Greatest Metal Vocalists of All Time!!

After not blogging for some time, I wanted to come back with something that was thought-provoking, comprehensive, and a bit controversial. I've spent the better part of the day coming up with my list of what I think are the best rock/metal vocalists of all time.

I came up with a list of 100 people, which was surprisingly difficult to assemble (many were left off the list), listed in descending order from greatest to 100th greatest. I paid no mind to popularity, albums sold, length of hair, sexual orientation, etc. but instead rated them on their soulful ability to hit the notes, passion, and their ability to emote the music. Many of them have octave ranges of either 4 or 5 and can crack glass without effort. Although I can handle the second on the rare occasion I do the dishes, I can't compete with them with my paltry 0.43 octave range, even though the gene is in my blood, as quite a few gentlemen and gentlewomen from my ancestral land of Finland (you'll notice their names when you read them)--the homeland of Progressive/Symphonic Metal--make the list.

I case you haven't heard of some of the guys, look them up on Wikipedia. Check some of these guys out when you have a chance. They're all worthy of a download and listen. Feel free to disagree and/or come up with your own list, but don't knock 'em 'til you've tried 'em.

Let the discussion begin . . .

My top 100 Rock/Metal vocalists of all time.

1. Bruce Dickinson (Iron Maiden)
2. Tony Kakko (Sonata Arctica)
3. Chris Cornell (Soundgarden, Audioslave)
4. Steve Tyler (Aerosmith)
5. Layne Staley (Alice in Chains)
6. Geoff Tate (Queensrÿche)
7. Hansi Kürsch (Blind Guardian)
8. Axl Rose (Guns ‘N Roses)
9. James LaBrie (Dream Theater)
10. Jon Bon Jovi (Bon Jovi)
11. Ronnie James Dio (Black Sabbath)
12. Ivan Moody (Five Finger Death Punch)
13. Timo Kotipelto (Statovarius)
14. Sebastian Bach (Skid Row)
15. Matt Barlow (Iced Earth)
16. Phil Anselmo (Pantera)
17. Glenn Hughes (Black Sabbath, Deep Purple)
18. Paul Rodgers (Bad Company)
19. Tony Harnell (TNT)
20. Corey Taylor (Slipknot)
21. Eddie Vedder (Pearl Jam)
22. Dave Mustaine (Megadeath)
23. Klaus Meine (The Scorpions)
24. Rob Halford (Judas Priest)
25. Miljenko Matijevic (Steelheart)
26. Glenn Danzig (Misfits, Danzig)
27. Ari Koivunen (Finnish singer)
28. Bon Scott (AC/DC)
29. Freddie Mercury (Queen)
30. David Draiman (Disturbed)
31. Jarko Ahola (Teräsbetoni)
32. James Hetfield (Metallica)
33. Maynard James Keenan (Tool, A Perfect Circle)
34. Robert Plant (Led Zeppelin)
35. David Coverdale (Deep Purple, Whitesnake)
36. Sammy Hagar (Montrose, Van Halen, Chickenfoot)
37. Ozzy Osbourne (Black Sabbath)
38. Russell Allen (Symphony X)
39. Lance King (Avian, Balance of Power, etc)
40. Philip Labonte (All That Remains)
41. Marco Hietala (Nightwish)
42. Myles Kennedy (Alter Bridge)
43. Lemmy Kilmister (Motörhead)
44. Jonathan Davis (Korn)
45. Mike Patton (Faith No More, Mr. Bungle, etc.)
46. Michael Sweet (Stryper)
47. Eric Adams (Manowar)
48. Ian Gillan (Deep Purple)
49. Andi Deris (Helloween)
50. Brian Johnson (AC/DC)
51. Geddy Lee (Rush)
52. Mark Slaughter (Slaughter)
53. David Defeis (Virgin Steele)
54. Tarja Turunen (Nightwish)
55. Amy Lee (Evanescence)
56. John 'Gio' Cavaliere (Black Majesty)
57. Serj Tankian (System of a Down)
58. Steve Perry (Journey)
59. Jeff Keith (Tesla)
60. Roy Khan (Kamelot)
61. Chad Kroeger (Nickelback)
62. Bono (U2)
63. Yama-B (Galneryus)
64. Howard Jones (Killswitch Engage)
65. Mikael Åkerfeldt (Opeth)
66. Robin McAuley (MSG)
67. Michael Kiske (Helloween)
68. Tim "Ripper" Owens (Beyond Fear, Yngwie Malmsteen)
69. Chuck Billy (Testament)
70. Jeff Scott Soto (Yngwie Malmsteen, Journey)
71. Joe Lynn Turner (Yngwie Malmsteen, Deep Purple)
72. Rob Zombie (Rob Zombie, White Zombie)
73. Wayne Static (Static X)
74. Pepper Keenan (Corrosion of Conformity)
75. Billy Corgan (The Smashing Pumpkins)
76. M. Shadows (Avenge Sevenfold)
77. King Diamond (King Diamond, Merciful Fate)
78. Randy Blythe (Lamb of God)
79. Anneke van Giersbergen (The Gathering)
80. Vince Neil (Mötley Crüe)
81. Chad Gray (Mudvayne, Hellyeah)
82. Blaze Bayley (Iron Maiden)
83. Dee Snider (Twisted Sister)
84. Don Dokken (Dokken)
85. Joan Jett (Joan Jett and the Blackhearts)
86. Ian Astbury (The Cult)
87. Alexi Laiho (Children of Bodom, Warmen)
88. Dennis DeYoung (Styx)
89. Anders Fridén (In Flames, Dark Tanquility)
90. Cristina Scabbia (Lacuna Coil)
91. Zakk Wylde (Black Label Society, Ozzy Osbourne)
92. Paul Di'Anno (Iron Maiden)
93. Roger Daltrey (The Who)
94. Steve Marriott (Small Faces, Humble Pie)
95. Zack de la Rocha (Rage Against the Machine)
96. Lajon Witherspoon (Sevendust)
97. Tom Araya (Slayer)
98. Ryan McCombs (Soil, Drowning Pool)
99. Shagrath (Dimmu Borgir)
100. Josh Todd (Buckcherry)

