Were the Lunar Rovers faked?

[Abaddon]

It was never "shown that they weren't too small", not only are the motors ridiculously small for such a large mass but the wheels are ridiculously large, I have (admittedly) belatedly done some calculations.

1/4hp @ 125rpm = torque 10.5 lb.ft, tyre radius 15inch, mass fully loaded 400lbs.

So the torque is worked out from the 10,000rpm of the motor via a 80:1 harmonic drive.
T = 5252 x HP /  rpm

F = 10.5/1.25= 8.4

OK, with you so far. This is the Linear Force equation and 1.25 is the radius in inches.

8.4/400 = 0.021

So this is supposed to be acceleration=linear force/mass.

0.021*32.18 = 0.68 ft/s/s  or  0.2m/s/s (more than 13s to make it to 10kph)

Now I'm lost. 0.021ft per second * 32.18 feet per second. You appear to be multiplying the acceleration of one wheel on the LRV by the gravitational acceleration on Earth.

Is that right and why? What is this 32.18 figure if not the gravity of Earth?

It's a fraction of G, I will probably describe it wrong so I will give a demonstration instead.

If the force equalled 400lb and the mass to be accelerated was also 400lbs then it would be 400/400 = 1, then to convert 1g to ft/s/s you would multiply it by 32.18.

Fail. Force/mass=acceleration, not some dimensionless number.

Or, if F/m was 200/400, it would equal 0.5, then 0.5*32.18 equals 16.09 ft/s/s (or 0.5g)

At last you admit that the last line is acceleration multiplied by acceleration. Can you see that the result of such a multiplication cannot be another acceleration?
 

This formula is not restricted to earths gravity, it is still a=F/m, it's just what you have to do when using lbs.

Which renders your calculation invalid. What you have done is multiply ft/s/s by ft/s/s and goten ft/s/s as a result. Surely you can see how wrong this is?

[Abaddon]

And your reason for leaving the unit 'g' off in your original post was what, then?

Your presentation of equations is sloppy to say the least.

I second the motion for terribly sloppy presentation. Here is why:

It was never "shown that they weren't too small", not only are the motors ridiculously small for such a large mass but the wheels are ridiculously large, I have (admittedly) belatedly done some calculations.

1/4hp @ 125rpm = torque 10.5 lb.ft, tyre radius 15inch, mass fully loaded 400lbs.

No Back-up for the figures.

F = 10.5/1.25= 8.4

Formula used not identified, units used not identified, "1.25" number not identified. Now I could figure out which formula was used and where "1.25" came from from the context, but I am willing to bet not everyone could.

8.4/400 = 0.021

A number divided by a number equals a number. Again, no formula, no units. Again, I could figure those out from the context, but not necessarily everyone could.

0.021*32.18 = 0.68 ft/s/s  or  0.2m/s/s (more than 13s to make it to 10kph)

and again, your result is in the meaningless terms of ft²/s⁴, the square of acceleration.

[smartcooky]

Oh dear anywho.

I leave this thread alone for the weekend, and come back to find that you are STILL insisting that you are right and hundreds of professional, qualified aerospace engineers with many years of experience behind them, are wrong.

Do you truly believe that your argument from incredulity carries more weight than the combined physics and engineering expertise of the last fifty years?

You finally come up with some figures, but they don't prove your point. What they prove is what others here have been saying all along; that you lack any understanding of basic high-school physics. You have been told over and over, by people who are far more qualified than you appear to be, that YOU ARE WRONG. What will it take for you to accept this fact?     

[Mag40]

0.021*32.18 = 0.68 ft/s/s  or  0.2m/s/s (more than 13s to make it to 10kph)

and again, your result is in the meaningless terms of ft²/s⁴, the square of acceleration.

Not good, but he arrived at the correct result. The 32.18 should have been applied to the 400lb(mg) in the f=ma (ie. 1/32 of a slug) -

8.4/(400/32.18) rather than multiplied against the acceleration.


Still pointless though, we now have figures that show the LRV accelerates to 10kph in 6.5 seconds with one astronaut onboard and just under 13 seconds fully laden. But as has already been stated, so what.

[beedarko]

It's a fraction of G, I will probably describe it wrong

Welcome back to the discussion, Anywho. 

Would you kindly address my question(s) from a few pages back?  It appears you may have missed the original post, as I do not yet see a reply.

http://www.apollohoax.net/forum/index.php?topic=393.msg14505#msg14505

Thanks.

[ka9q]

If the force equalled 400lb and the mass to be accelerated was also 400lbs then it would be 400/400 = 1, then to convert 1g to ft/s/s you would multiply it by 32.18.

Or, if F/m was 200/400, it would equal 0.5, then 0.5*32.18 equals 16.09 ft/s/s (or 0.5g)
 
This formula is not restricted to earths gravity, it is still a=F/m, it's just what you have to do when using lbs.

