The highly anticipated Wolfram|Alpha site came online over the weekend, and here are some first impressions:

They need a little work on the html — this was Firefox, but it looked the same on Konqueror.

It’s a very nice, highly knowledgeable calculator for pure math:

For physics and engineering, it’s not so impressive. As a test, I tried to do the calculation I did last week to point out that the amount of energy in a year’s production of antimatter from CERN was equivalent to less than a teaspoon of gasoline. For that, I used Google and a calculator, and it took less than 5 minutes.

how much gasoline has the energy equivalent of the antimatter CERN produces in a year?

? Wolfram|Alpha isn’t sure what to do with your input.

CERN annual antimatter production

? Wolfram|Alpha isn’t sure what to do with your input.

? Wolfram|Alpha isn’t sure what to do with your input.

antimatter | matter consisting of elementary particles that are the antiparticles of those making up normal substances

(but nothing about the energy equivalence)

Which tells me all about the GDP of the nations which are members of the CERN consortium, but nothing about the experimental facility.

OK, so it’s back to Google. The query *cern grams antimatter year* points me at the same page I referenced before, where we find

At CERN we make quantities of the order of 10

^{7}antiprotons per second and there are 6×10^{23}of them in a single gram of antihydrogen. You can easily calculate how long it would take to get one gram: we would need 6×10^{23}/10^{7}=6×10^{16}seconds. There are only 365 (days) x 24 (h) x 60 (min) x 60 (sec) = around 3×10^{7}seconds in a year, so it would take roughly 6×10^{16}/ 3×10^{7}= 2×10^{9}= two billion years!

so half a nanogram per year. That’s a whole nanogram of mass-energy for a year’s output, since you also count the mass of the matter it annihilates with. Back to W|A:

energy equivalent of nanogram of antimatter

? Wolfram|Alpha isn’t sure what to do with your input.

? Wolfram|Alpha isn’t sure what to do with your input.

OK, so we actually know E=mc^{2} even if W|A doesn’t.

1 nanogram * c^2 where c=speed of light

This works, among other answers we get 89.88 kJ. Call it 90 kJ.

Input interpretation:

electric utilities | volume | Perot Systems | volume | RBOB gasoline futures (NYMEX:RB) | volume

Nope, didn’t want stock market information, and it didn’t give me any information even about that.

? Wolfram|Alpha isn’t sure what to do with your input.

45 000 W s/g (watt seconds per gram)

OK, now we’re getting somewhere! let’s put them together:

(1 nanogram * c^2 where c=speed of light)/(gasoline heat of combustion)

? Wolfram|Alpha isn’t sure what to do with your input.

Alright, so I have to type the numbers and the units back in.

Unit conversions:

2 grams

11.1 millimoles of glucose

0.0111 moles of glucose

Good — 2 grams of gasoline. There’s light at the end of the tunnel. (Note, BTW, that you can type exactly the same expression into Google and get the same result.)

Input interpretation:

convert 2 grams to teaspoons

Result:

grams and tsp (teaspoons) are not compatible.

Corresponding quantity for 2 g:

Volume V of water from m=rhoV:\n | 0.41 tsp (teaspoons)\n | 2 mL (milliliters)\n | 2 cm^3 (cubic centimeters)\n | (assuming water density ~~ 1000 kg/m^3)

This is promising, but it assumed I meant water. I thought it was supposed to have a context mechanism, but if so, it isn’t working here…

? Wolfram|Alpha isn’t sure what to do with your input.

? Wolfram|Alpha isn’t sure what to do with your input.

volume of 2 g of gasoline in teaspoons

? Wolfram|Alpha isn’t sure what to do with your input.

? Wolfram|Alpha isn’t sure what to do with your input.

Result:

730 kg/m^3 (kilograms per cubic meter)

1 teaspoon * density of gasoline

Unit conversions:

3.6 grams

20 millimoles of glucose

0.02 moles of glucose

Finally. 2g of gas where a teaspoon is 3.6. But Google and a calculator is faster.

AnonymousJune 14, 2009 at 12:38 pmWolfram makes me a sad panda š