# Mathematical challenge: throwing cargo overboard for dV

## Recommended Posts

Not sure if this is the right section, but I figured it's a challenge involving spaceflight...

Difficulty level: solving this will probably require knowledge of basic physics, spaceflight physics, orbital mechanics, algebra, calculus

So the story is this: you are in a cargo hauler in orbit around a small asteroid, and it has run out of fuel. All types of propellant on-board have been depleted, and it seems there is no way to push the ship anywhere. The air/oxygen reserves are not great enough to consider venting air as a viable option. So you have decided there is one option left: to throw the cargo overboard. You figure if you throw individual cargo containers overboard at the right part of your orbit and in the right direction, you just might be able to push the ship out of orbit of this asteroid.

Now the asteroid is itself in high orbit of a planet with a moderately thick atmosphere, and the cargo ship is fully capable of withstanding a descent from orbit to surface with no power or fuel. So all you need to do is nudge your cargo ship a little bit further than out of orbit of the asteroid. We're also going to assume patched conics are real, so you can gradually throw cargo overboard over the course of several orbits and not have to worry about gradually falling out of orbit. Specifics of the asteroid orbit are not given, as the exercise is to find how much dV you can come up with, and we'll just assume it's enough to get the job done.

The total mass of the cargo is 10 tons, and the mass of the ship minus the cargo (including you and your snacks) is 1 ton. The cargo can be launched out to space in units of any mass you choose. Your physical strength and mass, for purposes of this calculation, has not been given specific values. You may use your own strength/mass or any other standard you like.

What is the ideal mass of each unit of cargo being thrown? How fast could you throw it?

How much dV could you muster by throwing cargo overboard?

##### Share on other sites

Ideal mass is whatever you can throw the fastest, which would mostly be determined by the shape and size. Ideally they would be balls, somewhere between golfball and baseball size.

If you can throw at about 60mph (~27 m/s) that gives your throwing arm engine a specific impulse of about 2.75s (27 / 9.81)

With a total starting mass of 11 tons, dry mass of 1 ton, and an Isp of 2.75s, you'll have delta-v of about 64.7 m/s (9.81 * 2.75 * ln(11 / 1)) and a very sore arm

If you happen to be a professional pitcher and can throw at 100mph instead, you'll have around 107 m/s to work with

Edited by zarakon
##### Share on other sites

I think ideally you would want each unit to have a mass of the limit of 1/infinity. Or a infinitely small mass. Assuming you can transfer all of your kinetic energy into the tiny units of mass they would gain an infinite amount of speed. After you throw all the units of the 10 ton cargo in tiny pieces it would given you infinite Dv.

##### Share on other sites
I think ideally you would want each unit to have a mass of the limit of 1/infinity. Or a infinitely small mass. Assuming you can transfer all of your kinetic energy into the tiny units of mass they would gain an infinite amount of speed. After you throw all the units of the 10 ton cargo in tiny pieces it would given you infinite Dv.

But you have no way of transferring all of your kinetic energy to it. You can only throw/kick/flick an object at whatever speed you're able to move the relevant body part.

##### Share on other sites
But you have no way of transferring all of your kinetic energy to it. You can only throw/kick/flick an object at whatever speed you're able to move the relevant body part.

Ok first I made a mistake when I put limit of 1/infinity. I shouldn't of put limit but that doesn't affect my answer.

Now to answer your question. Even if all your energy didn't transfer and only a little bit say .000000000000001 Newtons transfer my theory would still work. And about the speed you can move what ever body part. When you hit a golf ball with a club the golf ball will travel faster than what the the head of the club traveled at. Conservation of momentum.

##### Share on other sites
When you hit a golf ball with a club the golf ball will travel faster than what the the head of the club traveled at. Conservation of momentum.

True. In a perfect elastic collision between a massive object and a much less massive object, the relative speed between them will remain nearly unchanged from before to after the collision. This means that if an infinitely massive golf club is moving at speed X and collides with a stationary golf ball, the result will be the club still moving at speed X and the ball moving at speed 2X.

So the best course of action would probably not be to throw the cargo out, but rather to hit it out with something like a baseball bat or club - probably actually some kind of pipe or tube in this situation. The extended length of the bat could allow the end of it to swing faster than you could actually move your arms, and you could get up to double that speed as your exhaust velocity if the cargo is stiff or bouncy enough.

But if the scenario is limited to throwing, then your exhaust velocity is determined by how fast you can move your arm/hand.

##### Share on other sites

Ok, so far. And the direction?

