LOST JEEPS :: View topic - 4 way air hose

Huh Yea...lets dont let this get out of control. I will put together a parts list. But just to be sure the main purpose wasa to air down fast and have the same pressure.

And to prove it works.

A = pi * r^2 *4 (1 airhose * 4 chucks) First of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and

BAF = BAC + CAF = CAB + BAE = CAE.
ABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.

Similarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.

The square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get

c2 = (a-b)2+2ab
= a2-2ab+b2+2ab
= a2+b2

This proof is ascribed to Leonardo da Vinci (1452-1519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45o. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45o.) Now, area(ABHI)+area(JHBC)=area(ADGC)+area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem.

The math is here guys so no need to argue the math facts.

So Yea this proves it for sure...it is way too much faster. later...Clint

Author:	skrracing [ Sun Sep 18, 2005 2:47 pm ]
Post subject:	4 way air hose
Great idea man. I stole the 4 way air hose idea from Quinn a few years back and man it really works great. This is a great way to air down and air up tires and they always come out with equal tire pressure. Here is the entire system Here is the gage and shut off valve just shut off the value and read the pressure. Here is one of the ends. Pay the extra few bucks and buy some good air check ends. I bought some CHEAP ones at Harbor Freight and they did not last long at all. By the way ...this is a picture of the older cheap ones. Here is one side of the system Be sure to use good rubber hose and make it long enough to fit your Jeep. You will notice that there is a short side and a long side. Hope this helpes. Later...Clint

Author:	JJsTJ [ Sun Sep 18, 2005 4:29 pm ]
Post subject:
Good way to kick it off Clint. That air hose setup is great for airing up or down. Clint almost always finishes airing up or down before most of us are done w/ two tires because of that contraption.

Author:	Guest [ Sun Sep 18, 2005 4:37 pm ]
Post subject:
Man, talk about lazy. I like lazy. Interesting ingenuity none the less.

Author:	Tokyojoe [ Wed Sep 21, 2005 10:52 pm ]
Post subject:
Clint, Do you or someone on here have a write-up for it, handy? Parts list and where they found the parts? Cost? I think I saw Bill Tolle with one at Moab also.

Author:	dirtykj [ Thu Sep 22, 2005 12:07 am ]
Post subject:
How much faster does it air up anyway? You would think that splitting the air pressure four ways is going to reduce air volume going into each wheel accordingly, and that in sum total, you ought to still spend the same amount of time if you fill each on in turn. Obviously you say this is not the case, so do you have actual fill-up figures for parallel and sequential fill-up?

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4 way air hose http://www.lostjeeps.com/forum/phpBB3/viewtopic.php?f=59&t=2002	Page 1 of 2

Author:	Burchie [ Thu Sep 22, 2005 12:49 am ]
Post subject:
210, 220. what ever it takes.

Author:	Tokyojoe [ Thu Sep 22, 2005 12:55 am ]
Post subject:
Burchie, Think about this: How much time do you waste checking and rechecking your tire pressure as you air up/down? That takes up a lot of time that you could still be airing up/down Knees and back get tired of squating? Find one or two tires more/less PSI than the others? Like the idea of all 4 tires being the exact same pressure?

Author:	LoganSix [ Thu Sep 22, 2005 8:48 am ]
Post subject:
Burchie wrote: 210, 220. what ever it takes. Nice "Mr. Mom" quote. For those who didn't catch it. If you have a good compressor, I think it would definitely cut down on the time for airing up. And having all the tires at equal pressure is nice, even if the EVIC says they're not. It would be nice to have a list of parts and places you possibly could get the parts. An estimated price range would also be handy for each project in this section.