Predicting with confidence

If you've ever read time off of a clock, then you're familiar with something mathematicians call modular arithmetic. You might also think that modular arithmetic is science of counting all the possible ways you can configure an office using cubicles, and you'd be clever, but you'd be wrong. Modular arithmetic is simply ordinary arithmetic (adding, subtracting, multiplying, dividing, exponentiating) performed on numbers whereby you reduce the result to an equivalent number based on something called the modulus of the arithmetic. What does this have to do with a clock?

Well, let's say that it's 8:00 in the morning. You have to be at an important place in 5 hours, or 1:00 in the afternoon. Without much thought, you just did the following arithmetic operation:

8 + 5 =1

Writing it like that makes if feel strange and dirty and absolutely false, but yet it's true. . . in modular arithmetic. If you were in the military, or were one of those strange non-military people who prefer "military time," or a 24-hour clock, you would say that your important destination must be arrived at at precisely 1300 hours, or 13 o'clock. Because most people use the 12-hour clock, you're likely to get a few strange stares from people when you announced it, and you'd probably have to "do the math" for other interested, mathematically illiterate bystanders.

But it's exactly because clock values repeat them self in regular intervals of twelves that clock "arithmetic" is simply modular arithmetic with a modulus (abbreviated simply as "mod") 12. Because 8 + 5 really is 13, we can rewrite the above statement in modular form as

13=1 mod 12

We actually would read this as "13 is congruent to 1 modulus 12," which is a fancy way of saying 1300 hours is 1:00pm to most of us. Now that we've established this new math, let's say that we were big dorks, and one day we decided to not only abandon telling time on a 12-hour clock, but we wanted to keep track of the elapsed hours over a one year period. Imagine the strange looks we'd get if one afternoon 4 months later later, some stranger asked us for the time and we responded with "1841 o'clock . . . . on the DOT!" Would there be anyway to figure out the answer quickly without subtracting out full days at a time? Well yes.