I would say, you throw the cargo anterograde, every time you pass the asteroids orbit ahead of him, so your periapsis moves against his velocity vector in orbit. Just like you leave a moon. Right?

And maybe you divide the "thrust" over several orbits, to throw only in the right moment. Like leaving Kerbins orbit for Duna with ions that would require an impossible 1h burntime, or so.

##### Share on other sites
But you have no way of transferring all of your kinetic energy to it. You can only throw/kick/flick an object at whatever speed you're able to move the relevant body part.
True. In a perfect elastic collision between a massive object and a much less massive object, the relative speed between them will remain nearly unchanged from before to after the collision. This means that if an infinitely massive golf club is moving at speed X and collides with a stationary golf ball, the result will be the club still moving at speed X and the ball moving at speed 2X.

So the best course of action would probably not be to throw the cargo out, but rather to hit it out with something like a baseball bat or club - probably actually some kind of pipe or tube in this situation. The extended length of the bat could allow the end of it to swing faster than you could actually move your arms, and you could get up to double that speed as your exhaust velocity if the cargo is stiff or bouncy enough.

But if the scenario is limited to throwing, then your exhaust velocity is determined by how fast you can move your arm/hand.

Opps my bad I was assuming you could kick it or or just hit it with your hand. So it just really depends how fast you can throw it for how much DV you can get.

##### Share on other sites
Opps my bad I was assuming you could kick it or or just hit it with your hand. So it just really depends how fast you can throw it for how much DV you can get.

This.

Ek=1/2mv^2. The only limit is how fast you can chuck the cargo.

If you can rig up a catapult to chuck your 10 tonnes of cargo at 100 m/sec, that's 316 m/sec for your 1 tonne ship. If you can do it at 200 m/sec, it's 1,265 m/sec.

The limit is simply how fast you can throw the cargo.

Best,

-Slashy

##### Share on other sites

If your cargo is golf balls... regulation USGA maximum weight is 1.62 ounces. Are the tons long, short or metric?

Long = 1016.05 kilos or 2240 pounds = 22123.45679012346 golf balls

Short = 907.185 kilos or 2000 pounds = 19753.08641975309 golf balls

Metric = 1000 kilos or 2204.62 pounds = 21774.02469135802 golf balls

x10 the number of balls for this exercise.

##### Share on other sites

The maths from zarakon are right. Given a throwing speed of v, and the rocket equation (notice that a rocket engines launch particles/gases at high speeds), DV=v*Ln(Initial-mass/Final-mass).

GoSlash27 is also right, a mechanical artifact (like a long bow, or a ballista, or an sling, or a golf club) might achieve more throwing speed than a biological arm.

I read an arrow from a long bow might be as high as 66-100 m/s.

Therefore a ship of 10 tons, might accelerate in the vacuum up to 100 m/s throwing around 6.32 tons of arrows from its cargohold.

Another limiting factor is how much energy is available for the operation (stretching a bow, loading an spring, swinging a club spends so much energy). The max throw speed might not be the best energy / (speed * mass). In other words: If throwing at double speed consumes three times as many snacks, the best solution might depend on how many snacks you have.

Your own debris might also be a concern. Since you will be ideally impulsing at a given point of the orbit (looking for gravitational slingshot), you might be performing one/few launches at the optimum spot every orbit. If you throw one object from an orbit, the resulting orbits of you and the thrown object will be different, but will coincide in the throwing spot. If the resulting periods happen to be armonic, then the object will hit you after so many orbits. Hopefully, the expelled object might collide with the surface, or you might happen to doing a few throws every orbit (not just one at the very precise instant), so you might be a bit higher every orbit at that point. In a bad scenario, the resulting orbit might be a 2:1 armonic, and although you might raise your apo several meters, the impulse spot will just stay 1 inch away, and you will have arrow striking at you at 100 m/s the orbit after (hope it is a blunt hit).

Edited by fjarandag
Not second orbit, projectile turns twice but vessel once
##### Share on other sites
Not sure if this is the right section, but I figured it's a challenge involving spaceflight...

This section seems like as good a place as any, there is definitely a type of challenge to it and it's really unique. So a lot of people have already touched on specific impulse with numbers for a biological or mechanical arm producing between 3 and maybe 20, but we haven't touch much on thrust. While the idea of continuous thrust for a the mechanisms we are talking about is kinda silly we could certainly get an average. Thrust is pretty simple and is: T = v*dm/dt or basically how fast you can throw stuff times how much stuff you can throw in a given time.

