Nomad Tour Van- weight porting?

while I am well aware of golf clubs, sole grinds, groove types, etc...

the concept of altering the wedge MOI and COG with weight ports is interesting...

BUT!!

without specific details,  it's hard to guesstimate how much effect this really has...

anyone ever tried one of their clubs?

is the trajectory really that much lower with the same spinning power?

What's a nomad tour van?

 

Sorry, I haven't done extensive testing on weighting in my wedges. By and large I prefer my wedges to carry a +2-3 SW to my irons, but it's more feel than anything. One of these days perhaps I'll attempt a controlled experiment.

Let's do a little math. Based on the pictures, the diameter of each port is no more than .75 centimeter. It's probably even smaller, but let's go with .75 cm. As for depth, we'll even go a full cm. Therefore, the volume of each port is:

PI * r^2 * L = pi * (.75/2)^2 * 1.0 = ~.44 cm^3

Assuming that the club head is made of steel and that the inserted weights are tungsten, the mass removed from the clubhead is as follows:

density of steel: 7.85 g/cm^3
mass of removed steel: 7.85 g/cm^3 * .44 cm^3 =  ~3.45 g

density of tungsten: 19.25 g/cm^3
mass of inserted tungsten: 19.25 g/cm^3 * .44 cm^3 = ~8.47 g

net increase in mass: 8.47g - 3.45 g = ~5 g each

Since there are two ports, there is a net increase of ~10 g.

If you've ever held a 10 g object before, you'll know that you can barely feel it in your hands let alone on a clubhead that typically has a mass of ~300g, but let's continue on and do some rough MOI calculations.

Assuming that the reference axis of rotation is exactly in the middle of the two weight inserts, let's say that each weight is 3 cm from the reference axis. For the sake of simplicity, we'll treat the weights as a point mass and calculate the MOI as:

I = m * d^2
where m = mass and d=distance from reference axis

So, the MOI for the removed steel is:

I = 3.45 g * (3 cm)^2 = ~31 g-cm^2

Since there are two ports, the total MOI of the removed steel is ~62 g-cm^2.

Note: Yes, MOI is additive.

The MOI for the added tungsten insert is:

I = 8.47 g * (3 cm)^2 = ~76.23 g-cm^2

The total MOI for both tungsten inserts is: ~152 g-cm^2.

So the additional MOI due to the tungsten inserts is ~90 g-cm^2.

Now, what's the significance of a 90 g-cm^2 increase in MOI? I have no idea. Considering that the MOI of most clubs is on the order of 2000 - 5000 g-cm^2, I would guess that this increase in MOI is insignificant to the forgiveness of a wedge.

Sheesh. Ask a simple question... 

If that's the case, then the difference between the original, unadultered wedge and a "ported" wedge is still less than the difference between the original wedge and one with tungsten inserts. Since the density of tungsten is 2.5X the density of steel, the difference in mass between the removed steel and hypothetical tungsten insert is far greater than the difference in mass between the removed steel and empty space.

Furthermore, it looks as though they do fill it with SOMETHING. This something has to have some mass. Maybe it's not as dense as steel is, but if that removed steel is only around 3.45g as I estimated, what difference does it really make?

BTW, I just read the reasoning behind this service:


LOL That's an absolute joke. The removal of that small a mass will not move the CG any significant amount. I would guess that the CG would move on the order of a fraction of a millimeter at most.

There was a clown on ebay, who ruined a set on Nike tour blades, by "porting" them.

Maroon.