How does sound technicians use geometry? I NEED AN ANSWER BEFORE 4-21-06.?
436127
depends on what you want to do with the sound, but generally they use geometry to create sound friendly environments. (feedback reduction,  good resonance reverb, noise reduction, and others)
596509
1) Flat and hard surface. By flat I mean the microtexture, so the sound doesn't get scattered, convex like a parabole is possible if the sound originates from the center of the parabole.\n2) There has to be a large distance between wall and origin for accurate meassurement. Or you would need very fast measurementdevices.\n3) Since sound is carried by a medium the speed of that medium has to be taken into account. If you measure both ways then the speed of the medium, assuming that it didn't change directions, on one measurement has an negative effect and on the otherone an equal positive effect. By taking the average this is levelled out.\nEach measurement involves two reaction times; the reactiontime to the lightflash and the reactiontime to the sound. If both are equal, the measurement would be exact. Assuming that there is no significant difference for the reaction to sound and the one to light the error made by difference in both reactiontimes can be both ways. So by increasing the number of measurements and taking the average you increase the accuratesse of your measurement. If the error in measurement due to reactiontimes would allways be one way (allways to high or allways to low soundspeed) then taking the average would have a negative effect on the accuratesse, because the measurement with the smallest error would allways be the most accurate.
587954
Measurements (In distance I'm assuming) would give you volume.\n\nMass / Volume = Density\n\nIf you have volume & density, you can rewrite the equation:\n\nMass = Volume * Density
207128
4th dimension is easy, since it's time.  Just imagine something moving in a 3d volume and you're imagining four dimensions, x,y,z and t.\n\nYou can imagine an object moving in five dimensions or more pretty easily, too.\n\nImagine two three dimensional volumes, one is x, y, z, the other is x, y, w.   Now imagine a point moving along a line in each volume, such that if the point is folowing some equation f(x,y,z,w,t) the x,y,z volume is tracking its coordinates in x,y,z and the x,y,w is tracking its coordiates over time in x,y,w.  You need to keep them coordinated in your head, which can be tricky, but if you get good at it you can also imagine rotating one or both 3d volume about any given axis and keep tracking the point.  So, you can rotate x,y,z into x,y,w, or z,y,w, or whatever, while still imagining the motion of the point.  And, there you are, imagining five dimensions simultaneously.   You can do this for any number of dimensions, keeping track of them via rotation of basis and/or multiple 3d volumes, but it gets a little tricky, sort of like keeping two or three songs running in your head at the same time, but with practice it can be done.   I can successfully keep track of up to eight dimensions this way with ease, any more than that takes serious concentration.  It can be trickier if using non-euclidian volumes, but in some cases, like the small curled dimensions used in string theory, it can actually be easier than dealing with many large dimensions.
971127
The sound intensity due to a point source radiating equally in all directions (spherically symmetric case) is proportional to 1/r^2, where r is the distance from the source of the sound:\n\nI = k/r^2, where k is a constant of proportionality.\n\nIf we know the intensity (I1) at one distance (r1), then the intensity (I2) at a second distance (r2) is given by:\n\nI2 = I1*(r1/r2)^2\n\nIn Q1, we are told that the sound intensity at 1 m is 10^-5 W/m^2, and we want to know the intensity at 5 m.  Plugging these values into the above equation, we have:\n\nI2 = 10^-5 W/m^2 * (1m/5m)^2 = (10^-5 W/m^2)/25 = 4*10^-7 W/m^2\n\nThe intensity of a sound wave is proportional to the square of the amplitude of the displacement of a particle in the medium through which the sound is propagating:\n\nI = m*x^2,\n\nwhere m is a constant of proportionality, and x is the amplitude of the displacement.  Plugging this into the equation above that relates intensity and distance, we have that:\n\n(x2)^2 = (x1)^2 * (r1/r2)^2\nx2 = x1*r1/r2\n\nThe amplitude of the displacement is inversely proportional to 1/r.\n\nFor Q1 we are told that x1  = 70 microns at 1 meter, and are asked what x2 is at 5 meters.  Plugging these values into the above equation, we have:\n\nx2 = 70 microns * 1/5 = 14 microns\n\nFor Q2, we are told that x1 = 10^-7 m at r1 = 3 m, and are asked what the amplitude is at 5 m.  Again plugging these into the above equation, we have:\n\nX2 = 10^-7 m * (3/5) = 6*10^-8 m
626555
http://www.ilstu.edu/~dlnelso/Geo201/drummer.htm
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
427025
Read this \nhttp://baseportal.com/baseportal/goodsounds/howto
245219
i,m a roofer i use basic geometry to figure areas of roof in order to buy material. a little simple addition and i can figuer out a price for the customer... then a little subtraction to get the job never hurts.... on occation i need a bit of basic algebra to make corections in sales tax.\n at home simple math is needed to balance check books and figure the hosehold buget. out in the yard its back to geometry to calculate home much seed and fertilizer to buy.\n on the computor things get a little "hairy"  when it comes to responding to question on "yahoo answers" category ; math.. but i try...
