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You are in a spaceship that has separate rockets pointed to fire vertically, and horizontally.
Suppose you fire your horizontal rocket using 1 unit of fuel, making your horizontal velocity equal to 1. After two seconds, you ship's position is (2, 0), and you have traveled a distance of 2.
Suppose you fire both your horizontal and vertical rockets, using .5 units of fuel for each, and thus 1 unit of fuel total, making your horizontal velocity equal to .5, and your vertical velocity equal to .5. After two seconds then, you ship's position is (1, 1), and you have traveled a distance of 1.414.
Why did you travel less distance even though you used the same amount of fuel?
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Because the two rockets are orthogonal.
There will be a component of force, for each rocket, in the direction you are traveling.
There will be a component of force parallel to this direction, and for each rocket these are equal and opposite.
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>>8065084
what
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>>8065089
The rockets are working against each other.
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>>8065070
If 1 unit of fuel (assuming it's perfectly converted into kinetic energy) gets you to velocity 1, then .5 units of fuel get you to velocity 1/sqrt(2), with a kinetic energy of (1/sqrt(2))^2=.5. With a velocity of <1/sqrt(2),1/sqrt(2)> you travel distance 1 every second, or distance 2 in 2 seconds.
>>8065084
no
>>8065103
nooooo
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>>8065109
>That's not vertical and horizontal.
Oh dear god. Please tilt your head 45 degrees.
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>>8065109
>That's not vertical and horizontal.
This is the calibre we get on /sci/ everyone
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>If 1 unit of fuel (assuming it's perfectly converted into kinetic energy) gets you to velocity 1, then .5 units of fuel get you to velocity 1/sqrt(2)
why though?
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>>8065117
Because fuel is linearly proportional to acceleration, which is proportional to the square of velocity.
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>>8065117
E=(1/2)*m*v^2
I'm using m=2 to make the math simpler so
E=v^2 or v=sqrt(E)
so if 1 unit of fuel gives E=1,v=sqrt(1)=1, then .5 units of fuel gives E=.5,v=sqrt(.5)=1/sqrt(2)
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>>8065123
>E=(1/2)*m*v^2
why?
I get that that's the integral, but I don't see why energy should be the integral.
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>>8065137
Energy is the potential to do work. You can think of work as a force (kg*m/s^2) times/integrated over a distance (m) yielding (kg*m^2/s^2 = J)
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>>8065070
the term to search for is "cosine losses"
A good example: the launch abort motors on the upcoming SpaceX Dragon2. Because the motors are slanted away from an already slanted side surface, the thrust loses a bunch of efficiency compared to motors aligned straight down mounted on the bottom of the capsule like Soyuz.
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