studiot

Exactly. So something is wrong with the problem, or with your reading of it.

So what does that make the angle opposite the side you have labelled as d in the bottom triangle? In other words what does that make angles alpha and beta?

Fair comment. Thank you.

How many right angles in a straight line?

Ask Mr Microsoft why Office has to be online these days. Another scenario is the overpowerful IT department that insists everything is run from a central corporate server (which doesn't have to be on the web) but then the bean counters get offered an 'irresistable' deal on online (cloud) storage so buy that instead of paying to upgrade their own server.

I presume that those right angles at the centre are a mistake? This problem is known to surveyors and navigators as the resection problem. that is they determine their position by taking bearing sights on three objects of known position. One solution method is due to Tienstra https://en.wikipedia.org/wiki/Tienstra_formula Here is an online solver to play with http://mesamike.org/geocache/GC1B0Q9/tienstra/

Hello, Tar The situation of reaching orbit is unlikely as no normal gun can send a bullet at anything approaching escape velocity. Without reaching escape velocity or having the benefit of a sustained propulsion behind it any projectile will eventually return to Earth

This is homework help so you need to post both the actual problem and at least one of your failed attempts (perhaps in outline). Your jpg has not taken so I can't see the problem. There is a moderator online right now (swansont) so ask him for help getting your question posted.

You really should have provided more information. So I have offered an amendment to your OP and title, hope you don't mind. A bullet fired horizontally from a gun and a bullet dropped from the same height as the muzzle will reach the ground after the same journey time, ignoring the curvature of the Earth and wind resistance. This is because both bullets start with zero vertical velocity and are only subject to the same (vertical) distance to ground and (vertical) acceleration.

Hello pavel, good to see you back again, you often add something useful to threads. With regard to part (b) of your question the answer is not so simple. It depends upon what you are sampling. Here are two examples. 1) You have divided a cornfield into squares for testing yield. Placing all the tests sequentially in one corner may hit fertile or stony ground. 2) You are a buyer checking a box of apples for % rotten apples. A rotten apple is much more likely to infect neighbours than remote apples. In both of these sampleing sequentially may well produce different resutls from a proper spread of testing. Sampling theory developed from real life situations like these and it is obviously important to ensure that the sample is as representative as possible of the whole sampled population. This is why sampling theory is a huge area in its own right.

Thank you both for this entertaining game of ping-pong and also for introducing me to Bell's paradox, which I had not heard of before. +1 each I don't actually see the issue with Bells. Even in a purely Newtonian universe if you accelerate a compound object hard enough and long enough you will break it. The pair of spaceships and the gossamer thread constitute a compound object. Just as a bag of blood in a centrifuge does. What happens to the bag of blood if you overspin? It demonstrates the Principle of Equivalence very nicely.

In order to have a loop you require a continuum that can contain such a loop. Closed timelike curves can occur in the geometry of spacetime, which is the continuum in which timelike curves of any description occur. You do not therefore need extra dimensions. For a continuum with one totally independent time dimension, there are not enough dimensions to accomodate loops. So you would have to postulate a continuum with additional dimension(s). This is just the geometry. In order for an inhabitant of either conintuum to traverse the loop you would also need an equation of motion that included the loop in its solutions.

In my first post and several times subsequently I asked you very politely to define radius. I expect you have seen the type of fairground ride where a seating capsule or capsules is rotated on the end of an metal arm or arms of fixed length L about a central hub. So here is my question to you about your claim: What is the radius of rotation of this mechanism ? I am not seeking numbers, symbols in a formula will do. Please note this forum requires you to answer this question fully in a way that I can use to calculate the radius. "If the radius is zero then the angular momentum is also zero by definition." There is where you are showing lack of understanding. Consider a rotating sphere with non zero radius. Every point in that sphere has zero radius but is a mass point. The overall sphere has an angular momentum which is the sum of all the individual angular momenta of all the mass points about the central axis of that sphere. The overall mass of that sphere is the sum of the masses of all the mass points. If, as you say, all the mass points on the axis of that sphere have zero angular momentum they contribute nothing to the angular momentum of the sphere, but they do contribute mass to the mass of the sphere. How can this paradox be resolved?

The radius of a point particle is zero, by definition of a point particle. Point particles may possess angular momentum. You didn't address my second point. What is the radius of a twisted space curve?