Can we please, please, please use SI units? This confusion between pounds of mass and pounds of force is an excellent example of why the customary English system of units ought to have no place in modern science and engineering.

[ka9q]

Do you have a reference for any of this?

I certainly do:

NASA TM X-66816, LRV Operations Handbook Appendix A (Performance Data)

Figure 2-8 on page A-11 (pdf page 20) graphs the radial deflection of the tire as a function of applied load. The nominal deflection is 1.75 inches (4.45 cm) at a radial load of 63 lbf (280 N). The bump stop is at a deflection of 3.25 inches (8.26 cm) with an applied load of 135 lbf (600 N).
 
Given that the rover's mass is 210 kg and could carry a payload of up to 490 kg (including two suited astronauts with a mass of 96.2 kg for each suit and perhaps 80 kg for their bodies) it should be obvious to anyone with your vast engineering knowledge that such tires on earth would nearly bottom out under the rover's own weight, and would most definitely bottom out if even one unsuited adult human sat on it. And while hitting the bump stop occasionally won't immediately destroy the wheel, it is something you definitely want to avoid doing unnecessarily. Especially with flight hardware on earth before launch. (Obviously you have never worked in aerospace, so you cannot appreciate the care -- approaching obsessive-compulsive paranoia -- with which people handle hardware intended to go into space.)

I have never read any document that specifies the wheels as the problem, and the two NASA quotes I supplied suggest that the structural problems go beyond the wheels

Again, someone with your vast engineering knowledge should not need explicit, formal statements from NASA public relations persons. You claim to be able to do your own computations, so do them.

[ka9q]

By the way, this same document (NASA TM X 66816), as you might expect from its subtitle (Performance Data) contains page after page of specifications and graphs showing the performance and limitations of the vehicle in its intended operational environment: the moon.

It even includes graphs showing exactly how much force is needed to move the control handle a certain distance.

Of special interest are the many graphs showing the allowable c.g. locations, the maximum slopes on which the rover will remain stable under various loading conditions, the limits to dynamic stability in sharp turns, etc, etc, etc.

Even a non-engineer can appreciate that an awful lot of work went into this thing. And since these documents have been made public, doesn't it seem likely that if they were phony some engineer would have already discovered it by now? And that his discoveries would be quickly confirmed by his peers?

[twik]

If the force equalled 400lb and the mass to be accelerated was also 400lbs then it would be 400/400 = 1, then to convert 1g to ft/s/s you would multiply it by 32.18.

Or, if F/m was 200/400, it would equal 0.5, then 0.5*32.18 equals 16.09 ft/s/s (or 0.5g)
 
This formula is not restricted to earths gravity, it is still a=F/m, it's just what you have to do when using lbs.

Can we please, please, please use SI units? This confusion between pounds of mass and pounds of force is an excellent example of why the customary English system of units ought to have no place in modern science and engineering.

This is a clear indication that anywho doesn't know a lot about physics. S/he is completely oblivious to the notion of units, and having them work out to give you the concept (force, acceleration, velocity) that you are looking for. If the units don't cancel out correctly, you haven't done the math correctly. So, when s/he uses pounds for both weight and mass, it indicates a basic lack of understanding.

[anywho]

Not good, but he arrived at the correct result. The 32.18 should have been applied to the 400lb(mg) in the f=ma (ie. 1/32 of a slug) -

8.4/(400/32.18) rather than multiplied against the acceleration.


Still pointless though, we now have figures that show the LRV accelerates to 10kph in 6.5 seconds with one astronaut onboard and just under 13 seconds fully laden. But as has already been stated, so what.

Those ridiculously slow figures don't even allow for rolling resistance, frictional losses, slopes, or uneven terrain where the weight might not be evenly distributed.

It's a bit of a joke really.

[Allan F]

What's the difference between rolling resistance and frictional losses? How much loss?

[Luckmeister]

It's a bit of a joke really.

Maybe that's part of your problem. If you took this subject more seriously, you might actually learn something.

Once again, you smugly dismiss rebuttals with generalized remarks that prove nothing.

Now I'll ask you once again:
Was the entire Apollo program a hoax? If not, what missions were real?

[Tedward]

Well, learned more on the motors from this recent bit. You got around to it anywho, digesting the replies to see that I did things wrong as well, just like yourself.

I shall start from the beginning on a clean sheet when time allows to see what mine looks like then.

[ka9q]

This is a clear indication that anywho doesn't know a lot about physics. S/he is completely oblivious to the notion of units, and having them work out to give you the concept (force, acceleration, velocity) that you are looking for. If the units don't cancel out correctly, you haven't done the math correctly. So, when s/he uses pounds for both weight and mass, it indicates a basic lack of understanding.

Well, to be perfectly fair anyone (including me) can easily foul up the units when we're forced to work with the English system. So let's see anywho redo everything in SI. If s/he still gets it wrong, then we can confidently conclude that s/he hasn't a clue. :-)

[Andromeda]

Please don't call it the "English system".  We use SI!