GoSlash27

High t/w, good Isp, and dirt- cheap. It's way out of balance with the other engines. Best, -Slashy

http://s52.photobucket.com/user/GoSlash27/slideshow/KSP/Lotek%20II I cleaned up the Lotek a bit. Less parts, a little more DV, and now remote guidance for rescue jobs. Orbital DV for this run works out to 917 m/sec. http://wikisend.com/download/318490/Lotek2.craft Best, -Slashy

Looks to me like your sig has the answer: "It needs more wings". I don't know how it works in FAR, but in stock he's got enough wing for about 80 tonnes of aircraft trying to lift 216 tonnes. Best, -Slashy

100 m/sec should be plenty of airspeed. I'd sort out why the plane isn't flying at that speed before worrying about moar boosters or longer runways. Best, -Slashy

Kerbals do not have mass when they are inside a cockpit, but they do have mass when on EVA or in an external command seat. Likewise, ships that you are not controlling at the moment exhibit weird physics. When you go EVA, your kerbal can kick around a multi-ton spaceplane pretty effortlessly. That's probably what you're seeing. Best, -Slashy

What Pecan said. Tylo isn't a big deal. Eve *is*. Tylo lander Eve lander Best, -Slashy

Penn State and Iowa State both have excellent aerospace engineering programs. Best, -Slashy

Somebody please double-check my math on this. According to my screen shot, I had a mass of 5.19 tonnes and 106 units of o2 in orbit. O2 burns faster than fuel, so I'm only going to count what I could have burned in orbit (ignoring the leftover fuel) and ignore the monopropellant. I had 106 units of o2. An FL-T100 tank weighs .56t full and .06t empty, so it's F+O weighs .5t. That's 55 units of o2 and 45 units of fuel weighs .5t. (106/55)*.5t means I was carrying .96 tonnes of usable fuel and o2. Orbital mass = 5.19 Mass less fuel and o2 = 4.23 Isp of my 48-7S engines is 350 sec Therefore my orbital DV was 703 m/sec. Best, -Slashy - - - Updated - - - In theory it's 1km/ sec. I didn't get anywhere near that, though. I just climbed at 60* pitch and kicked in the rockets when my AoA hit 23*. Best, -Slashy

http://s52.photobucket.com/user/GoSlash27/slideshow/KSP/LoTek I'll have to crunch the numbers to see how much DV I had in orbit. Not sure how much this design will be worth; It's got 30 parts, so no room to add anything else. I knocked it together in sandbox, so i used the grass just to show that it's rough field capable. http://wikisend.com/download/976582/lotek.craft Best, -Slashy - - - Updated - - - Starhawk, Scary how similar our designs look! -Slashy

Xannari Ferrows, Everybody here is trying to tell you the same thing: What you have posted here is correct and nobody disputes it, but it's not a "proof". andrewas put it very simply and accurately: Not tryin' to bust your hump, that's just the way it is. Best, -Slashy

Oh, believe me I'm not. I have a congenital drama allergy I'm just trying to clarify what is "proof". In fact, an unwrapped circle *is* a triangle, but this assumes it to be so without proving it. A rigorous mathematical proof will assume nothing that is not proven, and will show that for any arbitrary value the mathematical model is correct. This demonstrates the concept (taking the posted assumptions as true), but it does not *prove* it. Sorry if I'm ruffling your feathers; I don't mean to -Slashy - - - Updated - - - ^ That is what I was looking for! This must be included as part of a proof, else we cannot assume that an unwrapped circle is a triangle or the areas are the same. Treating the "2pi" part of the circumference as a constant (K), we can show that for any radius r, the circumference is Kr. Therefore, the base of the triangle is a fixed ratio of the height for any arbitrary r. This proves that an unwrapped circle is, indeed, a triangle. Best, -Slashy

Aye... but you've merely restated what you've already posted. How do you know that an unwrapped circle makes a triangle? How do you prove it? -Slashy