108454
#include "ALLOC.H"\n#include "DOS.H"\n#include "CONIO.H"\n#include "STDIO.H"\n\n\nvoid main()\n{\n char *name;\n printf(" Enter the file name...:");\n scanf("%s",name);\nplaywav(name,1.0);\n\n}\n\n\nvoid playwav(char wavefile[14],float delaytime);\nstruct WaveData {\n  unsigned int SoundLength, Frequency;\n  char *Sample;\n};\n\n\nstruct HeaderType {\n  long         RIFF;      //RIFF header\n  char         NI1 [18];  //not important\n\n  unsigned int Channels;  //channels 1 = mono; 2 = stereo\n  long         Frequency; //sample frequency\n  char         NI2 [6];   //not important\n  char         BitRes;    //bit resolution 8/16 bit\n  char         NI3 [12];  //not important\n} Header;\n\n\n\n\n\nstruct WaveData Voice;         //Pointer to wave file\n\n\nunsigned int    Base;          //Sound Blaster base address\n\n\nchar            WaveFile [25]; //File name for the wave file to be played\n\n\n/*************************************************************************\n***\n** Checks to see if a Sound Blaster exists at a given address, returns\n**\n** true if Sound Blaster found, false if not.\n**\n**************************************************************************\n**/\nint ResetDSP(unsigned int Test)\n{\n  //Reset the DSP\n  outportb (Test + 0x6, 1);\n  delay(10);\n  outportb (Test + 0x6, 0);\n  delay(10);\n  //Check if (reset was succesfull\n  if ((inportb(Test + 0xE) & 0x80 == 0x80) && (inportb(Test + 0xA) ==\n0xAA))\n  {\n    //DSP was found\n    Base = Test;\n    return (1);\n  }\n  else\n    //No DSP was found\n    return (0);\n}\n\n\n/*************************************************************************\n***\n** Send a byte to the DSP (Digital Signal Processor) on the Sound Blaster\n**\n**************************************************************************\n**/\nvoid WriteDSP(unsigned char Value)\n{\n  //Wait for the DSP to be ready to accept data\n  while ((inportb(Base + 0xC) & 0x80) == 0x80);\n  //Send byte\n\n  outportb (Base + 0xC, Value);\n}\n\n\n/*************************************************************************\n***\n** Plays a part of the memory\n**\n**************************************************************************\n**/\nvoid PlayBack (struct WaveData *Wave)\n{\n  long          LinearAddress;\n  unsigned int  Page, OffSet;\n  unsigned char TimeConstant;\n\n\n  TimeConstant = (65536 - (256000000 / Wave->Frequency)) >> 8;\n  WriteDSP(0x40);                  //DSP-command 40h - Set sample frequency\n  WriteDSP(TimeConstant);          //Write time constant\n\n\n  //Convert pointer to linear address\n  LinearAddress = FP_SEG (Wave->Sample);\n  LinearAddress = (LinearAddress << 4) + FP_OFF (Wave->Sample);\n  Page = LinearAddress >> 16;      //Calculate page\n  OffSet = LinearAddress & 0xFFFF; //Calculate offset in the page\n  /*\n      Note - this procedure only works with DMA channel 1\n  */\n  outportb (0x0A, 5);              //Mask DMA channel 1\n  outportb (0x0C, 0);              //Clear byte pointer\n  outportb (0x0B, 0x49);           //Set mode\n  /*\n      The mode consists of the following:\n      0x49 = binary 01 00 10 01\n                    |  |  |  |\n                    |  |  |  +- DMA channel 01\n                    |  |  +---- Read operation (the DSP reads from memory)\n                    |  +------- Single cycle mode\n                    +---------- Block mode\n  */\n\n\n  outportb (0x02, OffSet & 0x100); //Write the offset to the DMA controller\n  outportb (0x02, OffSet >> 8);\n\n\n  outportb (0x83, Page);           //Write the page to the DMA controller\n\n\n  outportb (0x03, Wave->SoundLength & 0x100);\n  outportb (0x03, Wave->SoundLength >> 8);\n\n\n  outportb (0x0A, 1);              //Unmask DMA channel\n\n\n\n  WriteDSP(0x14);                  // DSP-command 14h - Single cycle playback\n  WriteDSP(Wave->SoundLength & 0xFF);\n  WriteDSP(Wave->SoundLength
953022
There sure is!  \n\nFirst, recall this: given a tetrahedron, choose any of its faces to be its base.  Now find the area of the base (remember all the faces are triangles).  Next, find the perpendicular distance from the fourth vertex of the tetrahedron to the plane that contains the base.  This distance is the height of the tetrahedron.  Now the volume of the tetrahedron is 1/3 X area of its base X its height.  \n\nNow, using three-dimensional Cartesian co-ordinates, consider the tetrahedron ABCD where A=(0,0,0), B=(27,0,0), C=(0,16,0), and D=(0,0,3).  Face ABC is a right triangle in the x-y plane.  The two perpendicular legs of ABC have lengths 16 and 27, so the area of face ABC is 1/2 X 16 X 27 = 216.  The distance of D from the x-y plane is 3, so the volume of ABCD is 1/3 X 3 X 216 = 216.  \n\nLet E=(8,2,0).  E lies in the x-y plane in the interior of ABC.  Join E to A, B, C, and D with line segments.  Triangle ABC is subdivided into triangles ABE, BCE, and ACE, and tetrahedron ABCD is subdivided into tetrahedra ABED, BCED, and ACED, and the heights of these three sub-tetrahedra are also 3.\n\nThe distance from E to the x-axis is 2, so triangle ABE has \narea = 1/2 X 2 X 27 = 27 and tetrahedron ABED has \nvolume = 1/3 X 3 X 27 = 27.  \n\nThe distance from e to the y-axis is 8, so triangle BCE has \narea = 1/2 X 8 X 16 = 64 and tetrahedron BCED has \nvolume = 1/3 X 3 X 64 = 64.  \n\nArea of ABC = 216, area of ABE = 27, and area of BCE = 64, so triangle ACE has area = 216 - 27 - 64 = 125 and tetrahedron ACED has area 1/3 X 3 X 125 = 125.  \n\nThat does it!
