Were the Lunar Rovers faked?

[Jason Thompson]

normally you might expect to be able to sit on something like a kids bike, and even ride it on a smooth surface, but possibly not be able to ride it over bumpy terrain without it breaking.

And engineering and physics often throw up things that are unexpected to the layman, or even to the physicist who just doesn't happen to have all the relevant facts to hand.

I used to work behind a bar. One night I dropped a pint glass from a shelf above my head. It hit the floor and bounced around a little but remained completely intact. Another night I knocked a pint glass of exactly the same type from a work surface into a sink, and it shattered. Now, 'normally' would you expect a glass that broke after a fall of less than 8 inches into a steel sink to survive a fall of about seven feet onto a concrete floor? Surely that glass had more momentum when it hit the floor than the one falling into the sink did, right?

Unexpected results are how science progresses. The fact that most science does not conform to everyday expectations is why we have specialists rather than laymen doing certain jobs in those fields, and they have to study them for years before they are allowed to work in the field.

[Not Myself]

Momentum is determined by weight x time.

How about that? Momentum in a vertical motion caused by gravity - an astronaut jumping up and coming down on the rover seat - is directly proportional to weight.


[One of you actual engineers please check my work - if I made a mistake please point it out and I'll gladly retract it.]

Not an actual engineer, but I feel qualified to go at least part way with this.

It all depends on what else you hold fixed.  If we're talking about someone jumping up and then landing on something, how high do they jump?  Just to simplify it, let's suppose they start off level with the thing they're landing on.  If they are on the moon, but take off by applying the same amount of force for the same amount of time as they would on earth, then they're taking off with the same velocity, the same mass, and the same momentum.  And they're coming down with the same momentum.  Here, the lower weight is exactly offset by the greater amount of time spent falling - they're going to go higher in the moon's weaker gravity, take longer to get to the top of the jump, and take longer to fall back down.  So you have a smaller weight, multiplied by a larger time, giving the same momentum.

If on the other hand, the astronaut on the moon jumps with less force for less time than he would on earth, then the smaller weight will be offset less by the time spent falling, or not at all, if the time spent falling is the same.  For example, suppose that the astronaut shoots for a maximum height of 20 cm above the seat, on earth or on the moon.  On earth, it will take two tenths of a second to fall from this maximum height onto the seat (ballpark).  On the moon, it will take the square root of six times as long, or just under a half second.  So you'd land on the seat with one-sixth the weight, after spending the square root of six times as long falling, so 1/sqrt(6) times as much momentum - about 41% as much.

For purposes of determining whether you're going to break something when you land on it, I'm not sure if we should use momentum or kinetic energy.  But, mass is invariant to gravitational field, so if momentum is the same on the earth and the moon, then so is kinetic energy - if the astronaut takes off with a fixed momentum on the earth and on the moon (and therefore reaches a much higher height, taking much longer to do so, on the moon), then he comes down with the same momentum and the same kinetic energy in both places.  If he jumps the same height above the seat before coming down, the momentum will be 1/sqrt(6) as much on the moon, and the kinetic energy will be 1/6 as much.

[JayUtah]

You have computed the "roll stability factors for the lunar rover" and yet you question the premise that it would be many times easier to roll on the moon?

Yes, I dispute your claim.  I have asked you to support it by showing the proper computations and engineering study.

Why do you ask for numbers when you have already done them, you should know that it either is or isn't "many times easier to roll on the moon".

Because I'm asking you to show your work.  Either you have done the work necessary to support your claim or you have not.  If you have, show it.

By questioning the premise are you indirectly saying it isn't many times to roll a vehicle in 1/6g than it is on earth?

Do not try to rewrite my disputation into some sort of affirmative claim.  You have claimed the LRV is impossibly unstable.  Show your work.

You're an engineer who disputes the premise that the vehicles are unbalanced with 3/4 the weight on one side?

I dispute your analysis of static weight distribution and your ad hoc formulation of "balance."  Show me the math.

Why don't you ask a truck driver? They would roll their eyes at anyone who thinks you can have a weight similar to that of the vehicle itself, load it entirely on one side, and then call the vehicle balanced. Most would probably refuse to drive the vehicle until it was properly loaded no matter what any engineer says, and that is on earth.

Supposition and question-begging.

[JayUtah]

Let me tell you something about him, he hasn't disputed one of my assertions so far.

Nonsense, I've disputed nearly all of them.

All he has done is ask for numbers in an attempt to infinitely complicate simple assertions...

No, I have asked you for proper engineering rigor in your claim.  You are making engineering claims but you aren't doing any engineering.  You're simply waving your hands and demanding that people believe your conclusions.

Calling your assertions "simple" does not prove they are correct.  Oversimplifying a problem and then finding fault with a proposed solution because of it is a common layman's mistake.

You can either show your work or you can't.  Clearly you can't, and clearly you somehow believe you don't have to.  Hence I see no reason why rational people need to believe your outlying conclusions.