Author:	Burchie [ Thu Sep 22, 2005 12:13 pm ]
Post subject:
Thank you Logan, I never said I didn't like the idea. I think it's a great idea and that would sell too. I was even fooling with my own home design last night. Last time I was on the trail, after I was done airing down my rig, I went over and helped someone else. I also enjoyed the conversation while doing it. I guess a situation is only as bad as you want to make it. -mark

Author:	skrracing [ Thu Sep 22, 2005 12:51 pm ]
Post subject:	Hope this helps
Huh Yea...lets dont let this get out of control. I will put together a parts list. But just to be sure the main purpose wasa to air down fast and have the same pressure. And to prove it works. A = pi * r^2 4 (1 airhose 4 chucks) First of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and BAF = BAC + CAF = CAB + BAE = CAE. ABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM. Similarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB. The square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get c2 = (a-b)2+2ab = a2-2ab+b2+2ab = a2+b2 This proof is ascribed to Leonardo da Vinci (1452-1519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45o. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45o.) Now, area(ABHI)+area(JHBC)=area(ADGC)+area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem. The math is here guys so no need to argue the math facts. So Yea this proves it for sure...it is way too much faster. later...Clint

Author:	Guyute1210 [ Thu Sep 22, 2005 12:53 pm ]
Post subject:
Ditto on the parts list, and such...

Author:	Burchie [ Thu Sep 22, 2005 1:43 pm ]
Post subject:
It's all ball bearings these days. -Fletch

Author:	skrracing [ Thu Sep 22, 2005 2:14 pm ]
Post subject:
http://www.mcmaster.com/ Here is the link to McMaster-Carr all fittings are 1/4 in. Angled safetyhead chuck barbed male 1/4" part # 6140A61 4 each 1/4" hose 50' t fittings 1/4" 3 each 4way cross 1 each 1 air pressure gage 1 shut off valve with handle hose barbed fitting to pipe thread 8 each You will have to buy a few otgher fittings but almost all of this can be picked up at the hardware store. I will get some close up shots of the parts and post them maybe later today.

Author:	LibertyN8 [ Thu Sep 22, 2005 2:58 pm ]
Post subject:	Re: Hope this helps
skrracing wrote: Huh Yea...lets dont let this get out of control. I will put together a parts list. But just to be sure the main purpose wasa to air down fast and have the same pressure. And to prove it works. A = pi * r^2 4 (1 airhose 4 chucks) First of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and BAF = BAC + CAF = CAB + BAE = CAE. ABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM. Similarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB. The square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get c2 = (a-b)2+2ab = a2-2ab+b2+2ab = a2+b2 This proof is ascribed to Leonardo da Vinci (1452-1519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45o. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45o.) Now, area(ABHI)+area(JHBC)=area(ADGC)+area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem. The math is here guys so no need to argue the math facts. So Yea this proves it for sure...it is way too much faster. later...Clint What? Did you say something? i wasn't listening start over.

Author:	The Staff [ Thu Sep 22, 2005 7:12 pm ]
Post subject:
Sorry man ...Just had to post some fact so no one was thinking we were pulling their leg.

Author:	FreedomKJ [ Fri Sep 23, 2005 1:21 am ]
Post subject:
Dangit Clint, you posted that and then I had to work it out myself!

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Author:	Kryptes [ Thu Sep 22, 2005 4:37 pm ]
Post subject:
My brain hurts......but I shall never doubt you again......

Author:	Tokyojoe [ Thu Sep 22, 2005 10:45 pm ]
Post subject:
That equation is missing something: ....add these to it --- k + j then r + u, and l + e + s = . . . . . . . . . . . . KJ RULES Thanks for the info Clint.

Author:	macgyver [ Fri Sep 23, 2005 3:14 am ]
Post subject:
dirtykj wrote: How much faster does it air up anyway? You would think that splitting the air pressure four ways is going to reduce air volume going into each wheel accordingly, and that in sum total, you ought to still spend the same amount of time if you fill each on in turn. Obviously you say this is not the case, so do you have actual fill-up figures for parallel and sequential fill-up? It will air up faster if you have a fast air source. Most good air sources can move more air than will fit through one valve stems. With four at once, more air is moving and using the capacity of the pump. With my York in my TJ I can do 35's from 10 to 30 in about 5 minutes this way.