size DOES matter

[reliefdvm]

I had purchased a box of Tat red label Hermosa. I really liked them, but in general I like the robusto size. So since all the red labels are supposed to have the same blend I planned on buying a box of the nobles when I ran out. I bought a couple while at the B&M and found that I really didn't like them as well. Just to be sure that my tongue wasn't just having an off day I smoked another Hermosa later and was amazed at how different the 2 shapes tasted. Not to say the robusto tasted bad just not as good - personal preference.

 

[SkinsFanLarry]

I agree with your thoughts completely, I have always thought robusto/corona/lancero sizes pack more flavor and strength than a churchill or double corona.

 

[njstone]

Just because they have the same "stuff" inside and out doesn't mean the actual blends are identical. Pete tends to "tweak" each vitola slightly differently soas to maximize the qualities of that particular size.

I like the Hermosa better than the Nobles as well--in general, the small the ring gauge on the Havana VI, the more I like it. I still like them all, though.

Victorias > Angeles > Hermosa > (the rest)

 

[kuzi16]

much of this comes from the wrapper to filler ratio.
wrapper to filler ratio is basically how much tobacco is filler vs. how much is wrapper. first we have to make a few minor assumptions:
1) the part of the cigar that is burning and the part that is just about to be burned is infinitely thin. so at one instant its not burning and immediately after, it is --making it two dimensional.
2) the wrapper has a width of 1/64 of an inch.

we have to make these assumptions so that we have:
a) a surface area calculation only
b) a relatively easy set of numbers to work with.

Cigar "A" has a ring gauge of 50. Cigar "B" has a ring gauge of 40. if we look at the foot of each of these cigars we will be looking at two circles within circles. (the inner circle represents the filler of each cigar). Cigar "A" has a diameter of 50/64 and that means a radius of 25/64ths. since Area = Pi times radius squared we can figure out the surface area of the foot of the cigar. so:

3.14 x (25/64 x 25/64 ) = 0.4791 (ish) inches squared.

that is the entire surface area of the foot of the cigar. we need the wrapper's surface area only. we get that by subtracting out the surface area of the filler. this is where assumption number two comes in. if the wrapper is 1/64th of an inch thick then the filler diameter must be 48/64ths of an inch or a radius of 24/64ths therefore the surface area of the filler alone is:

3.14 x (24/64 x 24/64) = 0.4415 (ish)

now we must subtract out the filler surface area from the entire area to get the surface area of the wrapper alone.

wrapper surface area = 0.4791 - 0.4415 = .0376

to get the wrapper to filler ratio you devide the wrapper surface area by the filler surface area:

0.0376/0.4415 = .08516(ish)

ok ... now for the cigar with a ring gauge of 40...

total surface area = 3.14 x (20/64 x 20/64) = 0.3066ish

filler surface area = 3.14 x (19/64 x 19/64) = 0.2767 ish

wrapper surface alone = 0.3066 - 0.2767 = .0299

Wrapper to filler ratio = 0.0299/0.2767 = 0.1080

on the smaller cigar (cigar "B") the ratio produced a larger number than on the larger cigar (cigar "A") 0.8516 is less than 0.1080

therefore smaller ring gauge cigars have a higher wrapper to filler ratio and therefore have a "wrapper oriented" flavor than the same cigar rolled with the same tobacco and rolled with a larger ring.

 

[reliefdvm]

I understand all the nuances of diameter and length. My astonishment is more that I personally could tell a difference.

 

[n.olson]

much of this comes from the wrapper to filler ratio.
wrapper to filler ratio is basically how much tobacco is filler vs. how much is wrapper. first we have to make a few minor assumptions:
1) the part of the cigar that is burning and the part that is just about to be burned is infinitely thin. so at one instant its not burning and immediately after, it is --making it two dimensional.
2) the wrapper has a width of 1/64 of an inch.

we have to make these assumptions so that we have:
a) a surface area calculation only
b) a relatively easy set of numbers to work with.

