formula for figuring out length of Jib?

P

PDX_DVX

Veteran
Hey, I am hopefully going to be attempting to build my own jib soon, and I was wondering if there was any way to figure out the optimal length of the arm on each side of the central pivot? Say, if I want to have a 9' boom, I need *x* amount of length behind the pivot for the weights to have an optimal counter balance situation? I am thinking of making the jib anywhere from 9-12 feet of actual reach, so that leaves 4-5 feet of counter balance arm at the most I am guessing? Does anyone have any experience with this?
 
I am not certain that I am following your question. If by "optimal counter balance" you mean a neutral balance, then the formula is simply the product of effective mass and length for both sides of the pivot point. Simple polar to rectangular coordinate conversion will define the motions.

Perhaps you can rephrase this to a specific question for a specific configuration and clarify your concerns?
 
Check out this page I put together of DIY projects....
http://abracadabravideos.com/DIY/DIYpage1.htm
The boom shots may be helpful. The boom is 9 feet overall. 6 feet in front of the pivot point and 3 feet behind. I use a DVX100a on it and counter balance with 10 lb. barbell weight and a sliding 10 lb clamp on weight. Be sure to use a heavy duty tripod. Good luck!
 
i am wondering if it is best to keep it split up into 3rds? Say, if the boom is 12 feet total, 4 of that should be behind the main pivot, giving you 8 feet of reach? I hope that gives you a better understanding of what I am asking. Great site BTW Abracadabra!
 
Sounds about right. You need the weight away from the pivot, or the thing might still tip.

I'm building mine with extendable ends...on both sides. :)
 
The length of the boom behind the pivot point is somewhat irrelevant. The thing to remember is that the shorter it is, the more weight will be required to counterbalance it. If the boom end is 8 feet, and the weight end is 4 feet, there will be a 2:1 ratio required to counterbalance it. Given an 8 foot boom and 4 foot counterbalance 10 lbs on the head will require 20lbs of counter weight. Actually you'll need a bit more as 4 feet of the boom's weight is not factored in there. Mine has a 10ft reach with a counter length of 4 feet. About 20 lbs is required to balance it without a cam, so 12 pounds at the head, results in about 50 to counterbalance it.

Flexy materials will create some bounce in the head. A 12 foot boom will require some serious material, or wire type suspension to stabilize it.

My setup is here: http://www.pana3ccduser.com/showthread.php?t=2694

And some more here with a dolly: http://my.tbaytel.net/guskers/dolly.html
 
Thanks, glad you liked it....The key is to keep the "boom" side as light and rigid as possible. It's amazing how much "flex" can occur at the end of an 8' boom. The last thing you'll want is the boom end bouncing when you bring it to a stop. Consider making the weighted end removable...I found that the pipe flange connected to the pipe was a good solution as it was heavy where it needed to be, aft of the pivot point and fit barbell weights easily and screwed together for assembly and knock down. Good luck!
 
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