What we'd really want to know in this case is the answer to the question 1841=? mod 24 which would give us the current military time, or, since we know it's in the afternoon, 1841=? mod 12, which would give us the current time p.m. The answer, as it turns out, will simply be the remainder obtained when dividing the given number by the given modulus, 1841/24 or 1841/12 here.

I turns out that 24 goes into 1841 76 times with 17 left over (1841 = 24*76 + 17). So 1841=17 mod 24. That means it's 1700 hours, or 5:00p.m. Using mod 12, we see that 12 goes into 1841 153 times with, you guessed it, 5 left over. So what time is it? Time to have that drink.

It turns out that MANY numbers are equivalent to the same number for any given modulus. For example, 5, 17, 29, 41, 53, 65, 77, 89, 101, etc. are ALL equivalent to 5 mod 12. For all such numbers that are equivalent to each other, we say they are elements of the same Residue Class.

Modular arithmetic works for other things too that occur in regular intervals. Take the world of competitive skate boarding. Imagine a shredder pulling off an AMAZING "7863" rotational trick. Since 1 rotation = 360 degrees and 7863 = 21*360 + 303, we know that so the skater did 21 full rotations plus an additional 303 degrees. Stated mathematically, we can write 7863 = 303 mod 360.

Modular arithmetic also pertains to the days of the week, with a modulus, of course, of seven. Perhaps you'd be interested in knowing what day of the week it will be exactly 1,000,000 days from today. Not that you'd be around to enjoy that day, seeing how its more than 2,739 years from now, but you're such a dork that you ponder these things. How could we do THAT calculation using modular arithmetic?

Let's start with today. Today is Friday (woooo hoooo! is it 5:00 yet?). From here, it requires a little more cleverness. Since 1 million is a power of ten, we will start the computations with the following, seemingly non-intuitve equation, and try to build up the left side, simplifying the result (mod 7) as we go.

10 = 3 mod 7

Feel free to verify this by dividing 10 by seven and finding the remainder. Now in any equation, even if it's modular, we are allowed to square both sides. Here we go.

100 = 9 mod 7

From here, we'll note that 9 = 2 mod 7, so we replace it in the above equation.

100 = 2 mod 7

We can now make a big jump to a million by cubing both sides. Since 100 cubed is 1,000,000 and 2 cubed is 8, we obtain the following.

1,000,000 = 8 mod 7

Now because 8 and 1 are in the same Residue Class, they are interchangeable. Performing the interchanging act we finally reach our desired equation.

1,000,000 = 1 mod 7

So how does that tell us what day of the week 1,000,000 days from today be? Well, the equation tells us that it will be the same day of the week that ONE day from today will be. Say it with me now, "If today is Friday, then tomorrow is . . . Saturday!"

Wasn't that a lot more fun than using a calculator? (Just say "yes.") Besides, our method works for numbers that are beyond the calculating capability of your calculator. If you don't believe me, tell me quickly what day of the week it will be 10 to the 100th power(that's called a "google" baby, a 1 followed by one hundred zeros) days from now.

If you you said "Wednesday, " you're calculator is lying to you.

Feel free to get back to me on that.

Answer For The Ages

Yesterday I posed a mathematical puzzle in which I challenged you to determine the ages for three children based on some loose, esoteric clues. Due to the overwhelming lack of response, I assuming that either no one read it or I was able to sufficiently stump you.

For those of you who DID give it a go, did you determine that the woman had twins? Did you figure out which one had blue eyes? Do you now know what the census taker ate for breakfast?

Let's explore now how the census taker, after a hearty morning meal of bacon, eggs, and grits, was able to complete his inventory of the house of the enigmatic mother of three.

For the first clue, the woman said that the product of ages of her children equaled 36. If we assume that none of her children are age "zero",as I used to muse about my 6-month-old daughter, and if we also assume that we're talking about integer ages of kiddos, then there are only a small, finite number of possible ages that fit the equation. A careful attempt can yield all eight possibilities. Here they are. Check my work. Make sure they're the same as yours.


