What about the people that follow Fords advice on the door pillar? I don't follow that, but I have read threads where some people do.

 

That's good "stuff" right there..very "funny".

 

How come whenever Gene makes a posts with a bunch of equations and rules to live by, it gets everyone wound up??
I don't read past the first or second sentence in any of Gene's post: 1-I don't have the time; 2- I'd have to sit down with pen and pad to do all the figuring he is, and there is never a pen close by...

I just read it because it makes for great "discussion".

Gene, I think most people here know how to check their tire pressure. The majority of us use our trucks for work, and maintain them properly; not just oil changes but tire pressure. It makes for major downtime when you blow a tire, plus the expense of new tires and repairs to the vehicle. Shoot, I could have checked all the air pressure in my semi truck given the time it took me to get through this thread.

 

As you can see by scanning the two threads below ...many Ford door-jam stickers contain blatantly "incorrect" information! Shame on Ford ...and I was at least one of the first to bring this situation to Ford's attention!

https://www.ford-trucks.com/forums/6...r-sticker.html

https://www.ford-trucks.com/forums/6...gawr-info.html

The following quote gives my recommendation and it's also the recommendation of all tire manufactures who must meet the "Tire and Rim Association Ratings" standards and the charts in the link give the "minimum" cold inflation pressure that a tire requires to carry a given load.

As is clearly indicated on page 7 of the above referenced document the "load and inflation tables" for "light truck tires" apply for a "maximum" highway speed of "65 mph". For higher speeds fill out the "speed work sheet" on page 5 to determine the "minimum" additional inflation pressure that's required for the "maximum" speed at which you intend to travel.

To use these tables correctly you should ideally measure the load applied to each individual tire but the facilities for doing this are few and far between and the only time I found one was at an RV convention where the line of waiting rigs was too long for me to satisfy my curiosity regarding the exact load on each of my tires.

The next best approach for estimating individual tire loads is to weigh each axel and divide by two and then include a margin to account for the fact that the load on a given axel is seldom evenly applied to each tire. This approach is especially tricky for someone like me where the 3 axels and 6 tires on my 5er are all confined to the same platform on a truck stop scale but I won't bore anyone with the analysis I came up with to estimate the maximum likely tire load based on the total weight on all 6 tires!

After all is said and done the worksheet combined with the load table gives the "minimum cold tire pressure" that's required for the "maximum load" that you intend to apply to the tires and for the "maximum speed" at which you expect to travel ...but now you've got to adjust your tire pressure to this value when the actual "ambient temperature" is equal to the "minimum ambient starting out temperature" that you except to encounter!

If the actual "ambient temperature" doesn't cooperate with your desired tire pressure maintenance schedule then use this equation which I gave in my first post... P2={(T2+460)/(T1+460)}(P1) psig ...and as an example of how to apply this equation lets assume it's fall and I want to adjust the pressure of my 5er tires to account for the 35*F "ambient starting out temperatures" that I expect to see along I-40 before arriving in Las Vegas but it's still a balmy 60*F morning at my current east cost location.

To address this situation let P1=80 psig be the value from the load table for the "minimum cold tire pressure" that's required for my load and speed ...and let T1=35*F be the "minimum ambient starting out temperature" that I except to encounter ...and let T2=60*F be the actual "ambient temperature" when I adjust my tire pressure ...and now P2={(T2+460)/(T1+460)}(P1)={(60+460)/(35+460)}(80)={(520)/(495)}(80)={1.05}(80)=84 psig ...is the correct pressure to put in my tires before leaving the east cost on a 60*F morning and then I'll be good to go all the way to Vegas! But what about the mountains that I'll encounter? Well going to a higher altitude increases the "gauge" pressure in your tires but it's colder at altitude and this decreases the tire pressure so there's no need to worry about adjusting your tire pressure just to go up and over a mountain!

Here's one final thought ...I personally think you need to apply an additional margin to that "84 psig" number above to account for the "measurement accuracy" of your tire gauge! From this link... Tire gauges ...a 100 psi tire gauge is accurate to +/- 2 psi from 25 to 75 psi and this decreases to a +/- 3% accuracy for higher pressures so based on this I'd say to include a 2 psi margin to guard against the gauge reading low. I personally compare the readings on two identical gauges and for what it's worth this same link also states...

..."Do you sell tire pressure monitoring valve caps? No. We have chosen not to sell these because the caps are not accurate (usually +/- 5 psi) and can leak, leaving tires flat."...

 

You're brilliant, but common sense says the difference between 80 and 84 psi is so negligible it's not worth fretting over. Then add in factors as you stated such as elevation change, temperature change, and direct sunlight and you're just chasing your tail. Are we accounting for that in the formula? Should I have 2 psi less on the side that's facing the sun?
On my TT it says max load at 50 psi, so that's what I set it at before I hit the road every day (well usually every day). On my truck if I'm heavy I run the max 80 psi in the rear. If I'm empty I run less. Easy, right?
I love the equations and formulas, but we're not talking about a space shuttle launch here.

