y=mx+b is about as much as I remember from math. But here is what I remember from reality:
Last summer a sax friend and I took about a 5 hour trip to the Woodwind and Brasswind factory so he could buy a mouthpiece and I could play some horns. I spent about an hour each on a gold and a yellow brass 103 with mechanical linkage. I noticed absolutely nothing out of the ordinary, uncomfortable, inefficient, slow, or otherwise noteworthy aspects. It simply played like any other valves that I've ever played previously. I also played a 103 with mechanical linkage at a workshop last year and had an indentical experience. It's a horn. It has a valves. You push them. Etc. I wouldn't hesitate to buy a horn with this linkage based on my experiences with it. I did notice how uncomfortable my Schmidt wrap is and that I want to dump it every time I touch another non-schmidt horn! Chris --- [EMAIL PROTECTED] wrote: > Schmid doesn't need marketing. That's solid physics for you and I can prove > it. Now a mechanical linkage or string produces torque on the valve. Torque > is > the following equation: > > T = F * r * sin (theta) > > Where: > > T = Torque > F = Force > D = Radius of where torque is being applied to > theta = angle between force and the angle of radial line > > Now with a force of say 1 Newtons (heavy fingers) and a radius of about .01m > you do NOT apply uniform torque on the valve with miniballs and here is why. > > When an open valve is depressed initially it is at 135 degrees to the center. > > This is in turn 45 degrees from the radial line. What would that Torque be? > It would be > > (1)*(.01)*sin(45) = .0071 Newton-meters. > > At the midpoint the line would be an even 90 degrees yielding > > (1)*(.01)*sin(90) = .01 Newton-meters > > At the end we would have an angle theta of 135 degrees > > (1)*(.01)*sin(135) = .0071 Newton-meters > > Now with strings what is the torque? Well the string always applies toque at > 90 degrees so it would always be: > > > (1)*(.01)*sin(90) = .01 Newton-meters. > > I rest my case. With strings you always have uniform torque and always the > maximum torque allowable. With mini-balls you always have variable torque and > > it'll never be but once per stroke at it's maximum torque. > > You could throw spring strength or viscosity of oil but that will not effect > the amount of torque needed to push the valve or the fact that mini-balls are > > uneven torque. > > Why would Schmid need to market this? He would stand to make money no matter > what you used, strings or mini-balls. Additionally he could make them however > > you wanted. > > What a silly comment... marketing indeed. > > I've tried mini-balls before, and they're such a pain to reassemble and take > apart. Like I said, I can restring a horn in a very short amount of time. > Once > you cut the proper length of string, tie the end, melt the other end (which > is a total of two minutes tops) re-stringing is a walk in the park. I can do > 4 > valves in about one-two minutes flat. Can you do that with mini-balls? > > -William > > > In a message dated 12/6/2003 1:55:09 PM Pacific Standard Time, [EMAIL PROTECTED] > writes: > > > Yeah, I've read it. The memory of it makes me smile everytime I wiggle > > my wonderfully fast, light, direct mini-ball operated titanium valves. > > > > You have to hand it to Englebert; not only does he make good horns, > > he's also a top marketeer. > > > > Stick to your strings brother ;o) > > > > All the best, > > Tom > > _______________________________________________ > post: [EMAIL PROTECTED] > set your options at http://music.memphis.edu/mailman/options/horn/tedesccj%40yahoo.com __________________________________ Do you Yahoo!? New Yahoo! Photos - easier uploading and sharing. http://photos.yahoo.com/ _______________________________________________ post: [EMAIL PROTECTED] set your options at http://music.memphis.edu/mailman/options/horn/archive%40jab.org