What would the proper "common sense" reaction be to a person who disputes the universal, informed belief of the entire engineering community, who offers only simplified arguments, who merely demands to be believed, and who is unwilling to show any depth or rigor?  Would the common-sensical approach be to consider that this unknown person above all others is knowlegeable and qualified to dispute a widely-accepted fact?  Or would the proper common-sensical reaction be to surmise that this person hasn't a clue what he's talking about and is therefore likely wrong?

Now perhaps you see the reason why rigor is required.

[JayUtah]

For purposes of determining whether you're going to break something when you land on it, I'm not sure if we should use momentum or kinetic energy.

Generally kinetic energy determines fracture or deformation behavior.

[JayUtah]

And engineering and physics often throw up things that are unexpected to the layman, or even to the physicist who just doesn't happen to have all the relevant facts to hand.

That's the point of science.  Engineering is just applied science sold for money.  We evolved science because intuition is often wrong.  Science is a set of methods that systematically attempt to avoid improper assumptions and thus develop objective, dispassionate models that predict actual behavior.  This is why showing your work is important.  Simply stating a belief and demanding that others accept it as the likely truth is the antithesis of reason.  We have to see the quantitative reason that leads to a quantitative conclusion so that we can tell whether it hides any assumptions or guesses that need to be investigated.

[Not Myself]

For purposes of determining whether you're going to break something when you land on it, I'm not sure if we should use momentum or kinetic energy.

Generally kinetic energy determines fracture or deformation behavior.

In that case, the difference (in my example where a mass is dropped from a given height about the rover) between the earth and the moon would be larger, a ratio of 6 for kinetic energy, vs. the square root of 6 for momentum.

[Tedward]

What makes this claim even more unusual is that, unlike the kids bike, the rovers are engineered to be driven over bumps, yet we are told they cant be sat on. It's a complete paradigm shift where something is engineered for dynamic loading but not able to be sat on.

I keep having problems with resolving this unless you divvy up the goodies. I see the evidence behind the rovers that there is back them up. It is there, it is documented. I can find out a lot, except what you claim. There is nothing, what am I missing?

Anyway, main reason for this. How are the motor workings out coming on? You said they were not up to the job as well.

[JayUtah]

Your assertions may be simple, but the engineering and the physics is not, no matter how much you assert that it is.

Indeed that's the classic layman's ploy -- "I must be correct because my line of reasoning is simple enough for me to understand."  What matters is whether the line of reasoning accurately captures the important factors.  Dumbing down a problem to where it fits within one's understanding is not a viable way to solve the problem, especially when you come up with a different answer than everyone else who has the appropriate understanding.

Engineering and science are done with numbers and computations.

More specifically, with the correct models.  A correct model appropriately captures all the qualitative relationships as well as the quantitative effects.  It is most often expressed in mathematical notation.  "Roll moment" applies in airplanes.  It also applies in wheeled vehicles.  But a vastly different model applies in each case.  If I want to compute a car's moment of inertia in the roll axis, I can certainly apply the airplane model to it because the car has a well-defined roll axis and a center of mass.  However that moment of inertia would be inapplicable to the determination of roll stability in a car, while it is perfectly applicable to the roll behavior of an airplane.

It is important to get the concepts right.  The concepts are embodied in the model, which then goes on to quantify the concepts toward a meaningful prediction.  If you want to talk about wheeled vehicle stability in a roll, you need to be talking about TTR versus SSF and other qualitative relationships that imply further quantitative computations.

"Several times more likely to roll" is just a guess.

Sorry, no. This is a science and engineering problem. No competent engineer would argue things the way you are.

Indeed.  Engineering in the real world is not just high school physics.  It's certainly not handwaving.  And you definitely must show your work.  These days the engineer of record by law must stamp and sign not only his drawings, but also his computations.  This is so his choice of model and his ability to work the implied computations is part of the public record of his work.

[Zakalwe]

Dumbing down a problem to where it fits within one's understanding is not a viable way to solve the problem, especially when you come up with a different answer than everyone else who has the appropriate understanding.

^Quote of the thread just there, folks.^

And it describes the majority of HB's thinking IMHO.

[Not Myself]

And it describes the majority of HB's thinking IMHO.

I won't disagree, but do not find this mode of thinking confined to HBs.  Maybe the percentages are different.

[gillianren]

Indeed.  Engineering in the real world is not just high school physics.  It's certainly not handwaving.  And you definitely must show your work.  These days the engineer of record by law must stamp and sign not only his drawings, but also his computations.  This is so his choice of model and his ability to work the implied computations is part of the public record of his work.

