h a l f b a k e r yNaturally, seismology provides the answer.
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I was impressed, as apparently were they, that the Mythbusters could build hovercraft strong enough to carry a person using diesel leafblowers for propulsion. As I recall 6 were required. I propose that the "craft" piece be discarded and that the blowers be used to generate lift for a personal jet
pack. Blowers would be positioned along the shoulder, behind the legs and on each hand, the latter being useful to steer. Perhaps an exoskeleton-like device might be useful to affix blowers. If the blower were unable to generate adequate thrust to produce lift, additional blowers would be added.
The Mythbusters also tried to make a jetpack, I understand, but I presume they went about it wrong, as they had not asked my advice first.
Williams_X-Jet
http://en.wikipedia...wiki/Williams_X-Jet
These things are jus soooo cute, and great fun too. [8th of 7, Aug 17 2009]
Specific impulse
http://en.wikipedia...ki/Specific_impulse
[8th of 7, Aug 20 2009]
[link]
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It's a problem with power to weight ratios those small 2-strokes don't have enough grunt. |
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For a ducted-fan man-lifting device, look up the "flying dustbin" that was developed in the 70's as a one-man air vehicle. It used the same gas turbine powerplant as the Tomahawk. Cute, but too expensive and short range for general issue. |
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C'mon, bungston, at least do the math. How many leafblowers would you need to lift, say, a leafblower? |
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Well, one, but that's the problem - no disposable lift for payload. |
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With a LOT of tinkering, a leaf blower can be made to hover in the ground-effect region, but that's about it, but they guzzle fuel and the engine burns out fast. |
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Hovercraft need very little power to float on a smooth surface -- there even exist videos of pedal-powered versions. |
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Jet packs, on the other hand, need a huge power to weight ratio to work; vastly more power than can be produced by a leafblow. |
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You couldn't get off the ground, of course, but by counteracting the weight of the leaf blowers, you could strap dozens of running leaf blowers to yourself and totter down the sidewalk. Perfect for the door to door leaf blower salesman. |
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Hmmm ... maybe not true hovering flight ..... but maybe some form of "assisted jumping" ? If the powepack could reduce the wearer's weight by 50 - 70%, for the same muscle power and effort they could travel a lot further .... |
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I think Jutta's point was that home leaf blowers can't even support themselves, so doesn't matter how many you use, you'll just get heavier. |
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/C'mon, bungston, at least do the math./ |
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I have tried. I think I am not up to it. |
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From wikipedia a watt:
1W = 1Js-1 = 1kgm2s-3 = 1Nms-1 |
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or (mine) 1 kg * 1 m^2 / 1 s^3 |
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Acceleration provided must equal gravity 9.8 m/s^2 |
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The rocketeer wants to ascend 1 meter in 1 second |
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20 kg * 9.8 m/s^2 * 1 m / 1 s = |
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686 kg * m^2 / s ^ 3 = 686 watts. |
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But this is pitifully small; less than 1 hp. And it took me a long time I am ashamed to say. My name is bungston and I am a physics illiterate. |
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I need edification. Would someone please work it out and show the math? |
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if you place the leaf-blower on end on the ground then stand on it, Newton in this case simply assures that air will seep around the edges of the blower. |
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It might be simpler to develop a wearable leaf and then get
somebody else to blow you. No, wait.... |
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I can't see a flaw in your calculations, bungston. Given that human powered helicopters have flown (albeit very briefly), the figure you calculate seems reasonable (to quote Wikipedia "Elite track sprint cyclists are able to attain an instantaneous maximum output of around 2,000 watts"). |
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Of course the calculated power requirement of 686W gets much bigger when you include the weight of the device, the inherent inefficiencies, and the fact you need extra thrust to lift rather than just hover. |
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If someone will be good enough to clean the whiteboard for us .... |
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The Williams powerplant used in the flying dustbin has this spec: |
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Thrust: 2,7 kN
Weight: 66,2 kg
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This engine can lift its own weight, a 70Kg pilot, plus structural and control components, and fuel. |
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On your planet, 1 Kg mass experiences approximately 10N downforce. To lift 1Kg against local gravity, a force of 10 + δP Newtons is required (since P=mf). |
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Using a Newtonian reaction drive (how crude !), the downward momentum of the working fluid (in this case, air) must equal the downward force of gravity, if the vehicle is to hover. |
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The dustbin and occupant have an all-up mass of 250 Kg, which requires a force of 2500 Newtons (2.5 KN) to "hover" it - since the maximum thrust of the engine is 2.7KN, the device can move upwards against gravity (although not very fast). |
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Now: consider the specific impulse of the engine (q.v) - <link>. The working fluid is air, and momentum (but not kinetic energy) is conserved. |
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In your own time, calculate the required mass and velocity of air to provide 2.5KN of disposable lift. Write on both sides of the paper, in ink, via a pen if possible. Do not bend, spindle, mutilate or fold. You may be asked questions later. |
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Turn over your papers now. |
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It is meant to be the Greek symbol "delta" (lower case), as in "delta-vee". |
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Lower case delta is used in calculus to signify an infinitesimally small variation in a variable, i.e. x, delta x. Delta x is very slightly larger than x. |
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Delta can also signify the differential of a parameter; if velocity is V, delta- V is the rate of change of V, or acceleration. |
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Thank you.