 

Here's one of my favorites ...Question: What's your problem is it ignorance or apathy? ...Answer: I don't know and I don't care!

Well when it comes to "tire pressure maintenance" either answer ..."I don't know" ...the correct procedure ...or ..."I don't care" ...neglect ...is equally dangerous to the operator of the effected vehicle ...and even though I properly maintain my tires my life is still put at risk by those who don't because I've got to avoid all those road "alligators" every time I travel!

I'll use the box of "air molecules" depicted below to give an "equation free" explanation of what's going on inside your tires.



Assume the bottom surface of the box is an individual tire's "contact patch" with the road. Each molecule that collides with this "contact patch" transfers an extremely small "pound force" to the road and the total "pound force" that's transferred to the road by all the molecules colliding with the contact patches of all the tires must equal the "pound force" weight of the truck in order to keep the rims of the wheels from contacting the road!

The "pound force" that's transferred to an individual tire's "contact patch" depends on the number of molecules that are colliding with the "contact patch" and this depends on the total number of molecules in the tire. The average "pound force" that's transferred to an individual tire's "contact patch" also depends on the velocities of the molecules that are colliding with the "contact patch" and this molecular velocity depends on the absolute temperature of the air in the tire!

If you inflate a cold tire with the correct total number of air molecules to support its load and then drive on that tire the temperature of the air in the tire increases and this increases the molecular velocity and this increases the "pound force" that's transferred to the road from the tire's "contact patch". Since the total "pound force" that's transferred to the road by all the "contact patches" must equal the "pound force" weight of the truck ...the areas of the individual "contact patches" must decrease as the tires heat up.

For a hot tire the truck's weight is supported by a smaller "contact patch" area than when the tire was cold. However if you now adjust the tire pressure when it's hot by removing some air molecules the tire will have too large a "contact patch" area when it cools down and the tire will be "under inflated" for its load and it will undergo excessive flexing leading to at least some "micro damage" when you start off the next morning!

When I raced sports cars I would routinely adjust my tire pressures when the tires where hot in an attempt to improve handling by finding an optimum "contact patch" area with the track but according to the laws of "Physics" trucks hauling or towing a load will tend to travel in a straight line no matter what and the worst thing you can do is to try and swerve to avoid an obstacle ...and this is coming from an experienced race car driver who on his first trip towing his 5er thought he could dodge a road "alligator"!

In closing here's some interesting facts. The total weight of all the air molecules in all the tires on a typical pickup is only 3 pounds force (lbf) and this amount of air is enough to support a truck weighing 8,000 lbf! The average weight an air molecule is only 1.06x10^-25 lbf and this means that all the tires on a typical pickup contain a total number of air molecules given by... (3)/(1.06x10^-25) ...which is 2.8x10^25 molecules ...and if even one of these molecules leaks your tires are under inflated!

Well come morning I'm off to dodge some more road "alligators"!

 

Rofl!

What I'm saying is that at the beginning of a trip you set your pressure picture perfect, but then if something changes in the environment, road conditions, or load then it's all for naught. And these things do change constantly during a trip. So the best you or anyone can hope for is a happy medium and being 'close'. Or am I incorrect in that assumption?
When you drive a race car there is an opportunity to pit and do all that maintenance. Do you constantly reformulate, check and adjust your tire pressure while traveling? Just curious.

 

{Well when it comes to "tire pressure maintenance" either answer ..."I don't know" ...the correct procedure ...or ..."I don't care" ...neglect ...is equally dangerous to the operator of the effected vehicle ...and even though I properly maintain my tires my life is still put at risk by those who don't because I've got to avoid all those road "alligators" every time I travel!} earnesteugene

This is great information. I've printed every page (sic) to line the bottom of my parakeet cage. As you originally brought up bodily functions, I would say that if you can pour pee out of a shoe, you can air up your tires safely.
If everyone else is putting you at risk, then maybe exercising greater caution and staying home is the only answer. I take offense at the thought that you are the only one with the proper answers to the right questions. Oh wait, didn't they just give the Nobel Peace Prize to someone with a similar attitude. Common sense please.I apologize for not getting the quote entered right. Obviously a side affect of improper inflation, either tires or ego.

 

I encountered so many of these "beasts" on my I-95 trip to Coca FL that I decided to do a "road alligator analysis" to see why they breed so rapidly! It turns out if you go 75 mph there's a 286 pound Force (lbf) trying to rip each 1 pound Mass (lbm) of tread from the tire's casing and to put this in better perspective ...there's about a 1 pound Mass of tread underneath the tire's contact patch area!