Since you have already demonstrated that you can correctly perform the arithmetic of fractions, I see no reason why you can't work out the answer to this for yourself. So are they the same? When you have done that you might like to consider this. Most people just use the formulae you quoted in your post#1 without thinking about the detail or the rules. The rules are: The % must all be calculated on the same base. The % must be mutually exclusive. This is why I made such a fuss about the difference between and and or in my post#7 To show this try calulating the following %. [math]\left( {100} \right)\left( {\frac{{200}}{{500}}} \right) = a\% [/math] [math]\left( {100} \right)\left( {\frac{{260}}{{500}}} \right) = b\% [/math] [math]\left( {100} \right)\left( {\frac{{300}}{{500}}} \right) = c\% [/math] [math]\left( {100} \right)\left( {\frac{{360}}{{500}}} \right) = d\% [/math] They are all reckoned on the same base in accordance with rule 1. I have already discussed rule 2 in post#7, read it again. now calculate [math]\left( {b - a} \right) + \left( {d - c} \right)[/math] What do you notice?

I don't know if that was meant as an answer to my post or not so I will give the benefit of the doubt. You have been told umpteen times You can't just add the percentages. You can certainly simplify something but not to the point where you state it incorrectly. A number is a number and does not change. You have several different numbers So 360 is a different number from 300. 300 does not got to 360 that is meaningless. Let us state your latest post (quoted) correctly. The difference between two numbers, when expressed as a percentage of the smaller one is given by the equation [math]\left( {100} \right)\left( {\frac{{{\rm{Larger - Smaller}}}}{{{\rm{Smaller}}}}} \right)[/math] You have correctly used this twice to obtain the % difference between 200 and 260 and between 300 and 360 in your first two calculations. BUT And I have already said this, The numbers for the third calculation are different again from any of the first two sets. As before you have performed this calculation correctly [math]\left( {100} \right)\left( {\frac{{{\rm{Difference1 + Difference2}}}}{{{\rm{Smaller1 + Smaller2}}}}} \right)[/math] Which is equal to [math] = \left( {100} \right)\left( {\frac{{{\rm{Larger1 + Larger2 - Smaller1 - Smaller2}}}}{{{\rm{Smaller1 + Smaller2}}}}} \right)[/math] because you have the [math]\left( {100} \right)\left( {\frac{{{\rm{Total}}\;{\rm{Difference}}}}{{{\rm{Total base population}}}}} \right)[/math] Does this help?

Although it is not clear what ions were in the water, here is a tragedy that demonstrates the conductivity of water with ions. http://www.bbc.co.uk/news/world-europe-40388161

Remember that long thread last year? Well the world record longest sniper shot had to be made using correction for the curvature of the Earht. http://www.bbc.co.uk/news/world-us-canada-40372403

Some more detailed indication of the intended use is needed. Is this for real life or for a story? Are you thinking of a cloaking fabric against IR detectors for military or quasi military purposes? Infrared devices with reflectors are used in surveying and navigationover distances of several kilometres. The reflectors are of the corner reflector type. So I suppose you could consider microbead corner reflectors coating the fabric, like sequins. Some more detailed indication of the intended use is needed. Is this for real life or for a story? Are you thinking of a cloaking fabric against IR detectors for military or quasi military purposes? Infrared devices with reflectors are used in surveying and navigationover distances of several kilometres. The reflectors are of the corner reflector type. So I suppose you could consider microbead corner reflectors coating the fabric, like sequins.

I've no idea if that is the same paper to be honest. There was only one Englishman who could have attended the lecture on November 25th, 1915 and that was not me. In case you missed it there was a war on. I see I said 'submitted' but the report in my source actually states 'presented' so I suppose if it is not in the text of the paper it would have been in Einstein's presentation speech/lecture. Ferreira : The Perfect Theory : page 21 : paragraph 3.

However others teach it, it certainly can't be taught from ignorance.

Diode - Resistor OR gates required no power source and were used in early logic boards.

Well I still have no real idea what the OP meant, but here are some thoughts. Storage (memory) is arguably the most important element of electronic (and even non electronic) computers. All those ones and noughts are totally useless if you cannot organise them, store them until you want them and then retrieve them. Pretty well every computer chip or module built of discrete components (semiconductor or vacuum tube) contains some form of storage. In order to perform these functions you require pulses and to generate these pulses you require active devices - that is semiconductor or vacuum tube devices. Most modern computer memory has to be 'refreshed' many times a second or the stored information leaks away. Again pulses are required.

Lovely post, Carrock. +1 But would someone please remind me what the actual OP question was?

Anyone has to learn to walk before they can run. I offered you the opportunity to work through the walking stage to reach your equation in considerably more detail than you will find in any single textbook or three. And the only running you did was to run away from this opportunity. So I will leave you with the the following thoughts. Your very very elementary equation is not always true. For instance it is not true in the case of point particles with mass, spinning on their own axes. These have angular momentum, but r = zero. Again it is true in the case where r refers to a plane curve. But it is untrue in the case of a twisted three dimensional curve. There are many other complicated cases to explore.