In a pear tree?

A little too early for the pear tree.

Hmm . . . I think I need to have a conference with Williams and Bartlett. Perhaps we'll form a concord, we may even find el dorado, but Gorham from the highland has a strange aurora and may influence Clapp so that he's no longer favourite.

LOL

djy,you are from a strong apple region. Last time I was there, in the autumn, there was a large exhibition at Berrington Hall (north of Leominster) of different apple varieties - all grown from around Herefordshire. Do they still do that?

Not sure if it's an annual event, but there is the occasional show where growers try to promote old varieties.

Sadly a great many old orchards are now being ripped up, it's being cheaper to import from France. This, in turn, is one of the reasons I no longer drink Bulmers Cider, it just doesn't taste the same as it used to.


PS
For those completely loosing the plot;

• Conference


• Williams


• Bartlett


• Concord


• El Dorado


• Gorham


• Highland


• Clapp's Favourite

Are all varieties of pear.

djy's got enough posts for 4 people that's for sure, and Anthony has enough for 8.

Or just over engineered . . .



. . . to ensure their survival?

only 8? I need to try harder

I meant for the 8 people's entire lifetime...:)

Now if we divide Anthony's total by eight, and the result of that by five and a half, we get an annual average that is almost identical to that of Marky's. Ergo, one Anthony is approximately equal to eight Marky's.

Too complicated for me... :D

Marky_Mark896

Posts = 1257
Member = 3 (years)
Annual average = 1257/3 = 419


Anthony

Posts = 18566
Member 5.5 (years)
Annual average = 18566/5.5 = 3376
Enough entires for 8 people = 3376/8 = 422


Conclusion

One Anthony approximately equals eight Marky_Mark896's

so in other words you agree with his 8 :)

yep...good job :)

Not necessarily.

Since we have no definition of a "standard" member we have no means of reference and hence cannot quantify the figure 8 in terms of such. Indeed, it has just been assumed that you are way above standard whereas in fact you could be the standard; and if that were the case the 8 argument is immediately rendered null and void. Therefore, until such time as a "standard" member is defined all I can say, with any degree of certainty, is that one Anthony is approximately equal to eight Marky's.