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#67575 - 12/24/04 02:15 AM Lock Physics of a rim shot
UNCGQ Offline
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Okay, I've wondered this and found nobody with an answer that has satisified my curiosity.

I have no background in physics but I very much like knowing exactly why things do the things they do.

I want to know from somebody (who sounds halfway intelligent about this, and pref more so than me) as to why a rim shot (if played with the bead the same distance from the edge of the rim and with the same force as a normal "natural" stroke) is that much louder than said natural stroke.

My hypothesis was that since the stick is intersecting the rim and the head simultaneously, the stick is transferring more force to the head (and more importantly losing less of the force to "vibration").
So, if you're somehow still with this dumb explaination, i think of it as take the amount of force transferred back to the stick (and hand) after the moment of impact of a "normal" stroke and compare it to the amount transferred back during the rim shot and i think that the amount transferred back is much less in a rim shot and therefore the head is absorbing more of that given amount of force that has to go somewhere.

(also, fit in this somewhere in support of this argument, compare the amount natural rebound of a stick or a natural stroke and a stroke that happens to strike the rim and head at the same time (and notice that the rim shot has much much less rebound).

As always, thanks so much in advance, let the guessing begin.

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#67576 - 12/24/04 02:22 AM Re: Physics of a rim shot [Re: UNCGQ]
drumline_YOUKNOW Offline
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i think, repeat, think, that because you hit the rim and the head at the same time, the sound reverberates through the whole drum (shell and all) making it louder. but that is just a guess.
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#67577 - 12/24/04 01:33 PM Re: Physics of a rim shot [Re: drumline_YOUKNOW]
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Its all has to do with Torque (T), moment of inertia (I), and acceleration (here, I will use "@"). Ok, so Torque equals Moment of Inertia times acceleration (T=I@). The moment of Inertia of a drumstick is I believe 1/3ML^2, where M is the mass in kg, and L is the length in meters. So, if your pivot point (fulcrum of your grip) is located farther back on the stcik, you will be able to exert more Torque on the stick and produce a higher acceleration on a point on the tip of the stick (think of this, say you have a ball on a long string, and a ball on a short string. To make them have the same acceleration--defined as a change in velocity which is a vector (takes into account angluar displacement and speed)--the ball on the long string has the greater distance to travel in a shorter amount of time, and so moves faster to keep the same revolutions per second as the short stinged ball)). Think of a drumstcik in that fashion. The farther away that the center of the circle is from the end (fulcrum to the bead--in reality, what you are doing is moving the stick in a partial circle), the faster the stick must travel to reach the head in a given distance/time and the higher the angular acceleration (acceleration around in rotational motion--circles). So, now that you have that, keep it in mind becuase I believe it plays a very important role in this theory. So when you play a regualr note at a certain height, there is a constant torque being applied by your hand to overcome interia and friction in the air and resistance and all that fun stuff, and the stick hits the head with a certain force. Now, what about when you play that same note but hit a rimshot? That pivot point changes at the last moment from your hand to the rim. Suddenly, you have a much much smaller radius at which the stick moves about the pivot point but it is moving at the same linear velocity/angular acceleration as when it was moving about a larger circular path. So, recalculate moment of inertia with this new length, and you will find that I is much less then it was originally. However, you were still applying that same Torque to the stick from the farther pivot point, so if you plug those values into T=I@ and solve for @ (angular acceleration), then you will find that the acceleration about that point is much greater then it was before (which makes sense because of my previous explaination on this--larger radius, higher angular acceleration). And because Torque also equals force times the radius (T=Fr), there is a smaller radius when you hit the rim, so when you divide Torque exerted on the stick about the pivot point (the rim)by the radius for solve for F, force, you will get a bigger value for force, or simply, you hit the drum harder.

Hope that makes sense. Now, there may be a few flaws in my reasoning here, so is any Physics major or something would like to come forward and clarify a few things that I may have missed, it would be appreciated.
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#67578 - 12/24/04 01:43 PM Re: Physics of a rim shot [Re: TenorTones]
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Now why do pings have a thinner sound? Go!
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#67579 - 12/24/04 02:23 PM Re: Physics of a rim shot [Re: hyperionmsu]
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Lol, just what do you mean my thinner? Thinner compared to what? Normal shots where you hit the rim much closer to the fulcrum?
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#67580 - 12/24/04 05:20 PM Re: Physics of a rim shot [Re: TenorTones]
hyperionmsu Offline
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Quote:

Lol, just what do you mean my thinner? Thinner compared to what? Normal shots where you hit the rim much closer to the fulcrum?



