# Interview question

1. Nov 30, 2006

### D H

Staff Emeritus
I like to toss a problem at interviewees to see how they think. My new question is so simple if looked at correctly, but so very counterintuitive that I caught a freshly minted PhD with it today.

A space vehicle with two thrusters is depicted below.

+=====v=========+
|.....|.........|
|.....+-----O...|
|.....|.........|
+=====^=========+

The thrusters (the 'v' and '^' in the diagram) direct their exhaust normal to the long axis of the vehicle. The two thrusters have the same Isp and generate identical force magnitude. The vehicle is perfectly symmetric about the long axis. A single fuel tank (the 'O' in the diagram) feeds the two thrusters and is located some distance from the thrusters along the long axis of the vehicle.

The vehicle starts in a quiescent state at rest wrt some inertial observer. The two thrusters are activated simultaneously. The thrust from each thruster quickly ramps up to a constant value; both thrusters ramp up in an identical manner. After firing for some time, both thrusters are shut down simultaneously. Both thrusters ramp down to null firing quickly and in the same manner.

Describe the behavior of the vehicle.

Last edited: Nov 30, 2006
2. Nov 30, 2006

### Danger

I'm no scientist by any means, but my first thought (and second) is (neutral colour answer--highlight it to read)
I expect that it would travel forward (to the right in your diagram) due to reaction thrust against the fuel tank.

3. Nov 30, 2006

### vanesch

Staff Emeritus
My intuitive answer at first sight:

A slight shift to the right ?

4. Nov 30, 2006

### DaveC426913

Danger's answer was my first thought too. But:
<that force is counteracted by the force directed against the fuel lines where they turn 90 degrees. Thus the rocket will go nowhere. Actually there will be a SLIGHT movement to the right while the centre of mass shifts as the fuel drains. The movement will stop once the engines stop. The craft will have moved less than its own length and will be stationary thereafter.>
Thus, Vanesch is correct. And if I hadn't bothered to be so verbose, I would have beaten him.

Also, here's your rocket with no dots:
+=====v=========+
|. . . . . |. . . . . . . . |
|. . . . . +-----O . . . |
|. . . . . |. . . . . . . . |
+=====^=========+

Copy this code and remove the spaces from the [ color] tags.
+=====v=========+
|[ color="#E9E9E9"]. . . . . [/COLOR]|[ color="#E9E9E9"]. . . . . . . . [/COLOR]|
|[ color="#E9E9E9"]. . . . . [/COLOR]+-----O[ color="#E9E9E9"] . . . [/COLOR]|
|[ color="#E9E9E9"]. . . . . [/COLOR]|[ color="#E9E9E9"]. . . . . . . . [/COLOR]|
+=====^=========+

Last edited: Nov 30, 2006
5. Nov 30, 2006

### Danger

That's a good point, Dave. I was thinking of radiused bends in the fuel lines (I always assume that, because I always design things that way ). It seemed to me that the thrust on the perpendicular pipes would be distributed in various directions and thus not counteract that upon the front of the tank. Is that correct? In any event, they are shown as right-angle junctions which negates that.

6. Nov 30, 2006

### George Jones

Staff Emeritus
Well, I've only thought a bit quantitatively about the question, but I'm getting a completely different answer - at the end of the day, in the original inertial frame, the ship is drifting to the left.

I'll try and think more about this when I get home, but I doubt that my daughter will let me.

7. Nov 30, 2006

### Danger

You'd bloody well better; you have me intrigued, and you know how impatient I can be...

8. Nov 30, 2006

### D H

Staff Emeritus
Most of you saw that it drifts to the right while the thrusters are firing. Only George saw that it reverses direction when the thrusters are deactivated.

Way to go, George.

During a design review, one engineer complained that the vehicle design is invalid. His complaint: The thrusters would plume the space station when the design vehicle approached the station for berthing. The designer, who really liked his design because of its very fine control capabilities, thought for a second. He said, "Ahah. My new design avoids plume impingement problems and still has very fine motion control when I fire both jets simultaneously. In fact, it has the exact same equations of motion as the first design!" His new design:

+===============+
|...............|
>-----------O---<
|...............|
+===============+

How does this vehicle have the same equations of motion as the first?

