Sorry folks! It looks like (as every year) I’m going to commence 2008 by breaking a resolution. I’m apologising to you about this because the resolution was to avoid weighing down these glimpses into the science of darts with the sort of mathematical calculations which would put off most right-thinking readers. Unfortunately, this time I’m going to fail in that laudable ambition, although hopefully only a bit.
My problem is that we’re going to look at the effect the relatively light shaft and flights can have on a dart’s MI, its transverse moment of inertia - which, as we discussed last time, is an important factor in governing how it flies. Unfortunately, I’ve found that it really is Mission Impossible to explain this properly without using a simplified numerical example. Hence, even if I’m not exactly going to indulge in screeds of advanced mathematics, I’m afraid I can’t avoid a quick calculation or two.
So let’s be honest, unless you’re REALLY into learning about the technicalities of darts, this uniBlog is going to be so unexciting that I’d advise you now to skip to the end where I sum it up and explain why the next one should be a lot more interesting!
Still with me, eh? Such dedication! Well, apology and boredom warning over, here goes. Let’s suppose we have a magic dart that doesn’t need a point to stick in the board, doesn’t need a thread to hold the shaft on to the barrel and doesn’t need a flight holder to hold the flight on to the shaft. All it consists of is a perfectly cylindrical barrel followed by a perfectly cylindrical (plastic) shaft followed by a perfectly square flight. And, for simplicity, let’s suppose they’re all 40mm long.
Now the CG (Centre of Gravity, remember) of a uniform cylinder or square is obviously half-way along. So, the CG of our magic dart barrel is 20mm from the front of the dart, the CG of the shaft is 60mm (20mm plus the barrel length), and the CG of the flights is 100mm (20mm plus the barrel length plus the shaft length).
When I talked about the CP (Centre of Pressure) of a dart in an earlier uniBlog, I mentioned a process called “taking moments”. Providing we know the masses of the barrel, shaft and flights, we can use that process now to find the CG of our magic dart. Let’s suppose the barrel is 25gm, the shaft 1gm and the flights 0.5gm. By “taking moments” the CG of the dart can be found to be 23mm from the front (23.019mm, to be precise - those of you who know about taking moments and have suspicious natures can check this for yourselves, everyone else can just take my word for it!).
So far, so good – the relatively light shaft and flights have only moved the CG of the barrel by 3mm. Now the CP of our magic dart would be somewhere near the front of the flights, about 80mm back from the nose, so the static margin, the distance between the CP and CG, would be around 57mm. And the weight of the shaft and flights has only reduced this key aerodynamic factor by 3mm or 5% - not much, in other words!
Now for the real resolution-breaking. As I’ve said, the MI is also an important factor in governing how a dart flies and the MI of both thin uniform cylinders and squares about their CGs is one-twelfth of the length squared multiplied by the mass. One-twelfth of 40mm squared is 133.3, thus, in the units we’re using, the MI of our magic barrel is 3333 (=133.3 x 25), the MI of the shaft is 133 and the MI of the flights is 67.
Again, as with overall weight and CG position, it’s beginning to look like the relatively lightness of the shaft and flights will mean the MI of the assembled dart is only slightly different from that of the barrel alone. But not so fast. The MI we’re interested for each component isn’t the MI around its own CG, it’s the MI around the CG of the whole dart. And to find that you have to add to each component’s basic MI its mass multiplied by the square of the distance between its CG and the dart’s CG.
If you haven’t quite managed to follow that bit of gobbledegook, don’t worry, what’s important is that the MI of the barrel, whose CG is only 3mm from the CG of the dart, will only change slightly to 3561, but the shaft MI will rise eleven-fold to 1500, and the flights MI will increase hugely to 3030. The MI of the shaft and flights combined is thus more than that of the barrel, even though they’re only 6% of the weight.
And there (Note: readers who did decide to skip to the end when advised should enter after this dodgy Shakespeare quote!) is the rub. Doing the sums shows that changing, say, 0.5gm flights for thicker ones of the same shape that weigh 0.8gm and a 1gm plastic shaft for a similar one in aluminium that weighs 2gm might only move our dart’s CG or balance point by just over 2mm, hence reducing the static margin by only 3.5%, but it will increase the MI by nearly 40%, hence slowing down the swinging pendulum of the dart’s yawing motion and greatly increasing the yaw wavelength.
Which shows the importance of considering a dart as a total aerodynamic unit and not just as a barrel with whatever shaft and flights look or even feel right. And that neatly introduces a hopefully much more interesting topic than moment of inertia calculations - Sigma darts, which I’ll be discussing next time.
Until then, given all this talk about MI, what can I say but Carry On Cruising!