New Center Lock Socket and Interesting Torque Question
Posted: Sun Jan 18, 2015 7:53 pm
Hello Lotus Community.
I'm posting this in "Tools" for lack of a better place, although the interesting question is engineering in nature.
First, the tool: I just purchased a wonderful socket from RD Enterprise to use on my center-lock wheel nuts:
The socket is a little pricey, but very well made and makes tightening and loosening my new Panasports a cinch without chipping their finish.
Now the question: I haven't invested in a torque wrench that goes up to the 180 ft./lbs. of torque required for the Elan center lock nuts. Right now I have what the colorful guys down at my local Ace Hardware tell me is a "cheater bar." Here it is:
So my engineer buddy tells me "you don't need no expensive torque wrench - all you need is to mark a few places on the cheater bar [which, as you can see, I have] and apply the correct amount of pressure."
He says I can impart 180 ft./lbs. of torque to my wheel nut by applying 180 lbs of pressure at exactly one foot back on the bar.
"Great," I say, "But I'm a lot less than 180 lbs."
So he says, "No problem, if you apply 90 lbs of pressure at 24 inches, you achieve the same result."
I say, "I'm not 90 lbs either. How in the world can I ensure I'm applying exactly 90 lbs.?"
He say, "Still no problem. How much do you weigh?"
I say, "about 135."
He smugly says, "So mark your bar at 18 inches (half way between 12" and 24") and you'll get 180 ft.lbs. of torque by applying your entire body weight of 135 lbs at the 18Inch mark - because 135lbs is exactly halfway between 180lbs and 90lbs. Problem solved, without having to buy another expensive torque wrench!"
Is my engineering friend correct? Is it really that simple?
Many thanks!
John
I'm posting this in "Tools" for lack of a better place, although the interesting question is engineering in nature.
First, the tool: I just purchased a wonderful socket from RD Enterprise to use on my center-lock wheel nuts:
The socket is a little pricey, but very well made and makes tightening and loosening my new Panasports a cinch without chipping their finish.
Now the question: I haven't invested in a torque wrench that goes up to the 180 ft./lbs. of torque required for the Elan center lock nuts. Right now I have what the colorful guys down at my local Ace Hardware tell me is a "cheater bar." Here it is:
So my engineer buddy tells me "you don't need no expensive torque wrench - all you need is to mark a few places on the cheater bar [which, as you can see, I have] and apply the correct amount of pressure."
He says I can impart 180 ft./lbs. of torque to my wheel nut by applying 180 lbs of pressure at exactly one foot back on the bar.
"Great," I say, "But I'm a lot less than 180 lbs."
So he says, "No problem, if you apply 90 lbs of pressure at 24 inches, you achieve the same result."
I say, "I'm not 90 lbs either. How in the world can I ensure I'm applying exactly 90 lbs.?"
He say, "Still no problem. How much do you weigh?"
I say, "about 135."
He smugly says, "So mark your bar at 18 inches (half way between 12" and 24") and you'll get 180 ft.lbs. of torque by applying your entire body weight of 135 lbs at the 18Inch mark - because 135lbs is exactly halfway between 180lbs and 90lbs. Problem solved, without having to buy another expensive torque wrench!"
Is my engineering friend correct? Is it really that simple?
Many thanks!
John