9. Nov 30, 2006

### D H

Staff Emeritus
The easy way to solve this is to invoke conservation laws. The conservation laws dictate what the answer must be. The causal actions that result in that answer are a bit harder to puzzle out, but that is often the case. For example, I don't have to model contact forces, bending, and flexing to know the final state after one spacecraft docks of one spacecraft with another. I just need to know their initial states and their configuration after the transients die out.

Back to the problem at hand: The fuel flowing from the tank to the thrusters has linear momentum. The momentum resulting from that flow has to be balanced by some other translational motion to keep the total linear momentum zero as seen by the inertial observer who saw the vehicle start at rest. The vehicle has to move to the right while the thrusters are firing.

The vehicle attains a constant velocity as soon as the fuel reaches a constant mass flow rate. Ignoring transients, the exhaust gases in the first configuration have longitudinal velocity equal to the vehicle's velocity. When thrust terminates, the only moving entities in the closed vehicle+fuel+exhaust system are the exhaust and the vehicle. The exhaust has net momentum to the right. The vehicle has to be moving to the left after thrusting terminates.

If you want to know what causes this behavior, you have to look at the transients. Imagine that you just closed the valve to a faucet a bit too quickly. That bang you just heard is momentum transfered from the no-longer-flowing water to the pipes.

Last edited: Nov 30, 2006
10. Nov 30, 2006

### DaveC426913

Why?
10 char

I'm not convinced of this. It sounds like the rocketry equivalent of attaching a giant fan to the stern of a sailboat.

Last edited: Nov 30, 2006
11. Nov 30, 2006

### D H

Staff Emeritus
Imagine you and a buddy are driving in a dune buggy at a constant velocity on the moon. Your buddy is firing tennis balls normal to your direction of travel. To you, those tennis balls appear to be moving in a straight line away from the vehicle. The balls maintain their longitudinal velocity (the velocity component in the direction of your travel) after being fired. The exact same concept applies to the streams of exhaust. The vehicle structure is moving; those exhaust streams maintain that longitudinal motion. The thrusters (configuration #1) merely add a normal component to that velocity.

12. Nov 30, 2006

### DaveC426913

You're saying that firing tennis balls normal to our forward motion is imparting a force to slow us down. How so?

Note that, if our dune buggy is floating in free space, you can no longer tell we're moving, and in fact, might not be. Yet I can still fire tennis balls out the sides. By your logic, I would start moving backwards. Or is it forwards?

Last edited: Nov 30, 2006
13. Nov 30, 2006

### D H

Staff Emeritus
I said nothing of the sort. I used this as an analogy. Regarding the tennis balls, I said "you and a buddy are driving in a dune buggy at a constant velocity". The point of this analogy is that the balls are moving along with the vehicle longitudinally while they move away from the vehicle transversely. This is the exact same situation in the first configuration of the vehicle. The stream of exhaust doesn't lose its longitudinal velocity just because it left the vehicle, does it? To think otherwise is to revert to Aristotelian physics.

BTW, the equations of motion for this system are

$$m_v(t)\ddot r_v(t) + \ddot m_v(t) l = 0$$

where $$m_v$$ is the mass of the vehicle, including the fuel remaining on the vehicle, $$r_v$$ is the velocity of the vehicle structure (not the vehicle center of mass), and $$l$$ is the vector from the point midway between the two thrusters and the fuel tank. The transients are the only thing that changes the vehicle's velocity.

14. Nov 30, 2006

### DaveC426913

Question:

Given this arrangment (where I've eliminated the fuel line factor):

+=====v=========+
|. . . . . |. . . . . . . . |
|. . . . . O. . .. . . . . |
|. . . . . |. . . . . . . . |
+=====^=========+

and a slight rightward motion (wrt to what I don't know), which way would the rocket move?

15. Nov 30, 2006

### D H

Staff Emeritus
By eliminating the fuel line factor you just eliminated the sole motive force. The vehicle moves with a constant, slight rightward motion (wrt to the same unknown what). Look at the EOM in post #13.

BTW, your vehicle will look better if you use a fixed-point font. Wrap the whole thing in a [ FONT="Courier New"] ... [ /FONT] construct.

Also, thanks for showing me how to hide the periods in my original drawing.

Last edited: Nov 30, 2006
16. Nov 30, 2006

### DaveC426913

OK, I wasn't sure if the fuel line factored in to your leftward force setup. It does.

