Some thoughts on “balance”.
Since there is no way to counter-balance the torque of a moving bait, such as a crankbait or spinnerbait, let’s concentrate on bottom contact baits.
First, how do we describe a balanced combo? Most would say that a rod / reel combo is balanced if, when you hold it as you would when fishing, the tip is not being pulled down.
What we’re talking about is torque. In this case, the rotational force we feel at our wrists. The most common scale we use is foot-pounds or pound-feet. Take your pick. The result is the product of the two terms. A one pound weight, two feet from the center of rotation produces the same result as a two pound weight, one foot out.
So, where’s the center? Depends entirely on how you hold your setup. When I’m fishing a baitcast rig, all four fingers are in front of the trigger. I’ve fished several times with Kent, and he holds his rods completely different. Only his forefinger is in front of the trigger. Who is doing it correctly? I am, of course. And so is he, of course. We are going to experience different balancing torques from the same rod / reel combo. Like the disclaimers all say, your results may vary.
Let’s run some numbers, and since I’m a simple-minded sort of guy, let’s use some simple numbers to start with.
First determine where the rod balances naturally. That will be our center of rotation. Now suppose we’ve tied a piece of line onto the tiptop guide and tied a weight to the line.
We’re working in foot-pounds so let’s use 4oz, or 0.25 pounds.
And let’s say we have 8 feet of rod in front of our center of rotation.
That gives us 2 foot-pounds of torque.
Keeping in mind the way gravity works, the result is valid when the rod is parallel with the ground.
How do almost all of hold the rod when fishing jigs or plastics? With the rod pointed up, not parallel with the ground, or with the water surface, is you prefer.
Keeping the weight the same, as we move the rod from parallel with the ground towards perpendicular, what is happening? The distance from center is becoming less. The nearer to perpendicular the rod is, the closer to zero the distance becomes.
If the rod is pointing straight up, the distance from center is zero and there is no torque. There is however, the weight of the rod, plus the weight of the sinker being pulled straight down.
Think about that for a minute, and then tell yourself again how adding weight to balance a rod makes it “feel” lighter.
Let’s look at some “real” numbers. My most “out of balance” combo is a 7’6” XH Duckett swimbait rod. With the reel mounted, and me loosely holding it where I hold this one to fish, I measured about 4 ounces of force at the tip.
I also measured right at 6.25 feet of rod from the center of my hand/wrist.
That gives a torque of 1.5625 ft-lbs when the rod is parallel with the ground.
I hold my rods about 10 degrees off perpendicular for some reason. That’s just where it feels right to me. I’ve measured this, as best I can, from some pictures.
At 6.25 feet, 10 degrees puts the rod tip about 1.1 feet from center. This gives a torque of 0.275 ft-lbs. Doing the calculations, I’d have to add 3.4 ounces of weight to the butt end of the rod to make it “balance”, in the sense that most people use the term.
This rod only weighs a little over four ounces. Let me see, do I want to turn a nice 4 ounce rod into a 7.4 ounce pig of a rod? Or do I think my wrist can stand the incredible strain of 0.275 ft-lbs of torque?