My only tire failure was 8 years ago when I "momentarily" hit 75 mph so I could get up the next hill without downshifting ...well downshifting is a lot easier than changing a tire on a 5er! After I switched to a 4.10 diff I could pull better in 4th so I never exceeded 65 mph and since the "Centrifugal Force" that breeds road "alligators" is proportional to MPH^2 the 286 pound Centrifugal Force at 75 mph reduces to... (286)(65/75)^2=215 lbf ...at 65 mph! I guess that extra ...286-215=71 lbf Force ...is like the "final straw" that broke the "camel's back"!

If you look at the two pictures below ...the tire's structure ...and the computer model of a tire impacting an object ...it's easy to see why "flexing" tends to separate the "tread" from the "casing"!





Now that I'm in Coca FL for the winter and a shuttle launch is scheduled for Feb ...the "excessive" cold weather flexing of a tire's internal components caused by an underinflated tire reminds me of those cold brittle shuttle O rings on the Challenger! If the cold brittle rubber internals of an underinflated tire even crack just a bit due to hitting some obstacles on the way from the RV park to the interstate then the bond between the tread and the casing has a much harder job of resisting the "Centrifugal Force" that's depicted below.



If you use D=2.63 ft for an LRE tire, M=1 lbm for the Mass of the tread underneath the tire's contact patch, and MPH=75 in the equation shown above you get... F={(M)(MPH^2)}/{(D)(7.48)}={(1)(75^2)}/{(2.63)(7.48)}=286 lbf ...and this 286 lbf Force is applied to each 1 lbm Mass of tread ...and this 286 lbf Force is generated by an acceleration of 286 g's which is an acceleration that's 286 times the "standardized" gravitational acceleration of g*=32.174 fps/sec!

The tires on my 5er are less than 2 years old but the ones on my Freightliner will be 6 years old next month and they're showing some signs of their age. Old age kills everything including tires! As a rule of thumb ...1 "human year" is equal to 7 "dog years" and also equal to 14 "tire years" ..and since I feel the effects of age in my 68 year old body with each passing day I'm sure my truck tires which are 84 "human years" old are approaching the end of their life ...but hopefully they won't become road "alligators" on my way back to Hagerstown MD in the spring which is when I plan to get new ones!

Well I don't know how to "measure" the age of your dogs but when it comes to your tires just check the DOT Code ... the DOT Code begins with the letters "DOT" ...and the "last four" numbers represent the "week" and "year" the tire was manufactured ...for example if your last four DOT numbers are "1902" it means the tire was manufactured in the "19th week" of "2002" ...and this means you're driving on a tire that's too old to be safely used and it will likely become a road "alligator" if it's not replaced!

Tire Failure and Tread Separation Explained: Free Case Review
..."Tire manufacturers have long known that tires more than six years old, regardless of tread depth, pose a substantial safety hazard to consumers. Tire age degradation has been an "open secret" within the industry, but the public has only recently started to take notice as the number of crashes caused by "aged" tires has grown."...

 

One thing I have found that breeds rubber gators is tire being 5 or more years old tend to shed their skin at highway speeds (60+ mph) I had three come uncapped at 4 years old
The hot roads in Fla also breed more rubber gators than most

Nothing scientific or mathmatical (I know that pie R round and cornbread R square) just experience

 

Nothing scientific or mathmatical (I know that pie R round and cornbread R square) just experience

God Bless The Simple Things!!!!

Barney
____

 

X2.....................

 

Yes excessive "tread temperature" is also an independent "killer" of tires because higher "tread temperature" weakens the "vulcanization bond" securing the "tread" to the "casing" and this is especially problematical if this bond has already been weakened by rolling a cold underinflated tire over some obstacles before reaching the highway and bringing the tire up to it's normal road speed temperature!

The following information on "tread temperature" is based on my 11 years of measurements and on measurements reported by others on RV forums and on data given by tire manufactures. The center tread is the correct place to measure "tread temperature" because that's where the maximum value occurs and "tread temperature" measured there typically runs 30*F to 50*F higher than the temperature of the tire's sidewall which I also measure at each rest stop and if another RV or 18 wheeler pulls in next to me I also measure their tread and sidewall temperatures!

At normal road speeds of up to 65 mph the typical "tread temperature" shouldn't exceed about 60*F above ambient air temperature and depending on sun angle and road surface temperature this means the typical "tread temperature" shouldn't exceed about 150*F to 180*F.

However in severe conditions when driving at speeds above 65 mph and when the direct sun has heated the road surface to 120*F or more and when the ambient air temperature is 100*F or higher the "tread temperature" on a fully loaded tire might easily exceed 200*F! If the "tread temperature" approaches 250*F the "vulcanization bond" securing the "tread" to the "casing" weakens considerably because during the tire's manufacturing process vulcanization occurs at between 250*F to 320*F!