I'm still not getting where the leftward force comes from.
It's a force that
- occurs after the thrusters stop
- cancels the slight rightward motion caused by the shift in CoG
- adds an additional component that starts it in a leftward direction

Last edited: Nov 30, 2006
17. Nov 30, 2006

### DaveC426913

Doesn't matter. No matter how you distribute them, everything cancels out except the exhaust component.

If you could do that, then you could create motion merely by shaping your vehicle like a wedge using the logic that the air would just slide off the slanted end while pushing on the flat end.

18. Nov 30, 2006

### D H

Staff Emeritus
This problem is an offshoot of our investigations into a discrepancy in our propagated orbit insertion position. We hadn't accounted for the flow momentum in our equations of motion. A vehicle casts off 90% of its mass to get to into orbit; that momentum needs to be taken into account.

Added bonus: It makes for an incredibly nasty interview question.

19. Nov 30, 2006

### DaveC426913

Well, I guess I won't doubt that you're right then...

Last edited: Nov 30, 2006
20. Nov 30, 2006

### D H

Staff Emeritus
Starting the flow of fuel requires a transfer of momentum from the vehicle to the fuel in the line. It is this momentum transfer that makes the vehicle proper start moving. That momentum is transfered back to the vehicle when the flow is stopped. However, the vehicle mass has been reduced in the interim. Instead of stopping the vehicle, the momentum transfer reverses the vehicle's direction of travel.

21. Dec 1, 2006

### vanesch

Staff Emeritus
My reasoning was simply:
the center of gravity of the entire system must remain at rest (because only internal forces act).
Now, if you look at the center of gravity before the thrusters are activated, it is at a certain point on the symmetry line of the vehicle. At the end of the day, the center of gravity of the fuel has been displaced, from the center of the tank, to the point on the symmetry line between the two thrusters, and it stays there. So in order to compensate, the vehicle (without the fuel) must shift a bit to the right (in the direction of the empty tank).

So I still claim that the overall motion is just a shift to the right:
when the thrusters fire, there will be a rightward velocity, which will drop to 0 when the thrusters will stop firing. The only thing that counts is the fuel flow from the tank to the thrusters.

22. Dec 1, 2006

### George Jones

Staff Emeritus
Sorry, I was too exhausted last night to do anything more than lurk. I hope D.H. statisfied your curiosity before your impatience killed you.

D.H. has already explained the puzzle, but here,in my own words, is the reasoning that I used. All motion is with respect to the initial frame of reference.

I used Newton's second law, which says that the (time) rate of change of the momentum of a system equals the total external force on the system. I took the system to be everything, rocket plus fuel plus exhaust. In this case, there is no extrenal force, as all the forces are internal. Since the external force is zero, the momentum of the system is zero (its inital value) throughout the entire process.

Once the fuel starts flowing to the left, the ship has to move to the right in order to keep the momentum of the system zero. The velocity of the exhaust has no horizontal component, i.e., is completely vertical (up and down) with repect to the ship. Because the ship is moving to the right, the exhaust moves (partly) to the right. After the thrusters are turned off, fuel no longer flows, but the exhaust is still moving a bit to the right. In order for the momentum of the system to be zero, the ship has to be drifting to the left.

D.H. - nice puzzle.

23. Dec 1, 2006

### vanesch

Staff Emeritus
Right !!

24. Dec 1, 2006

### D H

Staff Emeritus
Thanks, and congrats.

Nobody has responded to the second part of the puzzle, post #8, repeated herein.

During a design review, one engineer complained that the vehicle design is invalid. His complaint: The thrusters would plume the space station when the design vehicle approached the station for berthing. The designer, who really liked his design because of its very fine control capabilities, thought for a second. He said, "Ahah. My new design avoids plume impingement problems and still has very fine motion control when I fire both jets simultaneously. In fact, it has the exact same equations of motion as the first design!" His new design:

+===============+
|...............|
>-----------O---<
|...............|
+===============+

How does this vehicle have the same equations of motion as the first?

25. Dec 1, 2006

### Gokul43201

Staff Emeritus
Are there flow regulators on the exhaust pipes or does the flow go like the length of exhaust pipe?

Last edited: Dec 1, 2006