Elite Engineer Posted May 29, 2013 Posted May 29, 2013 I'm currently building a hovercraft, with two leaf blowers. I need to know the trust of each leaf blower to know how/where to position them on the hovercraft. Both are identical models (craftsman, 200 mph, 430 cfm/ each). I'm a bit stuck on how to convert the CFM to thrust to know how heavy someone can be for it to work properly. Thanks, EE
swansont Posted May 29, 2013 Posted May 29, 2013 You need to know the speed at which the air is being ejected, and the mass of a cubic foot of air (which should be easy to calculate) mass*speed is momentum, and momentum per unit time will be force, or thrust
michel123456 Posted May 29, 2013 Posted May 29, 2013 I'm currently building a hovercraft, with two leaf blowers. I need to know the trust of each leaf blower to know how/where to position them on the hovercraft. Both are identical models (craftsman, 200 mph, 430 cfm/ each). I'm a bit stuck on how to convert the CFM to thrust to know how heavy someone can be for it to work properly. Thanks, EE Two? I'd suggest to use three.
swansont Posted May 29, 2013 Posted May 29, 2013 Two? I'd suggest to use three. Without knowing the thrust? Why?
dimreepr Posted May 29, 2013 Posted May 29, 2013 (edited) Due to the design of a hovercraft thrust isn’t what you need to be calculating, the volume of the air under the skirt when inflated is what you need to work out. 860 cfm sounds enough, to me, for a 1 or 2 man craft. As for positioning, it’s not really important, except for balance, so I’d suggest apposing ends or sides. Edited May 29, 2013 by dimreepr
michel123456 Posted May 29, 2013 Posted May 29, 2013 Without knowing the thrust? Why? Without knowing the thrust? Why? for equilibrium. I'd expect each thruster to act like a "foot" and you need 3 feet to achieve stability.
swansont Posted May 29, 2013 Posted May 29, 2013 for equilibrium. I'd expect each thruster to act like a "foot" and you need 3 feet to achieve stability. You can put multiple vents in place, i.e. T-off the output of the blower and position the thrust points wherever you want.
Elite Engineer Posted May 29, 2013 Author Posted May 29, 2013 (edited) You need to know the speed at which the air is being ejected, and the mass of a cubic foot of air (which should be easy to calculate) mass*speed is momentum, and momentum per unit time will be force, or thrust I assume you're talking about F=ma, but I don't think it applicable b/c the air is exiting at a fixed speed of 200 mph (89.41 m/s), and not accelerating? You can put multiple vents in place, i.e. T-off the output of the blower and position the thrust points wherever you want. would that be used to increase the flow rate of air or to direct the output of air so the hovercraft could move laterally? Due to the design of a hovercraft thrust isn’t what you need to be calculating, the volume of the air under the skirt when inflated is what you need to work out. 860 cfm sounds enough, to me, for a 1 or 2 man craft. As for positioning, it’s not really important, except for balance, so I’d suggest apposing ends or sides. ok thanks! what about the area of the tube the air is flowing through? You need to know the speed at which the air is being ejected, and the mass of a cubic foot of air (which should be easy to calculate) mass*speed is momentum, and momentum per unit time will be force, or thrust I apologize for bombarding you with a million replies, but I applied your equation and this is what I got: Mass of one cubic foot air @ STP= 0.036kg Exit speed of air = 89.41m/s F= m*v = (0.036kg)*(89.41m/s) = 3.2 N does 3.2N sound right for a leaf blower? If so wouldn't I need about 20 leaf blowers to support a 75kg person? Edited May 29, 2013 by Elite Engineer
swansont Posted May 29, 2013 Posted May 29, 2013 I assume you're talking about F=ma, but I don't think it applicable b/c the air is exiting at a fixed speed of 200 mph (89.41 m/s), and not accelerating? F = dP/dt is the actual equation. (it reduces to F=ma under many circumstances). So for a volume of air starting at rest, the rate at which the mass is ejected and the speed will give you the force. would that be used to increase the flow rate of air or to direct the output of air so the hovercraft could move laterally? Yes, you could do both with that. What you'll want to do depends on how much thrust you get. I have an air lifter at work that runs off of a ~4hp blower. The air is sent into two pads that have several dozen small holes in them, and in that configuration it can lift over 1000 lb a few inches. Conceptually, it's basically an air hockey table turned upside-down. http://blogs.scienceforums.net/swansont/archives/689 If you want to figure out details, you'll have to solve some fluid dynamics equations http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html I'm sure there are specific sites out there for hovercraft design, too
Enthalpy Posted May 29, 2013 Posted May 29, 2013 I vote in the same direction as Dimreepr, because of the skirt. A hovercraft has a surrounding skirt that deforms according to the terrain and spreads the air throughput. So the resulting thrust is not located at the fans but spread over the pressure area. Nor does the thrust result directly from air acceleration at the fans, since air then decelerates from the fans to the undercarriage, and accelerates again at the leaks under the skirt. The lateral stability of a hovercraft demands a (lossy) sectoring of the undercarriage, so that lift decreases on the side where the skirt is too high to follow the terrain. Standard hovercraft design rather puts all fans in common, then sectors the total flow in the skirt. Some reasons for such a design: - The overpressure acts on a bigger surface - Only a skirt achieves pressure with a limited throughput - This throughput, hence the fan power, is way smaller than computed from what fans would need to create lift - This is what differs between a hovercraft and a helicopter, and saves big power. ----- Fun : the first human-powered helicopter (Gamera) has flown recently. It relied fully on ground effect. But a human-powered hovercraft is much easier (obviously these guys don't have the power-to-mass ratio of the young lady who flew Gamera): http://vimeo.com/18859973
Elite Engineer Posted May 30, 2013 Author Posted May 30, 2013 Due to the design of a hovercraft thrust isn’t what you need to be calculating, the volume of the air under the skirt when inflated is what you need to work out. 860 cfm sounds enough, to me, for a 1 or 2 man craft. As for positioning, it’s not really important, except for balance, so I’d suggest apposing ends or sides. shouldnt the blowers be positioned where there is the most weight, or will the pressure of air be evenly distributed?
