Decay images

[Paul Melarange]

Thanks David. Very interesting indeed. 11% on the compression side of the stem is not a lot but as you say has been heavily reduced. All we need now is for someone to do a PhD on the t/R ratio of reduced trees and come up with a new safety margin. I'm not putting my hand up though😉

 

Jake Andrews

 

No need, the information is already out there. It is explained by the 'tree statics model'.

[Jake Andrews]

No need, the information is already out there. It is explained by the 'tree statics model'.

Thanks Paul. I'll look into that.

 

Jake Andrews

[Paul Melarange]

Hi Jake

 

Here is a paper which might be of interest.

 

Best regards

 

Paul

How Hollow May a Tree Be_Neue Landschaft (11) 1996_Tuin & Landschap (15) 1998.pdf

[Alinicoll]

Just thinking out loud here (or thinking online as it were) but it seems to me there is quite a gap in the knowledge of the transition between a hollow stem where a t/R can be calculated and a tree where it has started to separate into semi autonomous functional units.

 

As the functional units separate then the residual wall at many points becomes 0, yet the tree could be perfectly stable.

[Nick Connell]

Can't seem to open the link, anyone else managed it?

[Jake Andrews]

I can access it Nick.

 

PM me your email address and ill send it over if you like. Im just in the process of trawling through it now.

[Jake Andrews]

Just thinking out loud here (or thinking online as it were) but it seems to me there is quite a gap in the knowledge of the transition between a hollow stem where a t/R can be calculated and a tree where it has started to separate into semi autonomous functional units.

 

As the functional units separate then the residual wall at many points becomes 0, yet the tree could be perfectly stable.

 

Would this not be calculated as an open cavity?

[Alinicoll]

Would this not be calculated as an open cavity?

 

That might depend how many units there were.

[Jake Andrews]

That might depend how many units there were.

hmmmmmmmm. see your point now.

[Paul Melarange]

Just thinking out loud here (or thinking online as it were) but it seems to me there is quite a gap in the knowledge of the transition between a hollow stem where a t/R can be calculated and a tree where it has started to separate into semi autonomous functional units.

 

As the functional units separate then the residual wall at many points becomes 0, yet the tree could be perfectly stable.

 

Hi Alasadir,

 

I hope all is good with you.

 

You are quite right regarding the issue of semi autonomous units. The tree statics model is based on the wood failure in compression (wood is weakest in compression).

 

Strength loss can be calculated using section modulus (the distribution of the material) where gaps (open cavities) in the residual wall are incorporated into the calculation.

 

However, the model does not allow for sheer failure or torsional failure. It is worth noting that a lot more force is required for a wood to fail in sheer as opposed to compression. That said, the risk of sheer is increased when the residual wall is below t/R 0.1 (10%) because the stem/cylinder can separate into semi autonomous units which start moving independently of each other e.g the tension side vs the compression side.

 

I hope this makes sense?