E-585: Fill & Download for Free

A2A.I have already shared my views regarding cutoff .Cut-off of TA( CBIC) — 585±2 URCut-off of 4600 gp — 610±2 URThese are my predictions, it has no link with other's cutoff and since these are only predictions, must be taken with a grain of salt.Your score as of now is 488.5/600. You can expect your Normalised score in the range of 520–525. You need minimum 60 marks in descriptive to assure yourself a post , rest God knows.All the best.

A telephone pole sized cylindrical rod (1 foot diameter 20 feet long) of solid tungsten weighs 12.9 metric tons. A solid cone the same dimensions weigh 4.3 metric tons.Recall that both cylinder and cone are defined by their diameter and length. A cone that fits inside a telephone pole has a lower drag than a cylinder and weighs 1/3 as much.An object moving between 7.9 km/sec and 10.0 km/sec (depending on the altitude it is coming in from) the cylinder could release between 403 GJ (96.3 tons TNT) and 645 GJ (154.2 tons TNT), whilst the cone could release 135 GJ (32.3 tons TNT) and 215 GJ (51.4 tons TNT).The M-388 atomic round has a 10 ton TNT yield to 20 ton TNT yield.AddendumSomeone asked me about elliptical orbits, and yes, all things being equal, you can nearly double the energy density of the system by flying highly elliptical orbits. If you loop around the moon so that you reverse direction of the spacecraft so that it is flying against the spin of the Earth, you can strike things with 2.2x the energy of an orbital device.Going further you can imagine sending a payload out to Jupiter, and swinging by Jupiter so that you fly a highly elliptical orbit against the direction of the Earth! You can increase your speed to 69.3 km/sec - and get a Davey Crockett yeild out of a 1.8 liter cone! Amazing stuff. If you launch a large number of these you can have one fly by the Earth every few minutes and send radio signals to the passing device giving it approved coordinates to strike targets anywhere on Earth.The same set up would work for Mercury Venus and Mars. Venus would have higher energy systems and Mars would have lower energy systems given their position relative to the Sun.Weaponised asteroid belt!What is the advantage vs Orbital system?We’ve sent probes past Pluto by way of Jupiter. So, we know what the mass requirements are.To put some numbers on it, if we use 3.8 km/sec and 63 metric tons on orbit - and you’ve got an orbital velocity of 7.905 km/sec at Earth’s surface. So, 63 tonnes and 7.905 km/sec is;E = 1/2 * m * V^2 = 0.5 * 63,000 kg * (7905 m/s)^2 = 1.968 TJ.This is equivalent to 470.2 tonnes TNT equivalent. 47 Davy Crockett rounds at 10 metric tons each.Now, let’s assume we’re going by Jupiter and reversing our direction at Jupiter. Jupiter’s semi-major axis is 5.203 AU. So, a Hohmann transfer orbit is a = (1+5.203)/2 = 3.102 AU (round up to account for factors you’ve discussed based on actual experience.V = 29.8 km/sec * sqrt( 2/1 - 1/3.102) - 29.8 = 8.798 ~ 8.80 km/sechyperbolic excess velocity to get to Jupiter. Now Earth escape velocity is 11.18 km/sec so you need to attainV = sqrt( 11.18^2 + 8.80^2) -7.905 = 6.323 ~ 6.4 km/secOkay, so, you’re rocket exhaust in vacuo is 3.8 km/sec. So, your propellant fraction is according to Tsiokovsky is;u = 1 - 1 /exp(6.4/3.8) = 0.814409 ~ 0.815and structure fraction is 0.065 so payload fraction is;p = 1 - 0.815 - 0.065 = 0.120So, 63,000 kg * 0.12 = 7,560 kg.So, far smaller than 63,000 kg. But what’s the kinetic energy?Note we’re rounding things up for propellant and that’s accounting for all the factors you’re talking about when you look at the actual performance of the dozens of deep space craft that use Jupiter sling shot to fly through the outer solar system and fly into the inner solar system and the sun.Gravity assist - Wikipedia (Gravity assist - Wikipedia)Now, an object at Jupiter will slow toV = 29.8 * sqrt(2/5.203 - 1/3.1015) =7.41828 km/secAnd Jupiter is moving atV = 29.8 * sqrt(1/5.203) = 13.06440 km/secThis is a difference relative