y https://m.youtube.com/watch?v=1OfxlSG6q5Y They used a shielded bimetalic thermostat to measure the radiant heat coming from the toast surface, instead of a timer, to establish doneness (consistently toasting despite heating element variance) and implemented the mechanical lowering & raising of toast without a single motor (which is why 60+ year old examples still work).
- bern4444 3y ago
  What if you only want 1 piece of toast and not two? Will it still automatically go down?
  - ethbr0 3y ago
    I believe they're weighted so that a single piece of toast has enough mass to trip the mechanism. Since they only use a thermostat on one side, you do need to use a particular side for single toast.
  - reichstein 3y ago
    Only one of the holes has a trigger. You fill that last, or only.
- idlewords 3y ago
  But how do you cancel without a cancel button?
  - ethbr0 3y ago
    If you make a toast-related error, by inserting bread into the toaster and then deciding you don't actually want it toasted, you unplug the toaster. As the elements cool, your bread returns to you, unharmed.
    - mcpeepants 3y ago
      > unharmed harmed to a degree that is proportional to the toast-decision-making latency of the operator

PartiallyTyped 3y ago

On a serious note, how heavy does an object have to be for gravitational lensing or bending to be noticeable with a naked eye? s/visible/noticeable s/noticeable/noticeable with a naked eye s/lensing/gravitational lensing

deleted 3y ago
[deleted]
leidenfrost 3y ago
Depends on how you define "noticeable". If you can measure a small enough distance, you can see the lensing effect of any object.
- idlewords 3y ago
  Comments like yours are what keep this site great.
- PartiallyTyped 3y ago
  Noticeable with the naked eye.
- LorenPechtel 3y ago
  Only in a Newtonian world--but there wouldn't be lensing in a Newtonian world. Once you consider Heisenberg and quantum mechanics you find your signal swamped by noise for smaller objects. Now, figuring out this limit is left as an exercise for the reader as it's way beyond my abilities.
  - leidenfrost 3y ago
    Haha you're right. I was mistaken.
ben_w 3y ago
That you'd be able to see it with the naked eye? α = 4GM/((c^2)b), where b is the impact parameter[0]. Apparently human visual acuity is 0.3 milli-radians, so if b = 1 meter, that's approximately "the moon" (in at most a 1 meter radius volume)… …assuming I didn't mix up my units in this formula I never used before, though it feels about right given the Schwarzschild radius of the Earth is ~ centimetres. [0] never heard of this before just now; I think it's the shortest distance between the central point and the path the light would have taken if it hadn't been deflected?
- adtac 3y ago
  The Schwarzschild radius [0], which defines the radius at which the escape velocity equals the speed of light, is given by 2GM/c^2. I don't know what the impact parameter is either, but given these two expressions, it sounds like b is dimensionless. I don't know how fast the radius of curvature drops off as a function of the Schwarzschild radius, but I'd imagine it's at least R^-1. So assuming a spherical cow^H^H^H oven with gravitational lensing dying out at ~100x the event horizon, we need a Schwarzschild radius the size of a tennis ball in order to still see the curvature a few metres out. The oven needs to weigh ~4x the Earth's mass for that. [0] https://en.wikipedia.org/wiki/Schwarzschild_radius https://en.wikipedia.org/wiki/Schwarzschild_radius
- treeman79 3y ago
  There is also Superman’s key. Made of neutronium. https://dcmovies.fandom.com/wiki/Fortress_Key_(All-Star_Superman) https://dcmovies.fandom.com/wiki/Fortress_Key_(All-Star_Supe... Last I tried to calculate (poorly), if you were to touch it. You liquify and be sucked into it just before contact.
  - ben_w 3y ago
    (Superhero physics is vague and unrealistic at the best of times). Assuming "dwarf star material" means "neutronium" (which is unstable at any level less than a neutron star) and the "half million tons" mass quote for the key… That's about the energy content of all the world's fossil fuel reserves being released in a 10 minute half-life.
adolph 3y ago
> On a serious note, how heavy does an object have to be for lensing or bending to be noticeable with a naked eye? 22 grams. That is how heavy my glasses are.
dav_Oz 3y ago
A quick search would tell you: >The angle of deflection (theta) is: theta = (4GM)/(cr^2) toward the mass M at a distance r from the affected radiation, where G is the universal constant of gravitation and c is the speed of light in vacuum.[0] The best resolution our eyes can offer is about one arcminute (1/21600 of a turn). Depending on your distance from the object, just plug in some numbers. Say at the earth-moon distance 384400 km the object must be about 24x the mass of the sun to bend the incoming light at one arcminute (~0,0002909rad). The sun actually bends light at about 2 arcseconds as seen from Earth; the focal point would be about 542x the distance Sun-Earth. [1] Alternatively the object of say 1m^3 volume at a distance of 10 meters will bend light by 1 arcminute if it weighs 3.27x10^16 kg, the density of about 1/10th of a neutron star. To conclude: one will be instantly overwhelmed by the gravitational forces before being able to see an object bend light with one own eyes. That's why this kind of extreme bending/lending is reserved for galaxy clusters. [0]https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_spacetime_curvature https://en.wikipedia.org/wiki/Gravitational_lens#Explanation... [1]https://en.m.wikipedia.org/wiki/Solar_gravitational_lens https://en.m.wikipedia.org/wiki/Solar_gravitational_lens

ethbr0 3y ago

I have a Sunbeam Radiant Control toaster that still works flawlessly. https://m.youtube.com/watch?v=1OfxlSG6q5Y https://m.youtube.com/watch?v=1OfxlSG6q5Y They used a shielded bimetalic thermostat to measure the radiant heat coming from the toast surface, instead of a timer, to establish doneness (consistently toasting despite heating element variance) and implemented the mechanical lowering & raising of toast without a single motor (which is why 60+ year old examples still work).

bern4444 3y ago
What if you only want 1 piece of toast and not two? Will it still automatically go down?
- ethbr0 3y ago
  I believe they're weighted so that a single piece of toast has enough mass to trip the mechanism. Since they only use a thermostat on one side, you do need to use a particular side for single toast.
- reichstein 3y ago
  Only one of the holes has a trigger. You fill that last, or only.
idlewords 3y ago
But how do you cancel without a cancel button?
- ethbr0 3y ago
  If you make a toast-related error, by inserting bread into the toaster and then deciding you don't actually want it toasted, you unplug the toaster. As the elements cool, your bread returns to you, unharmed.
  - mcpeepants 3y ago
    > unharmed harmed to a degree that is proportional to the toast-decision-making latency of the operator