As I've previously stated in other threads I do these "technical posts" employing "equations" as my "retirement hobby" because it forces me to think and thinking exercises my brain and exercising my brain will hopefully slow its degradation with age! Unlike "tires" which wear out and "weaken" with continued use the "brain" actually gets "stronger" the more it's used! I've also offered to help anyone who's interested to better understand how equations can be derived and used to solve problems.

So here's yet another "equation" I came up with to help explain why tires get as hot as they obviously do at higher road speeds! The engine has to expend some of its HP to overcome the "rolling resistance" of the tires and the following equation gives an estimate of the RRHP=Rolling Resistance HP that's required... RRHP={(GCW)(MPH^1.3)}/{150,000} hp ...where GCW is the Gross Combined Weight lbf load on all the road tires and MPH is the mph road speed.

Let's take my old 5er as an example and it had a GCWR=14,000 lbf and I was running close to its maximum combined two-axel GAWR=12,000 lbf and if this two-axel load split equally among the 4 tires my LRE tires were almost fully loaded ! At 65 mph the above equation gives... RRHP={(GCW)(MPH^1.3)}/{150,000}={(12,000)(65^1.3)}/{150,000}=18 hp.

This 18 hp of "rolling resistance" is dissipated as heat energy in the 4 tires so each tire must dissipate 4.5 hp worth of heat energy and since each hp equals 745.7 Watts this is 3,356 Watts of heat energy per tire which is also 11,450 Btu/hr of heat energy per tire and this is a tad more than twice the amount of heat energy produced by one of the standard size 1,550 Watt 13 Amp heaters that I use to heat my 5er!

A tire dissipates this heat energy by convection, conduction, and radiation and all of these mechanisms require a temperature difference between the tire and some cooler substance so the only way a tire can dissipate 11,450 Btu/hr of heat energy is to increase its temperature to a value well above the temperature of the ambient air and the road's surface!

If the road speed increases by a factor of 2 then from the above equation a tire must dissipate a factor of (2)^1.3=2.5 times more heat energy which requires an increase in "tire temperature" and to go along with this higher "tire temperature" the "centrifugal force" equation ...F={(M)(MPH^2)}/{(D)(7.48)} lbf ...shows that the force trying to separate the tread from the casing increases by a factor of (2)^2=4!

As is indicated on page 7 of this document... http://hmcclub.homestead.com/Goodyea...oad_Charts.pdf ...the "load and inflation tables" for "light truck tires" apply for a "maximum" highway speed of "65 mph". For higher speeds you need to fill out the "speed work sheet" on page 5 to determine the "minimum" additional inflation pressure that's required for the "maximum" speed at which you intend to travel. The reason tires need to be inflated up to 10 psi more than the inflation pressure stamped on their sidewall is to reduce their high-speed "rolling resistance" and this helps to limit the tire's temperature increase at higher speeds!

It turns out that since the "tangential tread MPH" that's used in the above equation is directly equal to the "straight line MPH" that's read on a truck's speedometer the "centrifugal force" equation I gave doesn't use the factor "Pi=3.14..." but if you want to relate the "centrifugal force" to "wheel rpm" you'd then need to employ the "Pi" factor.

You can also use this same equation ...F={(M)(MPH^2)}/{(D)(7.48)} lbf ...with M being the lbm Mass of the entire truck and D being the ft Diameter of a circular segment of a turn in the road to calculate the lateral cornering lbf Force that all the tires combined need to generate in order to steer around the turn. If the tires can't generate the required "lateral cornering Force" given by this equation you "spinout" and likely crash!

The "lateral cornering Force" that tire's can actually generate depends on the tire design and on the road surface condition and for truck tires the maximum total lateral lbf cornering Force generated by all the road tires is about 80% of the truck's lbf weight which is another way of saying the tires can generate enough cornering Force to steer around a "0.8 g" turn! So as long as the tires aren't overloaded they can generate a "g load" that's independent of the truck's weight!

This equation... g={(MPH^2)}/{(D)(7.48)} ...gives the "g load" that the tires need to generate and note that it's also independent of the truck's weight. Solving this equation for MPH gives... MPH=(2.735){(g)(D)}^0.5 mph ...and using g=0.8 as the maximum possible value for truck tires gives... MPH=(2.45){D}^0.5 mph ...as the maximum cornering speed for a given Diameter turn!

So if you're traveling along an interstate at 65 mph and you come to a clover leaf that involves a 400 ft Diameter turn you'd better get your speed down to at least... MPH=(2.45){D}^0.5=(2.45){400}^0.5=(2.45){20}=49 mph ...or lower otherwise you'll likely crash and no longer be around to read my "informative" and "entertaining" posts!!!