Space Travel..the alternate way..

Discussion in 'Alternative Theories' started by ZMacZ, Apr 16, 2014.

  1. ZMacZ Registered Senior Member

    ummm..not knowing by heart if ur formula in this case is correct/applicable...well I'm shocked...

    But since you like doing the math (which I personally don't..that's what computers oughta be for..) can you tell me something ?
    If you expend roughly 0.1% per hour of the total mass using the ion much speed would that yield for the last 50 % ?
    (so..a 500 hour burn ?..with all mass being expelled at 50 km/sec ?..thanx..)
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  3. Janus58 Valued Senior Member

    The equation is valid/applicable to any type of rocket when the velocity of the rocket is a small fraction of c.
    Computers are great for crunching the numbers, but if you really want to understand the physics behind the numbers, you have to grasp the mathematics behind it.
    I'm not sure what you mean by "how much speed would that yield for the last 50 % ?"

    Do you want to know how much speed the craft will have after burning the first 50% of its fuel, or do you want to know how much additional speed will be gained by burning the remaining 50% of the fuel?.

    In either case, the rate at which you use up your fuel is irrelevant, all that matters is the exhaust velocity. The exhaust velocity determines the speed you will reach after burning 50% of your fuel, the rate ate which you burn your fuel only determines how long it takes to reach that speed.

    Anyway, for either of the two possible questions above, all you need is the rocket equation I gave above.

    To get the speed after burning 50% of your fuel, just plug in Ms+ 0.5Mf for Ms on the bottom half of the fraction. You'll need to know Ms and Mf before you start.

    For example, if Ms= 1 kg and Mf = 1,000,000 kg, then your velocity after burning half your fuel is 34.66 km/sec.

    After burning all of your fuel, your final velocity will be 690.78 km/sec.

    This means that burning that last 50% of your fuel gains you an additional 656 km/sec. (which is the same answer you would get if you just plugged 500,000 kg in for Mf in the original rocket equation.)

    Another way of looking at this is that if burning 500,000 kg of fuel will get you up to 656 km/sec, doubling your initial fuel load only nets you an additional 34.66 km/sec. The reason for this is that in the early part of the trip the vast majority of the expended fuel is being used to accelerate the fuel the ship will use later.

    Here's a series of final velocities: starting with 1 kg of ship and 1 kg of fuel and then increasing the fuel load by 1 kg increments. Listed first is the final velocity, followed by the increase in velocity from the last increment.

    34.66 km/sec
    54.93 km/sec, +20.27 km/sec
    69.31 km/sec, +14.38 km/sec
    80.47 km/sec, +11.16 km/sec
    89.59 km/sec, +9.12 km/sec
    97.29 km/sec, +7.7 km.sec
    103.97 km/sec, +6.68 km.sec
    109.86 km/sec, +5.89 km.sec
    115.13 km/sec, +5.27 km.sec
    119.89 km/sec, +4.76 km.sec

    As you can see, each additional kg of fuel makes a smaller and smaller difference in the final velocity.

    Going from 3 kg of fuel to 4 kg makes results in a final velocity difference of 11.16 km/sec, while increasing from 9 kg to 10 kg only makes a difference of 4.76 km/sec.

    It is this decreasing change of final velocity as you add fuel that results in the extraordinarily huge amounts of fuel you would need to get even a small mass up to 1% of c with an ion engine.
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  5. ZMacZ Registered Senior Member

    thanx..but I guess I didn't think things through..
    since you need to speed up for getting there..and also needing fuel for deceleration..i though going 50% fuel would do for acceleration..and then using another 50% for deceleration..must have been thinking with my lower half..ofc you need less fuel for deceleration..the mass would be lower..
    I guess if I asked you what percentages would be needed for each (with Va + Vb = 0 ) you could do that too ?
    And do you perhaps also know the amount of micro matter outside the solar system ?
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  7. Janus58 Valued Senior Member

    it would depend on the situation, it isn't a fixed ratio. For example, in my last post, we note that a 1 kg payload needs 2 kg of fuel to get up to 54.93 km/sec. If you want to get up to this speed and then decelerate back down to a stop, you need a total of 8 kg of fuel, 4 times as much. If we look further down the list, we see that 10 kg of fuel will get you up to 119.89 km/s, but to reach this speed and then slow back down you need 120 kg of fuel or 12 times as much.

