All evidence to the contrary, I have started writing again. I teamed up with my pal Lou Kren to collaborate on an article for one of his ventures. I am my toughest critic – but I am very pleased with my work with Lou.

I also have no less than 5 book concepts in the works. At the rate I am going, it will take a couple years to get it all out there – but am close to getting the first one going. It will most likely be an engineering calculations-focused book.

I feel good overall – and I genuinely believe I am doing my best work ever.

That said, did you know that Vince Lombardi prepared his entire career to be a head coach? And no less than 20 years into his journey, he finally got the opportunity with the Green Bay Packers and helped put the NFL on the map. [my prediction is they will be in this year's Superbowl and walk away with the Lombardi trophy]

We are approaching the 25 year mark on my career. Does my dream job of building a championship-caliber die engineering and build team with owning the quoting method and die standards lurk on the horizon?

If I had THAT opportunity, the possibilities are boundless. I foresee a reversal of work going overseas back to my hometown with the right die engineering and build processes with my brand of modern die standards to guide the efforts.

I do not know if 2011 is my year or not, but I know this: I am having fun right now.

Forces in a flanging die are equal and opposite. So what happens when flanging occurs on one side of a cutting component? Unbalanced thrust.

The issue with unbalanced thrust is should the flanging component be heeled or keyed to counteract the lateral force.

The first step to this decision is to quantify the magnitude of unbalanced thrust. The equation for unbalanced thrust in a flanging die is:

• Fut = Stan • (Fb / t)

where:

• Fut = Force of unbalanced thrust (kN)
• Fb = Force of bending (kN)
• t = Material thickness (mm)
• Stan = Span between upper and lower flange steel entry radii tangents (mm)

For example, a flange die with a span between upper and lower flange steel entry radii tangents of 12 mm (in other words, the linear distance of unsupported material when the upper and lower steels are at initial contact), bending force of 75 kN with a stamping that is 1.2 mm thick has a lateral force or unbalanced thrust of:

• Fut = Stan • (Fb / t)
• Fut =12 • (75 / 1.2)
• Fut =12 • 62.5
• Fut = 750 kN

This means that 750 kN of lateral force is acting on the flanging component. Lateral force, or unbalanced thrust is perpendicular to the flanging vector and not necessarily horizontal.

The required slide return force for die mount cams needs to be calculated to determine if a cam pad or auxiliary pressure system is needed for the application. This applies to commercial cam units and homemade die mount cams.

The equation for die mount cam slide return force is:

• Fdsr = Fs + {(0.00981 • ms) • [0.20 • (cos β + sin β)]}

where:

• Fdsr = die mount cam slide return force (kN)
• Fs = stripping force (kN)
• β = work angle from horizontal (degrees)
• ms = working slide mass (kg)
• 0.20 = coefficient of friction
• 0.00981 = force due to gravity

NOTE: if there is a cam pad, then the stripping force (Fs) required by the die mount cam slide is equal to ZERO.

For example, assume a die mount cam has a working slide with 40 kg mass. The work angle is 10° from horizontal and 4.65 kN stripping force is required. There is no cam pad.

The required slide return force is:

• Fdsr = Fs + {(0.00981 • ms) • [0.20 • (cos β + sin β)]}
• Fdsr = 4.65  + {(0.00981 • 40) • [0.20 • (cos 10°+ sin 10°)]}
• Fdsr = 4.65  + (0.3924 • 0.3706)
• Fdsr = 4.8 kN slide return force

If the die mount cam slide return spring does not have the required force capability, then either a supplemental pressure system or a cam pad will need to be added to the cam.

We discovered that the cams were failing because the dies were engineered with no cam pad for stripping the cam steels. The slide return spring had to do all the work. In many cases, the spring was inadequate and the cam failed.

I personally contacted every supplier of commercial aerial cam units in the industry world-wide seeking an equation to quantify slide return force in aerial cams. The response was universal: there was none.

My personal mantra is find a way or make one. I had to make one.

I created an equation to quantify slide return force in aerial cams. This equation works regardless if there is a cam pad or not.

The equation for aerial cam slide return force is:

• Fasr = [Fs / cos(α + β)] + {(0.00981 • ms) • [0.20 • (cos β - sin β)]}

where:

• Fasr = aerial cam slide return force (kN)
• Fs = stripping force (kN)
• α = upper driver angle from horizontal (degrees)
• β = work angle from horizontal (degrees)
• ms = working slide mass (kg)
• 0.20 = coefficient of friction
• 0.00981 = force due to gravity

NOTE: if there is a cam pad, then the stripping force (Fs) required by the aerial cam slide is equal to ZERO.

For example, assume an aerial cam has a working slide with 40 kg mass with an upper driver angle of 30° from horizontal. The work angle is 20° from horizontal and 4.65 kN stripping force is required. There is no cam pad.

The required slide return force is:

• Fasr = [Fs / cos(α + β)] + {(0.00981 • ms) • [0.20 • (cos β - sin β)]}
• Fasr = [4.65 / cos(30° + 20°)] + {(0.00981 • 40) • [0.20 • (cos 30° - sin 30°)]}
• Fasr = (4.65 / 0.6428) + (0.3924 • -0.1541)
• Fasr = 7.17 kN slide return force

If the aerial cam slide return spring does not have the required force capability, then either a supplemental pressure system or a cam pad will need to be added to the cam.