Does anyone know how to calculate the centre of percussion for a door? I've found 1 webpage which kind of explains it but it isn't great. Thanks to anyone who helps! http://orca.phys.uvic.ca/~tatum/classmechs/class8.pdf (page 4)
any centre of percussion is given by: R1*R2 = k^2 R1 = distance of hinge from centre of mass R2 = distance of centre of percussion from centre of mass. k = radius of gyration from an axis of rotation passing through the centre of mass and parallel to the given axis of rotaion the proof is given on a forum initiated some time ago.. however the thread i believe is been removed to i dont know where... therefore i am attachin the page. comming to the solution of the problem.. i think(i am not too sure) the centre of percussion of the door ( length l, breadth b ) would be given by (assuming the door is turning about an along its length ) b/2*R2 = (b^2)/12 R2 = (b^2)/6 just out of curiosity... wat use is centre of percussion for a door in engineering... correct me if i am wrong ...it would provide bad mech advantage if a handle were attached on it... i presume it would just help lowering the normal forces acting on the hinge.
Thanks very much for your help, that was very useful. I missed the lecture and have been wondering how to do it for an exam. In terms of a door its a typical question to ask where a door stop should be placed in order to minimize the reaction on the hinges, ideally 2/3rds along the length of the door and half way up it, as i now know!
thanks guys that web page is pretty useful, didn't show up after 30 mins of scouring google for some answers
hi this is the 1st time i have read this topic and am glad i did now. the reason why i didn't read this topic is becoz it didn't ring a bell. can anybody explain the formulas? i have the same problem with the other member. i don't know where to put exactly the door stopper that won't cause undue pressure on the hinges. suppose my door is 6.5 ft high and 4.0 ft wide and 2" thick. where do i put the door stopper? your help would be highly appreciated
Assuming "a" to be the length of the door, "b" to be the breadth of the door, M being the mass of the door neglecting the thickness and that the door rotates about one of its lenghts, we get the moment of inertia of the door to be I=M[(b^2)/3] therefore the square of the radius of gyration k^2=[(b^2)/3] from the theory stated in the previous discussions we have, k^2=R1*R2, where R1=b/2 (distance between centre of mass and axis of rotation) and R2 are measured from the axis of rotaion therefore we get R2=2b/3 R2 is the distance from the axis of rotation of door and half way up the door where the stopper should be ideally placed therefore for a door of the description given, where a=6 ft,b=4 ft R2 must be 2.67 (ie 2/3 rd the width of the door) feet from the axis of rotaion and ideally half the way up the height of the wall. refer figure i think door stoppers are usually designed to be fixed at the bottom of the door. so that may cause a little reaction on the hinges as the door may have a tendency for rotation about an horizontal axis write back if u need help with the math and proof of various formulae.
hi sorry i saw your P.M. first. the solution i was looking is here. that is of great help. thanx for your time and explanation. now i can place those stoppers in their appropiate places.
Thanks haditya, a superb explaination. I did get your PM mozikluv, but was unable to post a solution as this is not really my field :huh:
hi dave & haditya, guys, guess what i did with my door using this percussion method. i didn't use the formula but rather base on the hinge reaction. "think door stoppers are usually designed to be fixed at the bottom of the door. so that may cause a little reaction on the hinges as the door may have a tendency for rotation about an horizontal axis" my door has three 6" hinge. so i place an improvised moving coil sensor on top of the hinge using clay to attch the coil. using the general area to place the stopper i let the door swing and let it bump and watch the sensor indicator. i kept trying until the the 3 indicator no longer lited and bingo thats the exact spot it should be placed. why did i use that approach? that's because i don't know the weight of the door. the formula considers the weight. :wub: am so glad we have members like B) HADITYA B) am also thankful to "goodbyegti" who presented this topic.