If we take zarakon's hunks the size of baseballs example we can say v = 27 m/s while it probably takes about 3 seconds to prep and throw a baseball sized object which weighs around 150 grams (so google has told me) thus the dm/dt term is 0.15/3 or 0.05 kg/s multiplied by the velocity of 27 m/s and we see a thrust of 1.35 Newtons... well that's basically nothing to a 10 ton ship. (I'm going to assume metric tons since that's what KSP uses and it keeps it simple) At the start the acceleration produced is 0.000135 m/s2 or put another way it's going to take us about 55.6 hours to burn our 10 tons of "fuel".

Like zarakon said a VERY sore arm.

##### Share on other sites

To add to the fun, whatever you use for the cargo, every piece thrown away imparts slightly more velocity to the ship than the previous one as the remaining mass of the ship+cargo decreases. You may want to alter the "burn" as the "fuel" amount drops.

Do you want a steady velocity change or a steady thrust that imparts an ever increasing velocity change? If among the cargo are some bungee cords, a slingshot with a calibrated scale (marked THRUST! with arrows labeled MOAR! and LESS!) would serve as a "throttle".

##### Share on other sites

this seems like you could test this in ksp by using a bunch of decouplers. i think i remember somebody doing this, deorbiting from LKO using a stack of decouplers.

##### Share on other sites

Pc=Ps

10(Vc)=1(Vs)

Vs = 10/1(Vc)

Vs = 10(Vc)

Your change in speed will be ten times more than the cargo's

Since the units of what you are throwing does not matter as seen in the equation above lets throw bases balls.

Fastest base ball throw is 169.1 km/h or 47.0 m/s

Vs = 10(Vc)

Vs = 10(47.0)

vs = 470 m/s

Really it depends on how fast you can throw something, but if you can throw the cargo in the size of base balls at 47.0 m/s you've got a change in velocity of 470 m/s.

##### Share on other sites
Pc=Ps

10(Vc)=1(Vs)

Vs = 10/1(Vc)

Vs = 10(Vc)

Your change in speed will be ten times more than the cargo's

Since the units of what you are throwing does not matter as seen in the equation above lets throw bases balls.

Fastest base ball throw is 169.1 km/h or 47.0 m/s

Vs = 10(Vc)

Vs = 10(47.0)

vs = 470 m/s

Really it depends on how fast you can throw something, but if you can throw the cargo in the size of base balls at 47.0 m/s you've got a change in velocity of 470 m/s.

That's only true if you do one 10 ton unit. For multiple units you have to use calculus since say for example you did units of a ton. The first ton thrown would would have an opposite force applied to the 10 ton ship and the change in velocity won't be that much but when you throw thy last ton the ship will gain a lot more velocity since it is only now 1 ton.

##### Share on other sites

Someone needs to make a particle accelerator engine that uses really large particles. More like a mass driver type of thing.

##### Share on other sites

What is the ideal mass of each unit of cargo being thrown?

* The ideal mass is infinitesimal. i.e. If your throwing speed (relative velocity between expelled object and source) is 100 m/s, and your Wet_mass / Dry_mass == 10. If you throw a single 9 ton container retrograde, the container will have previous_orbital_speed - 10 m/s and your empty ship +90 m/s. Whereas if you throw infinite/many objects of infinitesimal/tiny mass, then Tsiolkovsky/rocket equation is aplicable and you will get 100 * Ln(10/1) = +230.2 m/s

* However, presumably if projectile mass is smaller, thrust is smaller and energy/thrust is less efficient. You might be too old/exhausted before you complete the process.

* Maybe the question is not the right one. Everything depends on how much DeltaV do you need.

** If you needed for example 250 m/s, then you know you wont do unless you are able to throw cargo at 250 / Ln(10/1) m/s. If you are unable, consider that salvage ships might do something more meaningful with your stuff.

** If required throwing speed is affordable enough, you might consider throwing bigger chunks of cargo at lower speed to reduce manoeuvre time/etc

* Maybe you dont need to throw stuff overboard. Exploiting kerbal physics, you might accelerate/deccelerate raise/lower your orbit by slowly spinning and moving your Center of Mass prograde/retrograde /radial out/radial in. As Scott did with fuel pumps (

) but with cargo.

* Also, throwing so much debris into orbit might make you kondemned to keelhauling by a kerbal kourt.

##### Share on other sites

I am amazed noone has linked to this as yet.

Basically the exact same question, just from the 'go faster' side of things:

https://what-if.xkcd.com/85/

"Rocket golf". Note that it name-drops KSP within the first 3 paragraphs.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.