Yeah, gocks have that "fatter" sound, but pings have a real quick, thin timbre.
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#67581 - 12/24/04 06:33 PM Re: Physics of a rim shot [Re: hyperionmsu]
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Oh, ok. I think I know why this is, but don't quote me on any of this because its all jsut an assumption.

I believe it has something to do with the tension on the head, mainly because you get a very similar effect with normal playing as you move from the center to the edge. See, when you play in the center, the sound waves and vibrations travel across the entire head evenly, because the tension pulls out evenly in all directions. The sound will vibrate through the whole drum and head pretty equally becuase of that equidistance from all points on the rim. So, it takes longer for the sound/vibrations to travel that whole distance (something like that) so the sound is lower, has a lower energy level, and is deeper. Now, with a ping, or playing close to the edge, the tension is not even by any means. It will be pulling more toward the edge that is closest because of that distance from the center or toher edge (think of standing in the middle of one of those spinning playgroud things vs standing on an outside edge, same basic principle). So, the sound waves are going to be pulled more toward that edge as opposed to being allowed to resonate throughout the entire drum. This will create a thinner, more high-pitched sound because the waves will have more engery, being concentrated in that small location (due to the tension pulling all the vibrations to that point) as oppsed to being dispersed through the entire drum. I dont know, its just a guess. But it makes sense to me.

Also, consider some of the things I said previously about torque and force and all that. A Ping shot will have more energy and force behind it (or it should) then a gawk, providing they started with the same initial angular velocity, because of that shorter radius (I believe I explained that before). So that might come into play a little bit when determining that pitch... actually, that would makes sense and it would fit right in. Short energy waves have the highest engery and are the brightest/most high pitched. So I dont know, maybe that does fit in with my theories?
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#67582 - 12/24/04 07:28 PM Re: Physics of a rim shot [Re: TenorTones]
kevin_fu Offline

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ping vs gock.

think of when you have a ruler off the side of the desk and you flick the ruler and move it out and in. When more of the ruler is showing, the sound is deeper/fatter and when the ruler is pulled in towards the desk so that only a small portion of the ruler is moving, the sound is tight and high pitched.

Basically, when you make a ping, there is a smaller area of the stick to absorb the vibration (that is, Sound wave vibrations).
so think of this for pings:
small portion of stick (from bead to connection of rim) -> shorter distance for vibrations to travel -> short vibration/wavelength -> higher frequency -> higher pitched sound

gocks:
large portion of stick from bead to rim contact -> longer distance for vibrations to move -> lower frequency -> fatter sound.


As for why rimshots sound different than a regular stroke...
Well I would say it is a combination of stroking 2 different surfaces. You get the timbre of the metal/rim and the timbre of the head you are playing, and it is stacatto because the vibration of the sound is caught in between the bead of the stick and the portion of the rim it is touching. The sound vibrates very little through the drum, as it it absorbed by the stick and hands.

That's basics, rudimentary physics... I by no means am an expert, but I understand a little, and this makes sense to me.
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#67583 - 12/24/04 08:25 PM Re: Physics of a rim shot [Re: kevin_fu]
TenorTones Offline
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Yeah, I think thats pretty much what I was trying to say Kevin, but you put it in much more simple terms. Anywho, about the ping vs the gok, thats what I meant for the most part.
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Here's to the crazy ones, the misfits, rebels, troublemakers, the round pegs in the square holes, the ones who see things differently. They aren't fond of rules; they have no respect for the status quo. You can quote them, disagree with them, glorify or villify them. But you can't ignore them--they change things. They push the human race forward. And while some may see them crazy, we see genius. Because the people who are crazy enough to think they can change the world, are the ones who do.

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#67584 - 12/24/04 08:27 PM Re: Physics of a rim shot [Re: TenorTones]
kevin_fu Offline

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Yea, if I repeated anything that you said, I apologize. I didnt read your posts because it was a huge paragraph and I tend to skip posts that arent broken down.
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Patuxent HS - Lusby, MD | Fall '06
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