to Jupiter of 13.06440 - 7.41828 = 5.64612 km/secNow, coming ahead of Jupiter ((d) in the image above) the 7,560 kg payload is sent in the opposite direction and falls into the Sun - BACK TOWARD EARTH!As it comes by Earth it is moving at (according to Vis Viva)V = 29.8*(2/1–1/3.1015) + 29.8 = 79.79175 km/sec.And arriving at Earth with this hyperbolic excess velocity achieves at Earth’s surface;V = sqrt(11.18^2 + 79.79^2) = 80.57 km/secNow the energy contained in the 7,560 kgE = 0.5 * 7,560 kg * (80,570 m/s)^2 = 24.54 TJ.This is equal to 5858 metric tons of TNT equivalent (kiloton yield!). If we divide this into 10 metric ton increments, this is 585 warheads instead of 47.For for a Falcon Heavy launch we can have 47 Davy Crockett bomblets in LEO, or 585 Davy Crockett bomblets in the Jupiter orbit.Using Kepler’s law the orbit this device is in has a period of;3.1015^1.5 = 5.462 years -So, if we reduce this to 5 years EXACTLY - then we can calculate a changed semi-major axis we must achieve as we fly by Jupiter.5^*1/1.5) = 2.924Which means we’re moving inside the orbit of Earth, but coming back every 5 years exactly, which means the Earth is always there.Of course with 545 warheads, they spread out around orbits that slightly change their argument of perihelion - the orientation of the semi major axis, by controlling how each passes Jupiter - controlling their burnout velocity as they are shot to Jupiter.So, dividing 545 warheads into 5 years exactly, results in two warheads passing Earth every week - from a single launch.Ten Falcon Heavy launches bring a weapon into striking range every 8 hours with 5,450 in the system.Eighty Falcon Heavy Launches bring a weapon into striking range every HOUR with 43,600 warheads - sufficient to strike every town above 2,000 people.Note: These are kinetic vehicles and are NOT bannedhttps://www.rand.org/content/dam/rand/pubs/monograph_reports/2011/RAND_MR1209.pdfHere’s a cool reference!Jo Oztrander made the suggestion that we could extract energy from this system. That’s a better use of it. I did an analysis in responding to Jo. However, I thought it important enough to share in the main response.Enjoy! It shows anything that releases as much energy as a nuclear device can be used as a source of energy!You are basically slowing the Earth down and extracting energy from that motion. Of course you can also fly asteroids by the Earth and extract energy from Jupiter and restore that energy. That has already been worked out by NASA scientists.Nasa aims to move Earth (Nasa aims to move Earth)Um the best way to do this would be to arrange a technology of my friend Keith Lofstrom using his Launch Loop to pick up the kinetic energy and then toss the working mass back out dispensing with the rockets.More here about Lofstrom Launch LoopsA free-standing space elevator structure: A practical alternative to the space tetherSo, yeah, you toss stuff out past Jupiter and have it come back in the opposite direction with 12 the energy - but you don’t let it crash that’s wasteful. You merely stop it, reverse its direction and toss it back out again - electromagnetically.So, today humanity uses 5.528*10^20 J/yr of primary energy.That’s 17.53 trillion watts. If we have a 1 metric ton iron rod - we use a magnetic material now since we’re not using it to bombard anything - and so we can use loser densities. We have a soda can shaped cylinder - twice a high as wide - minimum surface area - that’s 127 liters in volume to equate to 1000 kg at typical iron densities. This is 0.3 meters in diameter and 0.6 meters tall.80.57 km/sE = 1/2 * 1000 kg * (80,570 m/s)^2 = 3.26 TJWe must toss this same thing out in the opposite direction with a speed of 12.4 km (and timed so the two streams don’t collide).SoE = 1/2 * 1000 kg * (12400 m/s)^2 = 0.07688 TJ.So we have a net gain of 3.19 TJ.So, to produce 17.53 TW of power requires we collect 5.65 per second.So, if we had 20 active catchers and 20 active tossers - as part of a network that provided global transit