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View Full Version : Bosch Colt - VS or not



Alex Berkovsky
03-10-2006, 4:38 PM
I am thinking of picking up a Bosch Colt laminate trimmer, but can't figure out why would one want a VS version. If you can't use a large router bits in one anyway, why the need to lower the RPM's?

Brian Hale
03-10-2006, 5:04 PM
I'd opt for the VS model if only to help reduce burning. It'll take 1/4" bits and some may be large enough to warrant a slower speed.

Brian :)

Roger Everett
03-10-2006, 5:44 PM
I always go for the VS. There are many times when you need to cut down on spd. to avoid burning. For instance Cherry is very easy to burn when routing a profile, and calls for a slower speed. Avoiding burning is a match of router spd. and feed spd., I find it much easier to run at a slower spd. most of the time, and not have to rush the feed spd. Even quite hard woods can be burnt when routing with a small bit at high spd.. Don't forget the smaller the bit the faster is the rim spd.
Roger

Bruce Page
03-10-2006, 5:51 PM
I have the VS Colt. The 35K single speed is too fast for many bits. It was a no brainer for me.
Amazon has them for $105 with free shipping.

Alex Berkovsky
03-10-2006, 6:49 PM
I have the VS Colt. The 35K single speed is too fast for many bits. It was a no brainer for me.
Amazon has them for $105 with free shipping.Bruce,
I actually already have ordered the kit, but it's backordered until early April. I see that Amazon has lowered the price on both VS and non-VS models. I will take the advise that stick to the VS kit.

Bruce Page
03-10-2006, 6:58 PM
Bruce,
I actually already have ordered the kit, but it's backordered until early April. I see that Amazon has lowered the price on both VS and non-VS models. I will take the advise that stick to the VS kit.
Alex, I think you will be happy with the VS.

Hoa Dinh
03-10-2006, 7:01 PM
Don't forget the smaller the bit the faster is the rim spd.
Please help. I'm confused :confused:

Doug Shepard
03-10-2006, 9:21 PM
Pick a point on the outer rim of 2 bits - both a small bit and a large bit. That point has a longer distance to travel on the bigger bit to make one complete rotation. Since the rpm on the router is the same for either bit it translates into a slower speed for that point along the rim of the larger bit. Don't know if I explained that as well as it could be, but I took a shot.

Hoa Dinh
03-10-2006, 9:32 PM
Since the rpm on the router is the same for either bit it translates into a slower speed for that point along the rim of the larger bit.
The point along the rim of the larger bit travels greater distance in the same amount of time. Thus its speed is greater than that of a point along the rim of a smaller bit. This is why you need to reduce the router's speed when swinging large bits (e.g. raise panel).

The formula is:


tangential or linear speed (in./min.) = angular speed (rpm) * 3.14 * diameter (in.)

Joe Chritz
03-10-2006, 9:32 PM
Doug hit it right on.

If a bit has a circumference of 1 foot and rotates at 1 PRM there is a speed of 1 foot per minute. Make the bit with a 2 foot circumference at 1 RPM and the speed is now 2 feet per minute.

Again over simplified but the bigger the bit the faster the cutting edge is moving at a given RPM.

Math make my head hurt. :confused:

Joe

Doug Shepard
03-10-2006, 10:33 PM
Doug hit it right on.
Joe

Nope. Hoa Dinh and you both explained it better. I was on the right track but got my logic reversed. Math makes my head jurt too. I ended up with a degree in Math/Comp Science. All it means is I've forgotten more math than most people ever learn. About all I ever have occasion to use is some trig and basic algebra. Calculus is just something my pets get on their teeth according to the vet.

Clint Malone
03-10-2006, 10:40 PM
Pick a point on the outer rim of 2 bits - both a small bit and a large bit. That point has a longer distance to travel on the bigger bit to make one complete rotation. Since the rpm on the router is the same for either bit it translates into a slower speed for that point along the rim of the larger bit. Don't know if I explained that as well as it could be, but I took a shot.

You got it exactly backwards.

The idea you are talking about is linear velocity versus angular velocity. The angular velocity will be the same for both size bits. The linear velocity will be higher on the larger bit because it has farther to travel in the same period of time.