Also: (pity the formatting of formulas doesn’t transfer over)

Unit

Currently, the article states

The unit of ACE is 104 kt2, and for use as an index the unit is assumed.

While this is what the sources say, I must point out that they are being sloppy with units. According to the article,

The ACE is calculated by summing the squares of the estimated maximum sustained velocity of every active tropical storm (wind speed 35 knots or higher), at six-hour intervals.

The process being described here is clearly numerical integration. Consider an equivalent calculation:

The XXX is calculated by summing the velocity of the car at one-minute intervals.

It is clear that XXX would then be an approximation to the distance travelled, and that its unit would not be “miles per hour” (for example, a car moving at a constant speed of 60 miles per hour for an hour would not have an XXX of 3,600 miles per hour), but “minutes times miles per hour” (the XXX would be 3,600 minutes×miles per hour, or 60 miles after simplification).

In the case of ACE, it is not distance travelled that’s approximated, but the quantity {\displaystyle \int v_{max}^{2}(t)dt} {\displaystyle \int v_{max}^{2}(t)dt}, where the integral is over the time a cyclone is a tropical storm. The SI unit would thus be {\displaystyle {\frac {m^{2}}{s}}} {\displaystyle {\frac {m^{2}}{s}}}, though a unit of kt2h or even kt2(6h) would be easier to work with.

I’m not quite sure how to proceed. The easiest would be to just relativise the claim that the unit is kt2 and make ACE, for purposes of our discussion, merely a number without any unit. I’d just add a note on this, but do not want to violate WP:OR.

This might be a bit of a minor issue; however, sloppiness when using physical (or mathematical) units is a real problem, and the treatment of units here is sloppy.

RandomP 16:08, 19 September 2006 (UTC)

It made more sense to me if I thought of it this way: the cyclone’s rate of energy use is being measured every 6 hours, because it appears to be assumed that the cyclone replenishes a proportion of its energy every some interval of time. This would be more like {\displaystyle \sum {\frac {d}{dt}}(kv_{max}^{2})t} {\displaystyle \sum {\frac {d}{dt}}(kv_{max}^{2})t} and the unit would be really be energy per mass - a velocity squared. I hope that makes sense? —AySz88^-^ 04:57, 22 September 2006 (UTC)

Interesting, I hadn’t thought of it that way. However, I do not see how it would change the unit. Of course you could introduce arbitrary constants to change the unit — but if that’s what’s happening, it needs to happen explicitly; furthermore, introducing a new constant whose value is six hours seems a bit redundant, since we’ve already got other constants only a factor of 4, or 6, respectively, away from that …

RandomP 12:23, 22 September 2006 (UTC)

{\displaystyle {\frac {d}{dt}}(kv_{max}^{2})} {\displaystyle {\frac {d}{dt}}(kv_{max}^{2})} has units of nm2/hr3. Then approximate the integral using the rectangle method and a delta-x of 6 hours, to get units of nm2/hr2, or kt2. The 6 hours isn’t an arbitrary constant, it’s just the resolution of the data available.

As for the k (above) that’s eventually tossed out at the end, I think this ends up with only a unit of kg if you derive the equation all the way from rotational energy (using assumptions which are something like that r and m don’t vary with time, and v varies linearly with r, etc. - sorry for the rather blurry description; I once derived the equation all the way but I can’t remember which assumptions I used…). The missing mass unit is reasonable since you wouldn’t want to pretend that the formula took mass into account. So the name of the quantity isn’t quite right - it’s accumulated energy per mass, not just “energy” - but the unit appears still correct. —AySz88^-^ 16:22, 22 September 2006 (UTC)

Oh, sorry. I misread. No, {\displaystyle {\frac {d}{dt}}(v_{max}^{2})} {\displaystyle {\frac {d}{dt}}(v_{max}^{2})} is simply not what’s being measured. At all.

6 hours is the resolution of the data; you suggested to give it some storm-related relevance as well (as I originally understood it, you essentially stated that the movement of a storm after six hours had passed was all “new” energy, with the old one having been used up).

Sorry, I still don’t quite see how your argument would work. What’s being measured is the average square-speed of the storm: assuming there’s a relatively constant mass of the storm (your k), that translates to energy-time available for destruction. A weird unit, certainly, but it does make sense.

Of course you can use another constant to get this back down to an energy estimate, but that constant is very unlikely to be anywhere near 6 hours.

RandomP 16:54, 22 September 2006 (UTC)

Now I’m confused with what you’re saying. If you say (using your words) that the energy of a storm is all “new” energy after 6 hours, is this not {\displaystyle {\frac {(kv_{max}^{2})}{t}}} {\displaystyle {\frac {(kv_{max}^{2})}{t}}}, an approximation of {\displaystyle {\frac {d}{dt}}(kv_{max}^{2})} {\displaystyle {\frac {d}{dt}}(kv_{max}^{2})}? —AySz88^-^ 17:10, 22 September 2006 (UTC)

But I’m not saying that. You did, and it appeared to me to be an unlikely coincidence, so I disagreed with it. (and no, that doesn’t look like an approximation of {\displaystyle {\frac {d}{dt}}(kv_{max}^{2})} {\displaystyle {\frac {d}{dt}}(kv_{max}^{2})} to me). RandomP 17:18, 22 September 2006 (UTC)

You folks commenting here are correct and the sources are wrong. The units on NOAA’s formulation of the ACE hurricane index are knots-squared times six hours.

As RandomP has correctly noted, the ACE hurricane index is a numerical integral. It is calculated by taking the sum of the maximum wind speed in knots, and squaring it, for six-hour intervals during the tropical storm’s lifetime. That sum only has meaning when you know that the time interval is 6 hours. If I took WRF output of Hurricane Katrina at one-hour intervals, squared the sum of the maximum wind speed in knots, would I get a comparable ACE index? No, because I have used 1-hour intervals instead of 6 hours. My total would be 6 times NOAA’s formulation.

The unit “knots-squared-six-hours” is mighty clumsy and confusing! A well-formulated index should enlighten, not obscure what’s going on. It’s unfortunate that NOAA’s formulation of the ACE index has come into usage without being vetted by a standards committee. Model output and direct wind measurements will not be forever tied to six-hour intervals.

In a web publication on hurricane metrics I have expressed the ACE index in knots-squared-days. Although “days” is not an MKS unit, at least it’s a standard length of time. Any knowledgeable person should be able to take hurricane wind measurements at any time interval and adjust their calculated ACE to days; the wind speed measurement should be multiplied by the number of days for which it applies (probably a fraction less than 1.0). To convert from NOAA’s formulation to “knots-squared-days”, simply divide by 4. Days are a reasonable measure of a tropical cyclone’s lifetime, and of course floating-point numbers should be used for days so that hours and minutes can be accommodated.

This is all explained in the web article. I also showed how to decompose the ACE index into intensity, number, and duration of storms. Various researchers have been exploring the changing nature and interaction of those hurricane characteristics.

CarlDrews (talk) 15:32, 28 November 2007 (UTC)