Astrodynamics looks like fun but I should probably get long division down pat first I think. |
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// δP //
Klingon emoticon. |
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I have learned that watts and newtons make me jittery. For purposes of determining feasibility, I will assert that to counter gravitational acceleration of rocketeer over one second and render rocketeer weightless, the rocket pack must over one second move down a mass equal to that of the rocketeer. |
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Rocketeer = 152 lbs.
1925 cubic feet of air at 0.08 lbs/ cubic foot
3326400 cubic inches of air at 1728 cubic inches / cubic foot |
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Assume blower with muzzle 12 inch radius moves a column of air with volume 3326400 cubic inches = weight of rocketeer. Column must move through blower in 1 second. |
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v (of cylinder) = 3.14 * 144 * height
height of air column = 7356 inches or 613 feet.
613 feet column traversing blower / 1 second = 418 mph. |
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Strong leaf blower might have muzzle velocity of 200 mph. 3 leaf blowers should do the trick. Extra power will generate lift. |
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Leaf blowers are made of titanium and carbon fiber for this purpose. Light as a feather! |
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You will only need one hand. |
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//418 mph // The power isn't additive, 3x 200 mph
won't give the necessary 418mph. |
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If the column of air is half as high it can go half as fast with the same volume of air still moved in one second. |
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[bungston] Your initial calculations would work for a relatively efficient lifting system, such as an electric winch. Using jet-like thrust is vastly less efficient and requires many times the power for the same result; most of the energy expended ends up as kinetic energy in the downward moving fluid rather than being used to increase altitude. |
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Just get a freaking leaf blower, mount it on a seesaw
(teetertotter) like device, and directly measure how much it can lift. I predict that it will be somewhat less than its own weight. |
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I've never had one take off. |
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Learn through experiment? My armchair will get cold! |
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There might be a clever way of patterning or weaving
the air flow to give a better air to air grip and
therefore greater thrust. I wish I knew the method.
I would be rich. |
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slowing down time in your immediate vicinity might allow you to "grapple" the surrounding air to a certain degree... or at least make a pretty decent space heater. |
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Go on, pick the hardest dimension to manipulate. |
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I have found that clicking the halfbakery link seems to speed up the minute hand, if that helps. |
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//if velocity is V, delta- V is the rate of change of V, or acceleration// |
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Oh, tut tut. Tut tut indeed. In that usage, Delta-V is still the change of V. Delta-V divided by Delta-T (where T is time) - now that gives you rate of change of V, otherwise known as instantaneous acceleration, normally "a", or "u" for some reason. |
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It's normal to use a lower case italic "d" to indicate derivatives (ie dV/dT for acceleration, or dX/dT for velocity) rather than deltas. Deltas are normally used for macro changes, ie a change over a given ammount of time, not micro ones. |
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(I write this knowing full well that [8th] is likely better schooled than I in calculus, and may well now pick the above apart. Brace for impact...) |
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I don't think bracing is necessary: there's a difference between a <Klingon emoticon> and a "d": dV is independent of t (or to be more exact delta-t = 0). |
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