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bio website ericmenze.com
location Minneapolis, MN
age 29
visits member for 2 years, 3 months
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I'm a Computer (Web) Programmer/Analyst based in Anchorage, AK and Minneapolis, MN. I use (among other things) ASP.NET, C# and SQL Server.

I build things. Bicycles, computers, websites, guitars, cars, motorcycles, sound sytems... lots of things.


Sep
13
comment Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
It should also be noted that $v_e$, by contrast, will exponentially relieve fuel requirements ($m_p = 1 - e^{-\frac{\Delta V}{e_v}}$. If we could accelerate the exhaust to 300 km/s instead of just 4.5 km/s (1% of c), then it would only take 3% of $m_0$ to get from the surface to LEO.
Sep
13
comment Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
Completely agreed, which is why I used the word 'ease' - it backs the percentage of fuel requirement down by some non-trivial amount, but it's still a large percentage of your starting mass.
Sep
13
revised Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
Corrected mass of the propellant from LEO -> GEO
Sep
13
comment Calculation of Distance from measured Acceleration vs Time
The model rocket will experience some amount of 'tipping' and may even reverse direction on descent, which will skew your measurements by providing smaller z accelerations than accurate, and then by adding to the position when falling, respectively. On a ground based vehicle x and y might be useful, but once you introduce suspension it really needs a gyroscope to track the roll, pitch, and yaw of the accelerometer to have much accuracy.
Sep
13
revised Calculation of Distance from measured Acceleration vs Time
Added sample excel contents
Sep
13
answered Calculation of Distance from measured Acceleration vs Time
Sep
13
revised Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
Fixed initial fuel mass to launch mass
Sep
13
comment Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
Out of curiosity, why did you use the Nabla $(\nabla)$ symbol? Is difference in gravitational potential energy a mathematical gradient, divergence, or curl, as mentioned in the wikipedia article on the symbol? Or did you just pick it not following a convention?
Sep
13
answered Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
Sep
13
comment Would a rocket burn more fuel to get from Earth's surface to LEO, or to get from LEO to GEO?
Thanks, your answer rocks. It should be noted that Space Guns or other forms of non-rocket space launches, including ion drives and the mechanism being planned by QuickLaunch, Inc. would all be, at least for some of their trajectory, fixed mass - so the calculation isn't completely irrelevant. I do feel I hijacked the question because of the interestingness of your fixed mass comment.
Sep
13
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
$$E_{in} = m_{weight}*g*h = 9.8*100*100 = 98000J$$, I didn't have room to include it above.
Sep
13
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
Let's take a system with an inelastic band, frictionless pulley, and no air resistance, sample problem above with a 100kg weight. The Force on the large weight side would be $F=m_{weight}*g$ and with the projectile side $$F=m_{projectile}*g$$. Now let's let the mass drop 100m. The total $$P_e = m_{weight}*g*h - m_{projectile}*g*h = 100*100*g - 10*100*g = 9000*g = 88200J$$ Now, this force gets the system moving $$K_e = 1/2*m_{system}*v^2 = 1/2*110*v^2$$ so $v=40m/s$. The energy now in the projectile is $K_e = 1/2*m_{projectile}*v^2 = 8000J$. $$Efficiency = E_{out}/E_{in} = 8000J/98000J = 8\%$$
Sep
13
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
Sure, F = m * a of the SYSTEM. The falling mass accelerates the system to some velocity. By what mechanism do you propose that all of the system's energy is magically concentrated into the projectile, and not just the large weight crashing into the ground? I suspect if you attempt an experiment of this nature you will find very little is transferred from the weight to the projectile.
Sep
13
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
No, it isn't. Think about how much kinetic energy is transferred into the ground by the large counterweight slamming into it - most of it. All of this energy would need to be transferred into the small one. In your pulley system example, this energy has not been removed from the counterweight.
Sep
13
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
In considering this design though, couldn't the drop tube be set to $90\deg$ and launch tube set to the desired angle? Would it be able to use any sort of hydraulic advantage by having the launch tube at, say, 1/2 the diameter so it moves twice as fast?
Sep
13
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
It sounds like this would require large, long tubes with tight clearances, and the pressurization of the gas would (under extreme conditions) heat the gas high enough to melt the tube. Government agencies have used a similar idea to launch scramjet engines, and run into the temperature limit often; as does QuickLaunch, Inc. I truly want a mechanical design that can be scaled up to 200,000kg dropped 91m being translated into a 10kg mass lauched at 6km/s. I don't think the temperature inside the cylinder of this design can be scaled to this.
Sep
12
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
Right, this may translate pe to ke in the weight, but not much of the falling weight's energy is translated into the launch of a projectile.
Sep
12
revised What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
Added sample
Sep
12
comment What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
I thought $m_{weight}$ and $m_{projectile}$ made it clear that they were different masses. Does the edit ask the question that I really want better?
Sep
12
revised What is the most efficient machine for translating gravitational potential energy of one mass into kinetic energy of a different mass?
Clarified Question