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Old 01-10-2011, 12:00 PM   #31
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Well, I guess that would totally depend on the center of gravity wouldn't it? If the camper was balanced on the two wheels; then no.
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Old 01-10-2011, 02:36 PM   #32
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Ahhh, you see my point. If taking two wheels off makes no difference, then neither would taking one off, correct?
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Old 01-10-2011, 02:43 PM   #33
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So, if we took off all four wheels, the camper would then be weightless and could be towed with a bicycle.

I give up. Still no word from Lockheed.
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Old 01-10-2011, 03:43 PM   #34
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"So, if we took off all four wheels, the camper would then be weightless and could be towed with a bicycle."
+++++++++++++++++++++++++
That might be worth a try!
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Old 01-10-2011, 06:17 PM   #35
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Quote:
Originally Posted by herk7769 View Post
You are probably right and I am "overthinking" it.
Seems counter-intuitive to me though.
Why it would not be 1/3 of the weight still escapes me.
I trust my sister's answer (when ever I get it) as gospel.
She got the looks and the brains in our family.
Herk,can you post a picture of your Sister? Youroo!!
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Old 01-10-2011, 06:40 PM   #36
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She is gonna kill me

Welcome to Facebook

She emailed me that they are working overtime on a project and will take a look at my question when things "cool down" ().
[Reentry joke]
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Old 01-11-2011, 08:19 AM   #37
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Wow, she did get the good looks in the family !!

Oops, sorry Lou......but us guys aren't supposed to be pretty.
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Old 01-12-2011, 12:21 PM   #38
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And I thought I was overthinking it

Still being batted around by folks much smarter than me but so far they all agree that I was wrong. It is not split 33% on each tire. What blows me away is that one of the mathematical proofs indicates it is possible that that the weight can be distributed between the two opposing tires. The cattie corner ones.
Thus there is 4000 pounds on two tires, one on each diagonal corner.

My son is in agreement with those that say 4000 on the sole tire on the side with the flat and 2000 split between the remaining tandem tires. (But what does HE know. He's a computer engineer)

My sister is too smart to answer me directly and bumped it down to a subordinate engineer.

I am still not convinced. I am a knucklehead.
I am the only one in my family who is NOT an Engineer of some kind.

************************************************** ***********
The wobbly chair - a 4 legged stool with one shorter than the other. When equally distributed weight is placed on the chair, the 3 normal sized legs will always be on the ground. Only when the weight shifts will the other leg touch. In this case, 2 legs receive 1/2 the weight, while the remaining leg receives the other half.

The Center of gravity shift is dependent on the springiness of the tire - It will shift slightly as the tire itself deforms.

Of course, I'm a computer science guy, not a mechanics guy :-)


On Jan 12, 2011, at 4:53 AM, Lou Amadio wrote:



What is the “wobbly chair” problem?

I just don’t understand why it does not become the “3 legged stool” problem.

I also don’t understand why the CG does not shift as a result of there now being a triangular shaped load supported by the three corners.

If the remaining tires were equidistant apart, would not the CG shift to the center of the Equilateral Triangle and as a result, the weight supported by 3 tires would be equal (in this case 2666 pounds; versus the rectangular shape CG and the resulting 2000 pounds per tire load)?

In this particular case the triangle formed is a Right Scalene triangle and why does the CG not move to someplace between the former (center of the rectangle) and later (the center of the Equilateral Triangle)?

Leticia’s friend’s opinion is stated thusly:

My rocket scientist friend thinks that if the CG was centered to begin with and stays centered after the flat, then:
<image001.png>
If the CG doesn’t move after A goes to 0, then he thinks D must be 0 also and B=C=4000lbs
A=0
B+C+D=8000
Left to right symmetry: B=C+D
Top to Bottom symmetry: B+D=C
Solving 3 equations with 3 unknowns….you get D=0 and B=C=4000 lbs each
I am not sure I am buying it, but he is kind of smart…..I still don’t know why the wheel doesn’t just rest on the ground like it does when you get a flat on your automobile…..


In this case my tag line might apply to me L


Qui si possano lavare il capo di un mulo,
ma alla fine, esso Ť uno spreco di acqua e sapone.
Sarŗ inoltre turbare il mulo.


Assuming the distribution of weight is equal laterally, then tire A would be supporting the weight of both A & B. The center of gravity won't shift, otherwise it will bounce between C & B, thus negating the original constraint of B not touching the ground. A & D act as a fulcrum.

The wobbly chair problem.

Also a factor is how much does b not touch the ground, and can this slack be taken by the springiness of the remaining tires?

I don't know if this truly represents the situation, as you do have an additional contact point - the 5th wheel.

Helps?




On Jan 11, 2011, at 7:19 PM, Lou Amadio wrote:


Question for a math whiz. Get Kathy to help if you need to.

The question is what happens when tire B goes loses air and no longer touches the ground.

GIVEN:
The trailer weighs 8000 pounds.
The trailer has 4 wheels each rated to support 2150 pounds at 50 PSI at 60 MPH.
The distance between A and C is 8 feet
The distance between C and D is 3 feet

The weight on each tire should be 2000 pounds. (Correct?)

ASSUMPTION: The flat tire B does not touch the ground.
FIND:

What is the weight being supported by Tires A, C and D?

Does the CG shift? If so; to where?

How does that affect the load on each remaining tire?
************************************************** **************************


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Old 01-12-2011, 12:35 PM   #39
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I looks like some serious ciphering going on there.
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Old 01-12-2011, 12:47 PM   #40
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Yea, and still no "Eureka! I understand now!" from me.
I am more confused than before.
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