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Speed of a falling object |
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| Nov27-10, 11:48 AM | #1 |
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Speed of a falling object
Hey Smart Guys,
At age 61, I am building a mechanical launching device (Trebuchet) and I need to know how fast my counterweight will be traveling. How fast will it be traveling after 1 inch How fast will it be traveling after 60 inch How fast will it be traveling after 120 inch Also, please provide the travel time to reach each point (1, 60 & 120 in.) The counterweight surface is flat, weighs 2000 lbs with a square surface area of 1296 Sq Inches. (3'x3') I don't do formulas very well, so plain English would be much appreciated! Thanks in Advance, Dale |
| Nov27-10, 01:21 PM | #2 |
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Force of gravity will convert to Kenetic energy mgh=1/2mv^2 solve for v (Devide both sides by m, multiply by 2 and sq rt both sides) v=sqrt/ 2gh EDIT: g=gravity at 9.8m/s and h is height in meters and v is also in meters/second When you find speed you already have distance v=d/t t=vd Multiply the speed in m/s and distance in meters to get the time |
| Nov27-10, 01:27 PM | #3 |
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There will be some energy lost to friction so the above analysis gives a high estimate. If you need better you will need to figure out a way to measure the velocity.
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| Nov27-10, 02:43 PM | #4 |
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Speed of a falling object
I'm sorry, but I don't understand these equations. Would you be kind enough to do the actual calculation. I flunked out of algebra.
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| Nov28-10, 12:46 AM | #5 |
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The answer depends on what kind of trebuchet you want to build. In addition, you need to specify things like the length, weight and shape of the beam and where the axle is in relation to both the counterweight and payload.
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| Nov28-10, 09:20 AM | #6 |
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Actually, I just want to use this information as a reference point. For now, all I need are the raw answers. Please, just disregard the Trebuchet design for now. If you are interested in this project, I would be extremely interested in providing you more detail a little later.
This is a new concept for a Trebuchet and there are many physics questions that will need to be answered. For now, just the basic math, please. I'm planning a 1 mile launch that has never been done before. |
| Nov28-10, 05:03 PM | #7 |
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http://www.google.com/search?&q=sqrt...inch%29+in+mph
You can easily change the distance to any distance you want to find. |
| Nov28-10, 05:50 PM | #8 |
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I copy/pasted that equation exactly as show into Excel and it did NOTHING.
Is '1 inch' suppose to be embedded into the equation? What value do I change. I'm am truly sorry for my ignorance. Can you give it EXACTLY as it should be put into an excel cell - for 1 in and 2 inches - then I will know how to modify the equation for any distance? the following was the last thing I tried. =sum(sqrt(2*(9.8(m/(s^2)))*1)) Obviously, I don't know what I'm doinh Thanks for your patience. Dale |
| Nov28-10, 05:54 PM | #9 |
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http://www.uni-siegen.de/fb11/nm/akt...chet_adams.pdf
This might be useful for you, in particular the model described in figure 5(c). |
| Nov28-10, 07:46 PM | #10 |
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In the absence of any design specifications the best you can do is set limits for speed and time based on a free falling counterweight.
For what it's worth, here's what you get for the numbers: 1 inch: Max speed = 2.3 feet per second, Min time = 0.072 seconds 60 inch: Max speed = 18 feet per second, Min time = 0.56 seconds 120 inch: Max speed = 25 feet per second, Min time = 0.79 seconds |
| Nov28-10, 08:48 PM | #11 |
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If you want to use excel the formulas should be: Code:
=SQRT(2 * 32.174 * 1) <- one foot, answer is 8.02 feet per second =SQRT(2 * 32.174 * 2) <- two feet, answer is 11.34 feet per second =SQRT(2 * 386.09 * 1) <- one inch, answer is 27.79 inches per second =SQRT(2 * 386.09 * 2) <- two inches, answer is 39.3 inches per second |
| Nov29-10, 03:04 AM | #12 |
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