And really, I'm not sure high school physics covers quite as much as gets thrown about here.  Oh, my own high school physics class was not typical in any way; we did less than I think most people did in theirs, though for excellent reasons.  However, I don't think my sister (whose experience was much more typical) learned all that much more than I did.  Certainly I don't think she learned enough to do the requisite work to prove some of the HB assertions.  Most of what we learned was completely irrelevant--the two things I specifically remember learning were lenses and circuits.  You can't extrapolate from that to any of the issues under discussion here.

[Glom]

Well someone is going to have run some numbers.

The LRV has a mass of about 200kg.  Each astronaut had a similar mass.  So the all-up mass should be about 600kg.

I'm not sure where the centre of mass would be.  For the empty LRV, it's probably about the chassis, maybe a little above, so about 50cm.  The astronaut sitting on top will raise that significantly, so I'm just going to go with 1m above the ground.

The wheel base is a little over 2m, so in order for the loading angle to be on the verge of tipping the vehicle over, it will need to be at least 45°.

So simple trig tells us that the lateral acceleration will be the same as the gravitational field at the point of tipping, which is 1.6m/s².

The LRV had a minimum turning radius of 3m.  Since centripetal acceleration = v²/r, the speed required to achieve this acceleration is 2.2m/s or about 8km/h.

So in order to roll the vehicle, they would have had to be doing more than half the top speed of vehicle while in the tightest turn possible.

And that's assuming the wheels gripped the surface successfully in such a situation.

Is anyone amused by anywho trying to saying at the same time that the vehicle wouldn't have enough traction and that the vehicle would easily roll over?  In order to roll over, you need traction or the vehicle will just slide.

[Noldi400]

Momentum is determined by weight x time.

How about that? Momentum in a vertical motion caused by gravity - an astronaut jumping up and coming down on the rover seat - is directly proportional to weight.


[One of you actual engineers please check my work - if I made a mistake please point it out and I'll gladly retract it.]

Not an actual engineer, but I feel qualified to go at least part way with this.

It all depends on what else you hold fixed.  If we're talking about someone jumping up and then landing on something, how high do they jump?  Just to simplify it, let's suppose they start off level with the thing they're landing on.  If they are on the moon, but take off by applying the same amount of force for the same amount of time as they would on earth, then they're taking off with the same velocity, the same mass, and the same momentum.  And they're coming down with the same momentum.  Here, the lower weight is exactly offset by the greater amount of time spent falling - they're going to go higher in the moon's weaker gravity, take longer to get to the top of the jump, and take longer to fall back down.  So you have a smaller weight, multiplied by a larger time, giving the same momentum.

If on the other hand, the astronaut on the moon jumps with less force for less time than he would on earth, then the smaller weight will be offset less by the time spent falling, or not at all, if the time spent falling is the same.  For example, suppose that the astronaut shoots for a maximum height of 20 cm above the seat, on earth or on the moon.  On earth, it will take two tenths of a second to fall from this maximum height onto the seat (ballpark).  On the moon, it will take the square root of six times as long, or just under a half second.  So you'd land on the seat with one-sixth the weight, after spending the square root of six times as long falling, so 1/sqrt(6) times as much momentum - about 41% as much.

For purposes of determining whether you're going to break something when you land on it, I'm not sure if we should use momentum or kinetic energy.  But, mass is invariant to gravitational field, so if momentum is the same on the earth and the moon, then so is kinetic energy - if the astronaut takes off with a fixed momentum on the earth and on the moon (and therefore reaches a much higher height, taking much longer to do so, on the moon), then he comes down with the same momentum and the same kinetic energy in both places.  If he jumps the same height above the seat before coming down, the momentum will be 1/sqrt(6) as much on the moon, and the kinetic energy will be 1/6 as much.

Reading through that (without moving my lips; I was so proud*) I believe I agree with everything you say. The single - hopefully simple - point that I was trying to get across to Anywho is that he is insisting that "momentum" is unrelated to "weight", in the case of an object whose motion is a fall caused by gravity, the two are directly proportional

* That is not intended to be insulting in any way. My physics is so weak that I have to go slowly, with frequent backtracking.

[Noldi400]

Why all this assuming the astronauts jumped into their seats anyway? They could barely move in their suits so how could they jump into their seats?

It's not an assumption. It was the usual technique.

From the Apollo 17 Lunar Surface Journal:

117:42:47 Cernan: Let's see if there is any life in this here baby. (Burps; Long Pause) Okay. Getting up and on.

[Gene is mounting the Rover, climbing into the left-hand seat. He will fasten his seatbelt, turn on the Rover power, and take a test drive.]

<snip for space>

[Gene's breathing can be heard as he climbs on the Rover and attaches his seat belt. His breathing can be heard primarily because he is leaning forward to get the belt. Getting on the Rover is a relatively simple matter of standing alongside and then jumping up and inward, aiming for the seat. A slight jump and a pull on the handhold does the trick.]

Here's a photo series of Jack Schmitt hopping in:







I think the somewhat limited mobility is what engendered this technique. No bending or twisting was required, as it would have been in trying to get in in a more 'conventional' manner.