dimreepr Posted May 30, 2013 Posted May 30, 2013 Apart from the passenger/s the most weight would be the engines so apposing them on the craft would evenly distribute the weight. Here’s a basic design to illustrate where I would suggest to put your blowers and a suggestion for an insert to help distribute the air more evenly and avoid direct thrust leakage.
Enthalpy Posted May 30, 2013 Posted May 30, 2013 shouldnt the blowers be positioned where there is the most weight, or will the pressure of air be evenly distributed? The very design of a hovercraft lets the whole area create lift, whose center is essentially independent of the blowers' location.
dimreepr Posted May 30, 2013 Posted May 30, 2013 (edited) Thinking about it you could use both engines, as illustrated, for both thrust and lift. By ducting the blowers output, at a guess, 75% of both to lift and 25% (a nice easy ratio for the pipe work) to a simple directional ramjet providing thrust and steering. Edited May 30, 2013 by dimreepr
Elite Engineer Posted May 30, 2013 Author Posted May 30, 2013 Apart from the passenger/s the most weight would be the engines so apposing them on the craft would evenly distribute the weight. Here’s a basic design to illustrate where I would suggest to put your blowers and a suggestion for an insert to help distribute the air more evenly and avoid direct thrust leakage. Awesome man thanks! This is a superb design, everyone on this thread has been contributing alot thanks!
J.C.MacSwell Posted June 5, 2013 Posted June 5, 2013 I'm currently building a hovercraft, with two leaf blowers. I need to know the trust of each leaf blower to know how/where to position them on the hovercraft. Both are identical models (craftsman, 200 mph, 430 cfm/ each). I'm a bit stuck on how to convert the CFM to thrust to know how heavy someone can be for it to work properly. Thanks, EE These are primarily ground effect vehicles so the thrust formulas don't really apply. It comes down to area and how much pressure can be maintained underneath the craft against the considerable leakage taking place.
dimreepr Posted June 5, 2013 Posted June 5, 2013 (edited) You’d think a person of your experience would have, at least, read the thread before replying. Edit/ come to think how is it a ‘ground effect vehicle’? Edited June 5, 2013 by dimreepr
J.C.MacSwell Posted June 6, 2013 Posted June 6, 2013 You’d think a person of your experience would have, at least, read the thread before replying. Edit/ come to think how is it a ‘ground effect vehicle’? It is supported by the ground underneath the vehicle, not the thrust from the accelerated air. Compare to a helicopter or plane which power or deflect air downward.
dimreepr Posted June 6, 2013 Posted June 6, 2013 By that logic a car is also a ground effect vehicle.
J.C.MacSwell Posted June 6, 2013 Posted June 6, 2013 By that logic a car is also a ground effect vehicle. There are ground effects on cars, planes, helicopters near the surface...I think the term comes into play when that becomes a significant effect in the design
dimreepr Posted June 6, 2013 Posted June 6, 2013 I realise that a car was a poor example, as many racing/road cars employ ground effect to increase traction (but that doesn’t make it a GEV), a bike would have been a better. This from my link, for clarification...“A ground effect vehicle (GEV) is one that attains level flight near the surface of the Earth, making use of the aerodynamic interaction between the wings and the surface known as ground effect.”
J.C.MacSwell Posted June 6, 2013 Posted June 6, 2013 I realise that a car was a poor example, as many racing/road cars employ ground effect to increase traction (but that doesn’t make it a GEV), a bike would have been a better. This from my link, for clarification... “A ground effect vehicle (GEV) is one that attains level flight near the surface of the Earth, making use of the aerodynamic interaction between the wings and the surface known as ground effect.” If that is the case, then I'm wrong calling a standard hovercraft a ground effect vehicle. It is supported by the ground due to an increased pressure, rather than thrust or deflection, but not due to an aerodynamic interaction.
howlingmadpanda Posted July 15, 2013 Posted July 15, 2013 Thrust is typically measured in newtons if you want to calculate thrust I suggest looking around here: https://www.grc.nasa.gov/WWW/k-12/airplane/thrsteq.html as there will be plenty of other equations you may want or need.
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