869849
Do you mean how does a sound track produce the word in unison with the lips pronouncing the same word?\n\nIf this is so its works in accordance to the old cinematic standard which was call 'fmpts'? or something where each frame has a minute, second, tenth of a second, interval mark on it.\nThe sound track is then recorded on the side to line up with the image.
880477
hi\nwe call the distance between tow points as geometry.\nu can name this distance as measurment.\nor sth like this
10261
All of these people worked on models to aid in acoustical design of ampitheatres (or any space where clearly hearing sound is important).  The three equations that you listed all take the same general form, and the latter two are subsequent improvements in Sabine's original work.\n\n"W.C. Sabine carried out a considerable amount of research in this area and arrived at an empirical relationship between the volume of an auditorium, the amount of absorptive material within it and a quantity which he called the Reverberation Time (RT). \n\nAs defined by Sabine, the RT is the time taken for a continuous sound within a room to decay by 60 dB after being abruptly switched off and is given by:\n\nRT = (0.161V) / A\n\nwhere V is the volume of the enclosure (m³) and A is the total absorption within the enclosure (sabine)."\n\n\n"Several different approaches have been used to derive equations which give values of reverberation time in better agreement with measured results from less reverberant rooms. One of these, the Norris-Eyring formula, assumes an intermittent decay with the arrival of fewer and fewer reflections. This gives the following formula:"\n\nRT = (0.161V) / (-S ln(1-a))\n\n"where S is the total surface area (m²) and a becomes the average absorption coefficient."\n\n"When the materials of a room have a wide variety of absorption coefficients, the best predictions are obtained by the Millington-Sette equation. This is simply a matter of substituting an effective absorption coefficient ae = -ln(1-ai) into Sabine's equation to give:"\n\nRT = (0.161V) / (sum(-Si ln (1-Ai))\n\nwhere Si is the surface area of the ith material, and Ai is its actual absorption coefficient.
1180581
When I was first trying to do trigonometry problems like this, I found it helpful to draw a picture.\n\nDraw a horizontal line to represent the ground, and then put a vertical line to represent the radio tower.  We want to know the height of the tower, so label it "x".  The angle between the tower and the ground is 90 degrees, so label it too.\n\nNow draw a dot (or a little stick figure with a protracter if you feel like it) to the side of the tower, on the ground.  Label the line between the tower base and the dot "50", as it's 50 meters from the tower.\n\nNow draw a line between the dot and the top of the tower, representing the person sighting to the top and measuring the angle.  Lable the angle between this sight line and the tower top "67 degrees".\n\nYou now have a drawing of the triangle with two angles and one adjacent side known.  This should be enough to get you started picking the right trig function to apply here.\n\nIf that's not enough to get going, here's another hint: Remember the unit circle with a triangle and the sides labeled with "sin(theta)", "tan(theta)" etc?  Are there any similarities to your tower drawing?  I always found being able to draw that unit circle on my scratch paper at the beginning of a problem set or test helped immensely.\n\nGood luck.
78504
geometry is the link.
56173
To make them interfere destructively, they will have to be half a wavelength out of sync.\n\n\nSpeed = 340 m/s\nThat means in one second, the soundwave travels 340 meters.\n\nFrequency = 1960 Hz\nThat means in one second, 1960 cycles are produced.\n\n\nCombine those two facts, and you can calculate the wavelength.\n\nIn one second, there will be 1960 cycles in 340 meters. So wavelength = 340 meters / 1960 cycles.\n\nWavelength = 0.1734m\n\nHalf of wavelength = 0.0867m