How do I jack up Flagstaff?

[Herk7769]

Originally Posted by Mickey

Thanks for the info. I think that in my health condition (old and fat and bad out of shape) the trailer-aid is going to become standard equipment.
Mickey

Since I am in that camp (OF&BOOS), soon I will be depending on AAA for my ALL my roadside service and screw that jack thingy. I will only change tires on the curb side of the road in any case. I leave the car zipping past side to the professionals.

[crocus]

If you use one of the drive-on jacking aids, that puts all of the weight on one axle. We only have 4000# rubber axles, so that would overload the one by a factor of 50%. While that probably wouldn't hurt a regular spring-type suspension, I wonder if it would damage the rubber axle?
I suppose it shouldn't, as shock loads while driving are probably far higher than a static overload, but it does make me ponder.

[Herk7769]

Originally Posted by crocus

If you use one of the drive-on jacking aids, that puts all of the weight on one axle. We only have 4000# rubber axles, so that would overload the one by a factor of 50%.

John,

While on the surface that might sound logical, it not really true.

With 4 wheels on the ground the CG is calculated by drawing lines that connect each tire (forming a rectangle). The CG is in the exact center of the rectangle. This can be easily found by drawing an X through the 4 wheels. Since the intersection is equidistant from each wheel, each wheel is carrying an equal amount of weight. (ie 1/4th of the total load)

With 3 on the ground the actual load on each wheel gets a bit more complicated. We need to find out the new CG location to determine the actual load on the wheels in contact with the ground. To do this lines are drawn connecting the wheels as before, but it now forms a right triangle with an 8 foot side and a side that connects the centers of the tandem.
Use geometry to determine the center of the triangle (draw lines bisecting the angles of the triangle and the CG is the intersecting point). The actual load on each wheel is then calculated using vectors to each wheel. If the 3 wheels are equidistant apart (8 foot x 8 foot say), the weight on each wheel would be exactly 1/3 of the total camper weight. Since they are not, the load of the missing wheel is shared unequally by the 3 remaining tires. (slightly more on the side with 2 tires - shared by the 2 tires equally; slightly less on the side with one tire)

While the tires are still "overloaded" per se, that load rating is calculated to be the maximum load at 60 MPH and 50 PSI (for load range C); not a static load. This is well within the capability of the tire for brief periods.

That is why you can drive on a blowout to a safe spot without immediately blowing all the others (just not at 75 MPH!). The slower as safely possible the better.

[MtnGuy]

Lou, I think we had this discussion on another thread, and I am still not convinced that the weight is equally distributed (or close) between the 3 tires left after a flat tire. Your explanation is thorough, but I just can't get it through my thick skull.

It seems to me that during a flat, the CG of the camper does not change......it is centered still forward and center of the axles just like it is when all 4 tires are supporting the weight. Using the bisecting lines at the vectors will give you the center of the triangle, which would indicate equal weight on all 3 of the remaining tires. But the CG of the camper itself shouldn't change......other than leaning towards the flat side which would actually add more weight to that side.

I am so confused.

[crocus]

mtnguy is correct.

[Herk7769]

Since most of us who have actually changed a flat (pretty close to 100% of us with campers here) do so when still hooked up, the whole dynamic is even more confusing. I think we are just going to have to say...

The tire is safe to support however you decide to change it. Whether 25% of the supported weight; 33% of the supported weight; 50% of the supported weight, or some bizzaro "other number"; as long as you slow down, pull over, and safely support the camper while changing the tire you will be safe. Changing the "other tire on that side" as some have suggested is not neccessary unless you suspect the reason for failure to be tire age or dry rot; then change them ALL.

[Herk7769]

Bumped this whole thread to my sister. She works at Lockheed space and is a "no kidding" rocket scientist. She works in reentry ballistics (nuclear weapon) area but knows a bit about mechanical engineering since she has that degree. No answer from her yet this morning, I am sure she is working on her latest Reagan Missile test center results. Will let you know her reply.

I did find this though and, while it sounds like double speak, is kinda what I was trying to explain.

Answers.com - How do you calculate the bending moments for a uniformly distributed loads with point loads

[crocus]

It actually does not require a 'rocket scientist" to figure it out.
The dry axle weight on my fifth is about 6200 lbs, 3100 lbs per side. Raise one wheel off the ground, and there is still 3100 lbs on that side. Nothing changes, except that there is now 3100 lbs being supported by one wheel instead of two. Thus my concern considering the axle rating for that one wheel is only 2000 lbs on my 4000 lb rated rubber axles.
I am also an engineer, top 3% of my class of '77.

[Herk7769]

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.

[crocus]

Just think of it this way, if you took both wheels off the one side, would there be no weight there?

[Herk7769]

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.

[crocus]

Ahhh, you see my point. If taking two wheels off makes no difference, then neither would taking one off, correct?

[Herk7769]

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.

[crocus]

"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!

[youroo]

Originally Posted by herk7769

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!!

[Herk7769]

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]

[MtnGuy]

Wow, she did get the good looks in the family !!

Oops, sorry Lou......but us guys aren't supposed to be pretty.

[Herk7769]

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 è uno spreco di acqua e sapone.
Sarà 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?
************************************************** **************************

[MtnGuy]

I looks like some serious ciphering going on there.

[Herk7769]

Yea, and still no "Eureka! I understand now!" from me.
I am more confused than before.