Although it would be highly unlikely that the woman would have a pair of one-year-old twins and a another 36 year-old rug rat, but that still leaves seven respectable possibilities. It's obvious why the man had to ask for the second clue. But with the extra hint that the sums of the ages totaled the address of the house across the street, how does that help us? The census take actually got to SEE the number of that house, while we were kept in the dark. Well, it turns out that we don't even need to know the number of the house!! So why have that clue at all, right? To answer that, we'll need to look at what the sums of all the above combinations yield.


Do you notice anything peculiar about the sums? You should see that all but ONE of the sums are unique. The sum of "13" actually appears twice! This means that the house across the street MUST have been house number thirteen. Duhhhhh! Right? Why's that? If it's not obvious to you, I'll explain further. This fact has nothing to do with the numbers of the mathematics of the sums AT ALL. It comes from a clue from the puzzle scenario. Remember that the census taker needed a third clue in order to conclude his detective work. Had the sum been any of the other combinations, he would have had his answer, but because 13 appears twice, he needed further clarification as to which of the remaining two options were the actual ages.

Even though the woman was put out at the man's final request, comparing his annoying unrelentingness to her eldest child, and refused to offer her assistance any more, she unwittingly provided him with all the information he needed.

Do you see it yet?

Her children now had possible ages of 1, 6, and 6 or 2, 2, and 9. Because the woman has an eldest child, we can infer that she has a pair of two-year-old twins and one, eldest child age nine!

A pretty nifty piece of mathematical detective work, huh.

Well, stop the presses. Just like Andrew Wiles's origianl proof of Fermat's Last Theorem was later found to contain a flaw that was overlooked by everyone, so too does my little scenario contain a tiny glitch. My wife immediately pointed out that the ages could be a 1-year-old and a pair of 6-year-old twins, since even twins, apparently, are born ONE AT A TIME, so that even though they are the same "age," there is an OLDER twin and still an ELDEST child.

D'0h!

Andrew Wiles eventually fixed the error in his proof, and now it stands as rock solid. I'm hoping I can reconcile this tiny inconsistency as successfully as he was able to. Perhaps I can go back to the original scenario and somehow cleverly work in the necessary condition that all ages be rounded to the nearest year, or maybe I'll just rewrite it so that the status taker never even offends the woman at all, with her simply telling him, "two, two, and nine. Goodbye."

A Puzzle For The Ages

Here's a delightful puzzle to help you get over the Wednesday hump, or maybe it will get you stuck there. Either way, delight yourself in mathematical detective work to see if you can figure it out.

A member of a census organization is going door-to-door collecting information. He comes to a house where a woman answers the door. After introducing himself, he asks her how many adults live in the house. "Just me," she replies, as the screaming and yelling of children pierce through from the living room. "I'm guessing you have children," he astutely asserts, "how many?" "I have three, and they're about to drive me out of my mind. How many of them do you want?" She asks back. "Um, ma'am, if you'll simply allow ME to ask the questions," he rudely replies. "Now if you could please tell me the ages of your three children, I'll be on my way." Feeling slighted, the woman becomes irritated and less cooperative. "Absolutely NOT," she retorts. "You're gonna have to work at it now Mr. Census man. Here's a clue. If you multiply all their ages together, you get thirty-six." She then slams the door in her face.

The census man pulls out his calculator from his pocket and punches a few numbers, then knocks on the door again. When the woman answers the door this time, she is even more irate as the sounds of obstreperous children spill out of some hidden room in the house. "Sorry to bother you again, ma'am, but could you please offer me one more hint as to your children's ages? "If it'll get you off my doorstep, you should know that the sum of their ages equals the address number on the house across the street." BAM, slammed the door.

Determined to get this information, the census taker runs across the street and peers at the number on its mailbox. He again pulls out his calculator, punches a few keys, then scratches his head a bit. With a determined temerity, he knocks on the woman's door once again. "I'm so sorry to bother you again, but I need to have just one last clue and I'll be our of your hair for good." (Screams emanated from inside the house) "You're just like my oldest," she said "unrelenting!" Then she retreated back into the house and was gone.

Now the census man may not be suave, but he's no dummy. From her last, subtle clue regarding her child's temperament, he is able to discern the ages of each of her three children. Can YOU figure it out?

Answer and explanation tomorrow. But you can't read tomorrow's answer if you haven't mentally sweated over it today.