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# KW of laser required to cut through a giant redwood tree in one second?

#### twinbee

##### New member
I know that high powered lasers are used to cut metal, and how popular high power lasers are around here. That's all well and good, but how many kilowatts of laser power would be required to cut a swathe in one second through say, a giant redwood tree similar to this one:

http://images.travelpod.com/tw_slid...ood-tree-in-mariposa-yosemite-los-angeles.jpg

Not just a thin pin-hole through the trunk, but slicing the whole trunk completely, like a light saber, so the tree would topple. In one second too. Assume a 1 millimetre diameter/thickness for the laser.

Impractical of course, but just curious.

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12500kW

#### twinbee

##### New member
12 megawatt. Wow - do you think that's fairly accurate? I'm assuming a 1mm laser width/point. I bet the goggles needed for that would be bulky. Nevermind the eyes, even standing by watching would be a burn risk to the rest of the body maybe?

For comparison, that military defense laser in the news recently was 'only' 50kW and has a presumably much higher laser diameter.

NIF (National Ignition Facility) on the other hand is much more powerful at 500 terawatt, but that's only for a few picoseconds.

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#### weeba2kv

##### New member
wrong , it takes 12501kW

#### daawood123

##### New member
wrong, it takes 12500 but depends on whether it has rained in the past week or not, if so, then 12501kw is needed

#### Zeebit

##### New member
No. It takes 1.21 Gigawatts.

#### twinbee

##### New member
Hmm, those numbers are quite different. Perhaps you and/or phenol could show some math to back up the claims

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#### phenol

##### Member
The mathematical justification behind that number is of course mind-dazzling, as is the task of putting it in written form here. That figure is based on empirical probabilistic estimations carefully considering the average rainfall over the last 1200 years and the average optical power it would take to disperse the green peace activists gathered around the tree target.

#### jArdi

##### New member
The mathematical justification behind that number is of course mind-dazzling, as is the task of putting it in written form here. That figure is based on empirical probabilistic estimations carefully considering the average rainfall over the last 1200 years and the average optical power it would take to disperse the green peace activists gathered around the tree target.
where do you get the math for that?

#### HitShane

##### New member
The mathematical justification behind that number is of course mind-dazzling, as is the task of putting it in written form here. That figure is based on empirical probabilistic estimations carefully considering the average rainfall over the last 1200 years and the average optical power it would take to disperse the green peace activists gathered around the tree target.
LOL at the tree huggers being cut in half to save a tree...

#### pschlosser

##### New member
The carbon content within the chemical makeup of the tree poses the biggest hurdle to slicing it like butter. As heat is applied by a beam, the oxygen and hydrogen are released from those starchy rings as water vapor. There is some loss of carbon to carbon dioxide, but much of the carbon remains behind and creates a charred insulator. Carbon has a seriously high boiling point.

It is anybody's guess how much output power it would take to slice a 40-foot diameter redwood tree. And that's if you aim high. It could be close to 60-feet at the base, and we all know how dense a redwood burl table is, particularly if you've smashed your shin or fallen against one at a party.

With that said, I imagine it's easier than slicing through 40-60 feet of steel. Even SciFi exaggeration in Star Wars Episode I took quite some time for light sabers to cut through the 5-10 foot thick blast doors leading to the bridge of a ship. Perhaps they took so long, because they were battery operated. :crackup:

#### Things

##### New member
There is no maths here, you can't calculate how fast a laser is going to cut something easily, especially something as inconsistent as the trunk of a tree.

"How long is this piece of string?"

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#### BCGanja

##### New member
The ones on the drones would do it.

#### RedDart

##### New member
Get a 5mW cheapie green ebay laser! I hear they're usually overspec :crackup:

#### twinbee

##### New member
Okay, the recent Slashdot thread "Is there a limit to a laser's energy" got me inspired to figure this one out once and for all. Here's the math, and it's on this tree - the 'General Sherman' - the biggest tree in the world:
General Sherman, the biggest tree in the world

Here's what a laser can manage in terms of cutting capability (on walnut hardwood - let's hope a Redwood's wood is similar, at least in the same order of magnitude):
' Tree felling with lasers: a big idea in 1965 - Laser Focus World

24.1m = circumference of the General Sherman tree (girth)
24.1m/pi = 7.67m diameter
3.84^2*pi = 46.32m^2 area
46.32m^2 / 7.67m = 6.04m average thickness if tree slice was a rectangle instead of circle
7.67m * 6.04m ' equivalent tree dimensions as rectangle for simplicity

Lasers can cut a wooden gear from 0.5 - in. - thick walnut hardwood at a speed of 85 in. per minute using power levels of 400 W (bottom)

85/60 = 1.417inches per second
1.417inches in metres = 0.036m per second at 0.5inch thickness
0.5inch in metres = 0.0127m thickness for laser described
6.04m / 0.0127m ..... General Sherman tree = 476x thicker than above laser tool can manage
0.036m /476 ..... Therefore instead of 0.036m/s, the 400W laser could only manage 0.0000756m/s on the tree
7.67m / 0.0000756m = 101,455x more powerful laser needed to cover 7.67m in one second
400watt*101455 = 40,582,000watt
40,582,000watt as megawatt = 40.6 megawatt

So the final conclusion.......:
40 megawatt of laser power required to slice through (and topple) a giant redwood tree in one second!

Phenol was actually pretty close with his estimate!! (off by a factor ~4).

If anyone can point to flaws in the math, please let me know!

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