^ This. Here's how I know that the area of a circle is, in fact, pi*r^2 I knocked it together as an explanation of a birthday card I gave a math nerd friend. Not actually intended as a proof, but it serves as one. I also have a proof of pi itself floating around somewhere. Not exactly the same way it was done originally, but it does work. I'll see if I can find it. *edit* So here's how we skin this cat: We start with a simple right diamond. The vertices have a length of 1, so the hypotenuse has a length of sqrt(2). The perimeter of that shape (carried around the center) is 4*sqrt 2. Now bisect that angle. We have now defined a new right triangle. the adjacent sides are sqrt(2)/2 and the vertex is 1. Supposing we wanted to create 1/4 of an octagon. We would simply project our bisector to 1. The length of this new segment is 1-(sqrt(2)/2), or (2-sqrt(2)/2). This defines a new right triangle, whose hypotenuse is sqrt((sqrt(2)/2)^2-(2-sqrt(2)2)^2). Throwing some algebra at it, it simplifies down to sqrt(2-(sqrt(2)). Since we have doubled the number of sides of our octagon, we would multiply this by 8 (or 2^3) to find our perimeter. We repeat the process. Bisect that angle, extend the bisector out to 1, and solve for the new right triangle. The hypotenuse is sqrt(2-(sqrt(2+sqrt(2)) and the perimeter is this same value multiplied by 2^4) Rinse and repeat. perimeter of a 32 sided shape: 2^5 (sqrt(2-sqrt(2+sqrt(2)))). And again. Perimeter of a 64 sided shape:( 2^6) (sqrt(2-sqrt(2+sqrt(2+sqrt(2))))) This process repeats indefinitely. As we double the number of sides, the ratio of the perimeter to the radius approaches 2 pi, and we can define a calculable value for pi: lim n->(1/0) [sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(...(2+sqrt(2n-1+sqrt(2n))))...)n-1)n)] This is a process that can be carried put with an ink pen and parchment, or an abacus. Simply find the square root of 2 to however many digits. Add 2. Increment a counter. Find the square root. Add 2. Increment the counter. Find the square root. Add 2. Increment the counter. Continue this process until the last step, then *subtract* 2, increment the counter, and find the square root. Multiply this number by 2^(counter) and *poof*: pi. And since our resultant value is, by definition, closer to pi than it is to the previous iteration, it can be proven that our result will differ from pi by less than f(n)-f(n-1). Once those 2 values differ by less than our desired margin, we can say that our result equals pi within the margin. And since we have already proven that the square root of 2 is an irrational number, we can conclude that the sum of any irrational number and a rational number is necessarily irrational. Likewise, the square root of any irrational number is irrational. Therefore, we have proven that pi will never repeat, no matter how many digits it's carried out to. Posted on my FB page back in the day when I first sorted it out. I was pretty jazzed about it (I'm easily entertained) Best, -Slashy

Oh, I'm not calling it into question. We know that it's true. I'm just saying it's not a valid proof. How would you mathematically prove that the area of your triangle is the same as the area of your circle? How do you know that they are exactly the same? Best, -Slashy

I see what you mean. The 26.16 is completely superfluous. What matters is that it is assumed that 1) the area of a triangle b=c h=r is exactly the same as the area of a circle r=r and 2) c=2*pi*r Neither of these assumptions are mathematically proven here, so it's a circular argument. Best, -Slashy

Unless I'm mistaken, this proof is circular logic. It assumes at the outset that a=pi*r^2 by setting the circumference to 26.16. Best, -Slashy

I have gone super- green! -Slashy

^ Foxster's craft isn't "cheaty", it's just efficient. Not sure he calculated the score correctly, tho'. Speaking of, the score formula seems a bit wonky. 10(time in seconds)+20(mean orbit altitude in??)+ vehicle mass in tonnes. If this is correct, then the tonnage will have negligible impact on the final score. If orbit altitude is in meters, then time to orbit will be negligible as well. Best, -Slashy

Hahah it's a "get off my lawn" contest! /whippersnappers

Velocity, As much as I am *not* interested in arguing, I feel the need to correct this for clarity: The Oberth effect is not dependent on gravity. It has nothing to do with being close to a point source or being far away from it. The reduction of gravitational attraction over distance doesn't factor into it. It doesn't matter whether you're in close proximity to a planet or in open space. Adding velocity to a higher velocity imparts more kinetic energy than adding velocity to a lower velocity. Now... as a practical matter, yes. We have our highest velocity at the periapsis of an orbit, which is why we choose to burn there. *BUT* that gravity has nothing to do with how or why the Oberth effect works. Burning at twice the velocity imparts 4 times the kinetic energy. This is true regardless of whether there's a planet nearby or not. Best, -Slashy

Old post, but clearly still useful RainDreamer's intuitive understanding is closer to the mark than LostOblivion's. LO, Forget about gravity. It has no bearing on how the Oberth effect works. Changing orbits is all about changing energy. We change our energy by adding or subtracting velocity, but the relationship between them is not linear. Doubling your velocity does not double your energy, it quadruples it. Therefore it stands to reason that the faster you are going when you make a velocity change, the larger the resultant change in your energy. Best, -Slashy

I'm too old-school to bother with "battery management". Here's my solution: Having a big battery means I can charge my phone whenever I get around to it. Best, -Slashy

True. You're standing by and allowing 4 people to die who you could have saved. No way of knowing that short of actually putting them in that situation. Besides, it wouldn't answer the original question. Best, -Slashy

I yanked that sucker. The decision to not act is still a decision, so the only choice I have is to either kill 4 people or kill one. Best, -Slashy

^ This is the key. If the new 1.0 aerodynamics reduce the DV requirement to an attainable level, then SSTO will be at least theoretically possible. Best, -Slashy