Is the Space Between Frames on Film Standardised?

Any film without sprocket holes is unlikely to have any standard frame size or spacing between frames. As other people have said, 120 film standardizes the film (& backing paper) size, but not the frame size or the spacing between frames. There are many different frame sizes for 120, and spacing is inherently somewhat variable: the camera can not actually know quite how much film it's pulled forward because the film can slide over the rollers. By contrast, any film with sprocket holes is very likely to have a fairly reliable spacing between the equivalent edges of neighbouring frames. The reason for this is that the camera is controlling the position of the film by a roller which engages with the sprockets on the film, and that roller is almost certainly going to advance by an integer number of sprocket-spacings. This is not quite always the case: there are cameras which do not seem to do this (which are a real pain to use), but almost all do. What this means is that the spacing between the (say) left-hand edge of a frame and the left-hand edge of the subsequent frame will be pretty constant (assuming the camera is being used in landscape). Anyone who has cut 35mm film up into strips will have noticed this. But this is not quite the same as saying that the spacing between frames is always the same, because the frames can be different sizes. Frames are nominally 24mm by 36mm, but they can vary in size significantly depending on the camera. Again, this is something that people who have made enlargements from 35mm will have discovered: you have to either adjust the mask, adjust the magnification (ie move the head) or both when changing to negs from a different camera.For 35mm still film there are a bunch of standards of course, described here. 35mm film for still cameras is 'KS1870' which means it uses Kodak Standard perforations and the spacing is 0.1870 in. There are 8 perforations between subsequent full-frame edges (and I think 4 for half-frame). This means that subsequent full-frame edges are spaced by 1. 496 in, which is 37. 9984 mm: 38mm in other words. For a neg size of 24mm x 36mm this gives a spacing of 2mm between frames.35mm film is well-standardised, of course, because it started as movie film and movie people really need to care about things like sprocket holes and frame spacing.Sheet film is less standardized. The two common extant varieties are 4in x 5in and 10in x 8in. There is a common film-holder standard (which has a name which I forget), but that standard is more concerned with making the film-holder fit the camera: the neg size between different models of film holder can vary somewhat. But there are other sheet-film sizes such as, for instance, whole plate and various fractions of a plate (half, quarter), where standards never really arose at all. Ilford have special yearly sales where they will cut sheet film to various odd sizes for people with these strange cameras (I really want a whole plate camera, although I can not afford the film for one realistically).

1. Is this how plastic eyeglasses frames always are?

It's at ease, the glasses cost 5$, nonetheless the Lens can even be further for those who occur to dont want the low rate plastic ****, they by and large moreover furnish certain supplies like anti reflective and distinct matters for an additional ammount. So its a comfortably trick to make you buy further then simply the glasses

2. Physical meaning of frames in general relativity

To calculate the gravitational acceleration you simply calculate the four-acceleration:$$ A^alpha = fracmathrm d^2x^alphamathrm dtau^2 Gamma^alpha_,,mu

uU^mu U^

u tag1 $$In your question you mention the geodesic equation, but the geodesic equation describes the freely falling observer i.e. the observer for whom the four-acceleration is zero. And indeed setting $mathbf A = 0$ in (1) immediately gives us the geodesic equation:$$ fracmathrm d^2x^alphamathrm dtau^2 = - Gamma^alpha_,,mu

uU^mu U^

u $$So by starting with the geodesic equation you are starting by assuming the acceleration is zero and you will never be able to calculate what you want.The gravitational acceleration $G$ felt by an observer, i.e. the proper acceleration, is the norm of the four-acceleration for that observer:$$ G^2 = g_alphabeta A^alpha A^beta $$and since this is a scalar we can use any convenient coordinate system to calculate it. The example commonly used for students is an observer hovering at fixed distance from a spherical mass i.e. the spacetime geometry is the Schwarzschild geometry. Since $u_r = u_theta = u_phi = 0$ the four-acceleration is simply:$$ mathbf A = (0, fracGMr^2, 0, 0) $$And the norm is then:$$ G = fracGMr^2frac1sqrt1-frac2GMc^2r $$Which is just the Newtonian expression modified by that factor of $1/sqrt1-r_s/r$.

3. what are some brands that offer light frames?

They have very light wire or even rimless frames, the problem is, if you have really strong lenses, they will be more obvious if you have very light frames. If you stick with small lense sizes and get the high index ones, it might not be a big deal, depends on your exact prescription. Some of the eyeglass places have a computer program that will tell you the approximate thickness your lenses will be if they know the prescription and the lens size. There are many on-line places to look at frames (even if you want to get glasses at a local shop, you would still have a list of frames you like). these have tons of nice designer frames. so you probably wo not find these exact frames in local eyeglasses shops