1,358 reputation
925
bio website ericmenze.com
location Minneapolis, MN
age 30
visits member for 3 years, 4 months
seen 4 hours ago

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.


Mar
23
comment How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Sure - in theory, I get that. However, that 'fast enough' is orders of magnitude more energy than it takes to just get to an orbital height; indeed this article reports that in getting to LEO, 93% is spent increasing the horizontal speed, and 7% is raising the altitude. Since I don't even have very good odds to get the 7% it will take to get to the altitude, let alone the 93% it would take just to fall at the same rate that the surface of the earth does, then it seems that any thrust in the horizontal direction will be wasted. permanent.com/space-transportation-earth-moon.html
Mar
20
comment How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Rockets usually take up their own oxidizer; for example Black Powder in model rockets and Liquid Oxygen in full scale ones, so they don't need 'air' to operate. I'm still confused as to how, say, a rocket engine is changing any of the horizontal kinetic energy from the 1 rev/day orbit, or utilizing it to move in the vertical direction.
Mar
20
comment How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Can you explain the process by which horizontal kinetic energy is converted to vertical potential energy, without a ramp or some similar structure? I don't think $KE_{horizontal}$ or lack thereof has any effect on the height an object will reach, unless we're leaving Earth's rotating reference frame or something.
Mar
20
comment Rocket propelled by a giant monochromatic laser
If photons have non-zero effective mass, then why wouldn't a mirror on the rocket and on earth give perpetual free momentum as the photons bounce back and forth?
Mar
20
comment How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Simplified for only the normal vector, and just reaching the Earth-Moon L1 from rest on earth's surface: $PE_{i} + KE_{i} + E_{rocket} = PE_{L1} + KE_{L1}$, so $0 + 0 + E_{rocket engine} = PE_{L1} + 0$, or $E_{required} = PE_{L1}$. I'm lost on how to calculate the Potential Energy at L1, since $PE = mgh$ doesn't really work (g = 0?). Your equation above doesn't seem to care about the mass of the object, a 100kg object surely has more PE at L1 than a 1kg object. Can you elaborate?
Mar
20
comment Rocket propelled by a giant monochromatic laser
From answers.yahoo.com/question/index?qid=20100521204409AAk1lkq, $m_{photon} = \frac {h*f} {c^2}$? Seems bizarre, and the conservation of momentum could be used to calculate it. How do you account for the Tsiolkovski Rocket Equation, based on the (rest) mass and velocity of the exhaust? Does it not apply in this situation?
Mar
20
comment How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
I think you're thinking of an orbital insertion; in my case I don't want or need any additional 'horizontal' velocity; unless it can be used to increase the vertical. I just want to reach the highest point I can.
Mar
20
comment How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Wouldn't launching off-normal (straight up) 'add velocity' at the expense of some altitude? What angle would maximize altitude from Earth's surface?
Mar
20
comment Would it be economical to add a counterweight to rocket launches?
Example: Take a fully loaded Saturn V rocket at 2,800,000kg, and add a counterweight that will reduce its effective weight for the first 10 seconds to 400,000kg and detach. It burns 2,100,000kg of fuel in the first stage in 161s, so 13,000kg/s of fuel is lost. Without the counterweight, the first 10 seconds would net a $\Delta v = 4,440m/s*\ln(2,800,000/2,670,000) = 211m/s$. With the counterweight: $\Delta v = 4,440m/s*\ln(400,000/270,000) = 1745 m/s$, or roughly Mach 5.1.
Mar
19
comment Would it be economical to add a counterweight to rocket launches?
I should add that I wasn't counting on the acceleration from the counterweights to be of much importance in and of themselves; the 63 m/s velocity difference would pale in comparison (I think) to how fast the onboard rocket engines could accelerate the lessened (or non-zero) effective mass. Is there a way to calculate this?
Mar
19
asked How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Mar
19
answered Rocket propelled by a giant monochromatic laser
Jan
28
comment Could the Heisenberg Uncertainty Principle turn out to be false?
The Heisenberg Uncertainty Principle is an unfalsifiable claim? All of (good) science is falsifiable. See the first paragraph: en.wikipedia.org/wiki/Falsifiability
Jan
13
awarded  Popular Question
Dec
15
awarded  Popular Question
Dec
8
awarded  Notable Question
Oct
28
awarded  Nice Question
Jun
25
accepted Could we make a trebuchet that could launch objects to a stable orbit?
Jun
25
comment Would it be economical to add a counterweight to rocket launches?
How about four steel cables, each holding 100 tons? Some bridge reinforcement cables are rated for more than this.
Jun
25
comment Would it be economical to add a counterweight to rocket launches?
Further, I do refer to wikipedia. Since you seem to be of the opinion that this has been answered therein, could you provide a wikipedia link that properly addresses this question?