Originally Posted By: teehjaeh57
I guess I'm curious what makes a tree a good candidate for syrup. Can one make syrup from other trees than Maple and Walnut I wonder? Cherry, persimmon, mulberry?

I just researched and found that unripe mulberries are mildly hallucinogenic!




A good candidate is a tree with a full canopy, or crown. That gives the tree more "storage" for sap. Trees growing in the woods are usually surrounded for other trees, and they compete for sunlight...making them tall, but not much crown. Maples spaced out along main street in town, or out in the open somewhere, usually have much larger crowns and are coveted by sugarmakers. Trees further up in elevation tend to receive more sunlight over a given period, and this may increase their output also.

Serious efforts thin the woods (known as a sugarbush), aggressively to allow for greater growth of desirable trees.

And yes, different species produce sap, or run, at different times. We've found that the black walnuts begin earlier, temperature wise, than the maples, but do not give as much sap. It makes sense to me, since walnuts are the last trees in our area to bud out in spring, and the first to lose their leaves come fall.

You can tap a few different species of trees and make good syrup. Obviously maple is the most well known, with Birch probably coming in second. Walnut is gaining interest on the maple forum I frequent, (yes, there really is such a place), but that may be just because I won't shut up about it. Sugar maples have the highest sugar content when compared to silver or red maples, but we don't differentiate in our woods, we tap em' all. Box elders make great syrup, and they are considered a member of the maple family.

I wonder if the hallucinogenic quality carries through into the syrup? Now THAT would probably be a good mover.

Pancakes on Sunday morning, trippin' on the couch in the afternoon....I could totally sell that.


"Forget pounds and ounces, I'm figuring displacement!"

If we accept that: MBG(+)FGSF(=)HBG(F1)
And we surmise that: BG(>)HBG(F1) while GSF(<)HBG(F1)
Would it hold true that: HBG(F1)(+)AM500(x)q.d.(=)1.5lbGRWT?
PB answer: It depends.