    The best way to look at it that if you need x amount of fuel to slow the ship back down at the end, It will take many times x amount of fuel to get up to speed, because you also have to accelerate the fuel that you are going to use to decelerate.
    The interstellar medium has an average density of ~ 1 atom/cc.
  8. ZMacZ Registered Senior Member

    Only 1 atom...but...are you sure ?..
    I mean I thought that it was even 1 stray atom per cubic meter ?

    Given that amount (1 atom/cc...traveling at 100 Km/sec.. atoms per sec for each square meter (using the 'scoop''..)
    Given a 'scoops' range being approx 1200 square meters...1.2 E+14 atoms..I'd say mostly H2 or He ...some heavier ones..but those don't really make the big a diff since there's only few of much weight would that be given a simple 2:1 ratio ? (2xH2 vs. 1 He..)

    Also..I came up with an idea to increase the starting speed to 10 KM/s..
    (moon based linear accelerated launch system, with fluid tank G absorbing for acceleration at 50G's...and yes my fav all-time president is Raygun)
    (FYI the launch system itself would be roughly 100 KM long...a lil more cuz you need to keep in mind that even fluid based G-absorbing has it's can't suddenly increase ur G's to the needs to build up just a lil..same with decreasing the acceleration..)
    Given a burn that will leave exactly enough fuel to decelerate to a dead stop at the end point...
    Given a thrust of 200Km/sec for all exhaust....expenditure @ 1 kg /sec
    Given a starting mass of kg, with 90% being fuel
    What would the speed be at the moment you end the acceleration burn ?
    Last edited: Apr 26, 2014
  9. ZMacZ Registered Senior Member

    omg..i just read back...

    ANTI-BALLISTIC foam...not anti logistic..what the eff was i on ??
  10. ZMacZ Registered Senior Member

  11. ZMacZ Registered Senior Member

    My version would not use their model though..

    I'd use.....


    Gyroscopes....but not regular ones...magnetic try and generate the same wave that that 'engine' does..

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    (and maybe as a result combine it with magneto plasma..using the strong electrostatics in the center to repel
    said plasma at even higher speeds than regular plasma thrusters....but that's the new theory for now..)

    And engine that works without propellant that can work without it too..
    The propellant would only be needed or used at the start..for an initial burst of speed....

    But hey..I'm a dreamer..

    - and will keep matter what anyone else thinks of it..
  12. ZMacZ Registered Senior Member

    I'm thinking this subject has no use to me anymore..

    Anyways, I'd so like a nice workshop..with a scrapyard..would save lots of time..
  13. DaveC426913 Valued Senior Member

    Sorry, explain how gyroscopes can propel a free-floating object?
  14. ZMacZ Registered Senior Member

    123 was a brainfart I think..
    You know the bicycle on a spinning stool ?
    you rotate the spinning stool with having leverage on any momentum part ?
    which indicated to me there's a way to create a force without it normally neccesary counter..
    the wheel starts spinning and when you try and rotate the wheel, in result the stool spins..
    so...from the perspective of the stool you apply a force that makes it spin without
    the normal needed counter-force..
    dependant on the construction if one were to apply it on 4 sides simultanously,
    it would create in a single upward force..but alas..
    somehow it does NOT function that way..or at least that's what they keep selling me..

    So..project gyroscope is in the fridge..
    Would have been really cool though to merely need electrics to create a force..
    (for electric space travel..electricity is abundant whereas fuel is not in space..)

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