| Author | Message | ||
     Tarek Khalil (Pharaohdisco) Member Username: Pharaohdisco Post Number: 41 Registered: 12-2003 |
OME762 and OME763 rear springs have the same length (16.5" DS - 16.1" PS). Therefore in a 3" OME suspension lift setup why use 1" spacers in the rear when fitting the OME762 rear springs and 1.5" spacers when installing the OME763 rear springs. Shouldn't the spacer thickness be the same when fitting either spring? | ||
| Matt Anderson (Disco01) Senior Member Username: Disco01 Post Number: 395 Registered: 10-2003 |
The OME springs are actually not much taller than the stock springs, yet becuase of their firmer rate they don't compress as much in use, meaning the vehicle sits higher. The stiffer the spring the higher it will sit, thus the need for the bigger spacer on the stiffer 763. Matthew | ||
| Tarek Khalil (Pharaohdisco) Member Username: Pharaohdisco Post Number: 42 Registered: 12-2003 |
Matt, something there I do not understand. A stiffer spring will sit higher, meaning it will provide more height, therefore when adding a bigger spacer on the stiffer 763, a larger lift will be acheived than when using the 762's with the 1" spacer and the height of both systems is not the same. | ||
| Matt Anderson (Disco01) Senior Member Username: Disco01 Post Number: 397 Registered: 10-2003 |
Ok Tarek.........I am an idiot. I guess I got it backwards, you would need the taller spacer with the 762. Sorry bout that. Matthew | ||
| Tarek Khalil (Pharaohdisco) Member Username: Pharaohdisco Post Number: 44 Registered: 12-2003 |
Matt, you are ofcourse right, the taller spacer, i.e. 1.5" should go on the 762 springs which are a bit softer. The Expedition Exchange website states on the OME springs page in their description for the 3" OME lift setup that the spacer thickness should be 1.5" for the 763 springs and 1" for the 762 springs. Is this a typing mistake or can the EE people explain? | ||
| Alan Bates (Alanb) Senior Member Username: Alanb Post Number: 516 Registered: 07-2002 |
The only reason to use the 1.5" spacers is if you carry very heavy loads. From EE website: Ho's Discovery Series I is fitted with Old Man Emu springs but has a 3" lift over the factory height. He is able to achieve this 3" of lift by fitting 764 springs in the front, 763 springs in the rear, and 1.5" Spring Spacers below the rear springs. 762 rear springs and 1.0" Spring Spacers would achieve a nearly identical height, but Ho fitted the 763 springs and 1.5" Spring Spacers because he carries very heavy loads on long trips. | ||
| Tarek Khalil (Pharaohdisco) Member Username: Pharaohdisco Post Number: 45 Registered: 12-2003 |
Yes, Alan I read that paragraph on the EE website and it says that " 762 REAR SPRINGS AND 1" SPRING SPACERS WOULD ACHIEVE AN IDENTICAL HEIGHT " as 763 springs in the rear and 1.5" spring spacers, which is incorrect as you will have a 0.5" larger height if you install the 1.5" spring spacers together with the 763 springs. Correct me if I am wrong. Ofcourse you could use this extra height to have the vehicle leveled when it is loaded. | ||
| Ho Chung (Thediscoho) Moderator Username: Thediscoho Post Number: 724 Registered: 02-2002 |
tarek, you'd achieve identical height if you didn't carry the same heavy load that i do. Ho Chung | ||
| Tarek Khalil (Pharaohdisco) Member Username: Pharaohdisco Post Number: 46 Registered: 12-2003 |
Ho, can you please elaborate. Do you mean that identical height will be achieved if you used 1" spring spacers for both springs, 762's and 763's and did not carry heavy loads. And if using 1.5" spring spacers with the 763 rear springs the vehicle will stand 0.5" higher with no load compared to a vehicle with 762's and 1" spring spacers with no load. | ||
| Ho Chung (Thediscoho) Moderator Username: Thediscoho Post Number: 728 Registered: 02-2002 |
LOL, let's try again, 762 + 1.0 + no load = 763 + 1.5 + lots of load. 762 + 1.0 + x load < 763 + 1.0 + x load. makes sense? Ho Chung | ||
| Will Selden (Ncrcwill) New Member Username: Ncrcwill Post Number: 36 Registered: 02-2004 |
"762 + 1.0 + no load = 763 + 1.5 + lots of load. 762 + 1.0 + x load < 763 + 1.0 + x load." Assume linear spring rate for 763 is 360 lb/in and 762 is 300 lb/in Then lots of load = (2) * {(1.5in)*(360lb/in) - (1.0in)*(300lb/in)} = 480 pounds Will | ||
| Tarek Khalil (Pharaohdisco) Member Username: Pharaohdisco Post Number: 47 Registered: 12-2003 |
Thanks for the detailed elaboration. | ||
| Bazzle (Bazzle) Member Username: Bazzle Post Number: 121 Registered: 09-2003 |
ROFL  Bazzle |
