OK, '640mm effective' or '640mm equivalent' are bad nomenclature.
The problem is that in the past there was only 35mm, so focal lengths were usually used instead of angle of view.
And the tradition has gone on of quoting an equivalent focal length for small sensor lenses, because it is easier for people to compare different cameras by using the 'equivalent' focal length - People have a good feel of what to expect from a 200mm lens compared to a 28mm lens.
Its too late to change it. Just get used to it - a 840mm lens no longer means a lens with a focal length of 840mm. It means a lens with the same angle of view that an 840mm lens would have with a 35mm image frame. This makes things especially confusing when the same lens might be used with a full-frame or APS-C size sensor. You can blame the journalists (so tedious to always say 'equivalent') or the camera manufacturers (what would sell better 5-100mm or 25-500mm?), but it isn't going to change things.
It is rather like using equivalent MHz as a CPU speed measurement unit.
But the point is that these smaller cameras can have amazingly small angle of view. Smaller than almost any DSLR lens. You can get a camera which has the same angle of view as an 840mm lens on a 35mm camera, and it will resolve more detail than than many older full-frame DSLRs (in the right conditions).
In good lighting conditions these can take excellent pictures. A DSLR will take better pictures in less well lit conditions, may focus more quickly and more accurately, and may take more pictures in quick succession. And a picture taken in good conditions with a long DSLR lens might resolve more detail than a good small camera. But there is not very much in it - and a small camera is certainly easier to carry around and handle. (And in Kuwait in daylight I expect the lighting conditions are quite bright).
With a small sensor you get a greater depth of field (for the same angle of view and aperture). But you get greater problems with diffraction - some cameras reach the diffraction limit at f5.6, so stopping down does not improve the image.
I am aware of medium and large format in addition to 35mm. But they were (are) always relatively specialised, and people who used them know what they are talking about and dont buy things based on meaningless paper specs, unlike many of the people who talk about 35mm equivalent focal length.
If you put your hand over the top of the light then you are receiving heat by convection and radiation. If you put it to the side then you are mostly getting radiation. The radiation is a mixture of visible light and infra red light (with a tiny amount of ultra violet).
The exact fractions of each type of energy depends on the type of light bulb. Halogen lights convert more into visible light, and less into infra red. The amount lost to convection and conduction also depends on the design of the light bulb - bigger bulbs might lose less through conduction and more through convection.
Wikipedia suggests that the amount of energy converted to visible light is likely to be much less than 4% (even for halogen bulbs). I dont have a figure for the proportion which is convection or conduction, but I think it is probably small in comparison with Infra Red. http://en.wikipedia.org/wiki/Incandescent_light_bulb
The section you are quoting is about heat engines. A light bulb produces no mechanical work (and also take no heat in from a hot bath). The equation simply does not apply. The desired output is Qout - it is not 'waste heat'. In this case efficiency is Qout/Win (Win being the electrical energy input to the system).
The section you should refer to is the 'Energy Conversion' section, which actually says that an electric resistance heater has near 100% thermal efficiciency.(Qin=0, Wout=0, Qout/Win=1). You might say that Wout is not zero if some of the energy escapes (through the windows) as light energy.
You are correct though in saying that a heat pump is more efficient than a resistance heater - in this case Qin>0, Wout=0, Qout=Qin+Win, and Qout/Win>1.
The above post is typical of a lot of errors that various people are making here. GW/h is not the same thing as GWh.
1GW/h is not a measure of energy. It might be a measure of the rate of change of power (it is rather absurb, but that is the point of the previous reply which was intended to be humorous).
The measure of energy could be in kWh, MWh or GWh.
So 6 equal sides of 9 sq/ft would make it 81 cubic feet.
I think you mean 27 cubic feet (assuming square sides).
It's true that an electric motor provides more torque than an internal combustion engine at low revs.
The shape of the toque curve is very different.
An electric motor can provide a lot of torque at 0 rpm, while an internal combustion engine can't even keep itself turning at very low revs.
This means that the power curve has a different shape. An electric motor has a much broader curve, so it is able to run with high power over a large range of speeds. So much so that it wont need a clutch, and may not need different gear ratios. It may still need some kind of gearbox to match the rpm of the motor to that of the wheels, but a single speed box is sufficient in many cases (possibly not if you want to get the max possible performance).
Wide tires dont give you extra contact area. The area of contact between the tyre and the road is determined mostly by the air pressure in the tyre. If you increase the width of the tyre without changing the pressure then you change the shape of the contact patch, but not its area (not much anyway). Wide tyres are useful because they are less affected by irregularities in the road surface and because they spread the load through a large area of rubber (so they dont overheat so quickly). The contact patch is also short and wide, which means that the front and rear edges of the contact patch are longer (and these edges carry a bit more of the load than the centre due to the bending of the rubber). An Ultra-wide (steamroller) tyre would not be useful. It would require some internal structure to transfer weight to the centre of the tyre (otherwise it would bend and just lie on the road surface in the middle with very little pressure). It would also cause cornering problems - how would you provide a differential? Wide tyres already have problems cornering due to the difference in road speed between the inside and outside edges - there is bound to be some slippage. Narrower tyres are usually more efficient. The only practical way to increase traction is to provide extra downforce (e.g. aerodynamic - which only works at speed) or to use all 4 wheels for traction (doing something to the materials of the tyre/road and the tread pattern also have some effect).
HA! Did you really try it?
137/206 works out as zero. As does 2/3.
So your program would pass the vote.
if (1/206 >= 2/3)
you need to do something like this
if (137.0/206.0 >= 2.0/3.0)
How can you do 'New Math' problems with an 'Old Math' mind? -- Charles Schulz