and movement beyond Earth as well.Now in an industrial economy wealth is a function of energy use per person. The formula isGDP = 965.27* BOE^0.8142Now Monaco is the nation with the highest living standard with a GDP per capita of 193,745.60 So this is a BOE ofBOE =(193745.6/965.27)^(1/0.8142) = 673.02Now if this seems like a lot, you haven’t seen the super yachts and private jets parked outside the homes of the 32,000 people who live in Monaco.Now each barrel releases 6.1 GJ. So, this is 4.105 TJ/year per person. So, for 8.5 billion people this translates to 3.49*10^22 J/yr. 1107 TW. Dividing by 3.19 TJ net this is 346.9/second.Now, if we have a 50 hz system that’s synchronised - then we need 7 active catchers and 1 active tosser. Out of dozens of launch loops that tie the world together and tie the world to the interplanetary colonies.The important detail is 350 cylinders per second being caught and tossed by the global network of launch loops - and the traffic being managed to avoid collisions with counter propagating streams - but these shouldn’t be any problem really.The orbits are 6years since the aphelion is 5.1727 AU and perihelion is 1.0000 AU. So, 350/second * 6 years = 66.23 billion tonnes of iron. World’s proven reserves of iron metal is 84 billion metric tons. So, we could conceivably do this with terrestrial resources. Of corse we could also do this with off world resources of iron as well.Iron Meteorite on MarsIron rich blueberries on Mars.Mars red colour comes from iron oxide and there’s plenty of iron on Mars. Furthermore, Mars could set up the system, test it out without risk to Earth, and then expand the system to Earth once all the details were worked out.Check out my Kickstarter campaign!Hyper Efficient Solar Laser- SpaceGold Miners

Here is a different and novel way to calculate a determinant that I recently discovered:Start with the matrix you want to calculate the determinant of, let's take the [math]3\times 3[/math] matrix example[math]M=\begin{bmatrix} 1& 2& -5\\ 7& 4& 10\\ 5& -9& 2\end{bmatrix}\tag*{}[/math]Then we create a layered pyramid of matrices as follows:Make a matrix entirely of 1s that is of order 1 greater than our matrix (I.e. order 4). This layer of 1s is the first level (or base) of our pyramid of matrices.The next layer is our [math]3\times 3[/math] matrix, the entries of which are each between four of the 1s in the layer below.[math]\begin{bmatrix} 1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{bmatrix}\tag{Layer 1}[/math][math]\begin{bmatrix} 1& 2& -5\\ 7& 4& 10\\ 5& -9& 2\end{bmatrix}\tag{Layer 2}[/math]The next layer is a [math]2\times 2[/math] matrix. Each entry [math]e[/math] in this is given by [math]e=(ab-cd)/f[/math] where [math]a,\, b,\, c[/math] and [math]d[/math] are the four entries in the [math]3\times 3[/math] layer beneath [math]e[/math], and [math]f[/math] is the entry two layers directly beneath [math]e[/math].By way of example the [math]2\times 2[/math] layer is:[math]\begin{bmatrix} -10 & 40 \\-83&98\end{bmatrix}\tag{Layer 3}[/math]because[math]-10=(1\cdot 4-2\cdot 7)/1[/math][math]40 =(2\cdot 10 - (-5)\cdot 4)/1[/math][math]-83=(7 \cdot (-9) - 4\cdot 5)/1[/math][math]98= (4\cdot 2 - 10\cdot (-9) )/1[/math]Then successive layers are formed similarly, hence the next layer is a [math]1\times 1[/math] matrix:[math]\begin{bmatrix}585\end{bmatrix}\tag{Layer 4}[/math]Since[math]585=(-10\cdot 98-40\cdot (-83))/4[/math].This top layer is the peak of our pyramid and the determinant we seek.[math]\det(M)=585\tag*{}[/math]This method can be used for any order square matrix.I'm not certain that it's easier than traditional methods for a [math]3\times 3[/math] but I thought I'd share anyway since it is more powerful for larger matrices as we can make intermediate calculations. :)Edit: Please see comments for a link to the original paper in which this method was introduced. Amongst other things it explains how to use row and column swapping in such cases where we appear to have to divide by 0s (ciphers).