Cigar "A" has a ring gauge of 50. Cigar "B" has a ring gauge of 40. if we look at the foot of each of these cigars we will be looking at two circles within circles. (the inner circle represents the filler of each cigar). Cigar "A" has a diameter of 50/64 and that means a radius of 25/64ths. since Area = Pi times radius squared we can figure out the surface area of the foot of the cigar. so:

3.14 x (25/64 x 25/64 ) = 0.4791 (ish) inches squared.

that is the entire surface area of the foot of the cigar. we need the wrapper's surface area only. we get that by subtracting out the surface area of the filler. this is where assumption number two comes in. if the wrapper is 1/64th of an inch thick then the filler diameter must be 48/64ths of an inch or a radius of 24/64ths therefore the surface area of the filler alone is:

3.14 x (24/64 x 24/64) = 0.4415 (ish)

now we must subtract out the filler surface area from the entire area to get the surface area of the wrapper alone.

wrapper surface area = 0.4791 - 0.4415 = .0376

to get the wrapper to filler ratio you devide the wrapper surface area by the filler surface area:

0.0376/0.4415 = .08516(ish)

ok ... now for the cigar with a ring gauge of 40...

total surface area = 3.14 x (20/64 x 20/64) = 0.3066ish

filler surface area = 3.14 x (19/64 x 19/64) = 0.2767 ish

wrapper surface alone = 0.3066 - 0.2767 = .0299

Wrapper to filler ratio = 0.0299/0.2767 = 0.1080

on the smaller cigar (cigar "B") the ratio produced a larger number than on the larger cigar (cigar "A") 0.8516 is less than 0.1080

therefore smaller ring gauge cigars have a higher wrapper to filler ratio and therefore have a "wrapper oriented" flavor than the same cigar rolled with the same tobacco and rolled with a larger ring.

wow that is really cool. i've never seen why i like smaller cigars explained mathmatically. kudos to you man

 

[kuzi16]

I understand all the nuances of diameter and length. My astonishment is more that I personally could tell a difference.

in that case...

quit doubting your taste buds. they know what they are doing.

 

[El Baton]

much of this comes from the wrapper to filler ratio.
wrapper to filler ratio is basically how much tobacco is filler vs. how much is wrapper. first we have to make a few minor assumptions:
1) the part of the cigar that is burning and the part that is just about to be burned is infinitely thin. so at one instant its not burning and immediately after, it is --making it two dimensional.
2) the wrapper has a width of 1/64 of an inch.

we have to make these assumptions so that we have:
a) a surface area calculation only
b) a relatively easy set of numbers to work with.

Cigar "A" has a ring gauge of 50. Cigar "B" has a ring gauge of 40. if we look at the foot of each of these cigars we will be looking at two circles within circles. (the inner circle represents the filler of each cigar). Cigar "A" has a diameter of 50/64 and that means a radius of 25/64ths. since Area = Pi times radius squared we can figure out the surface area of the foot of the cigar. so:

3.14 x (25/64 x 25/64 ) = 0.4791 (ish) inches squared.

that is the entire surface area of the foot of the cigar. we need the wrapper's surface area only. we get that by subtracting out the surface area of the filler. this is where assumption number two comes in. if the wrapper is 1/64th of an inch thick then the filler diameter must be 48/64ths of an inch or a radius of 24/64ths therefore the surface area of the filler alone is:

3.14 x (24/64 x 24/64) = 0.4415 (ish)

now we must subtract out the filler surface area from the entire area to get the surface area of the wrapper alone.

wrapper surface area = 0.4791 - 0.4415 = .0376

to get the wrapper to filler ratio you devide the wrapper surface area by the filler surface area:

0.0376/0.4415 = .08516(ish)

ok ... now for the cigar with a ring gauge of 40...

total surface area = 3.14 x (20/64 x 20/64) = 0.3066ish

filler surface area = 3.14 x (19/64 x 19/64) = 0.2767 ish

wrapper surface alone = 0.3066 - 0.2767 = .0299

Wrapper to filler ratio = 0.0299/0.2767 = 0.1080

on the smaller cigar (cigar "B") the ratio produced a larger number than on the larger cigar (cigar "A") 0.8516 is less than 0.1080

therefore smaller ring gauge cigars have a higher wrapper to filler ratio and therefore have a "wrapper oriented" flavor than the same cigar rolled with the same tobacco and rolled with a larger ring.

I never thought that any math besides general accounting would come in handy when dealing with smokes. I am going to have to try and make application of this newfound knowledge!

Thanks for sharing.