Inertia Machine
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Force Diagram and Maths
Ok, so Jiff has being crying out for it, here it is - a force diagram and some maths behind the average force over an entire cycle. The assumptions are:
1) The mass rotates through 360 degrees (2 Pi radians), and when it reaches it's original position it has the same angular velocity as when it started.
2) the forces are calculated from the point of view of the rotating mass. The force acting on the baseplate is equal and oppositte (good old Newton there).
The maths behind the y-axis force is similar and I'll leave as an excercise for the reader (I've spent way too much time typing equations into Word tonight).
Of course this really is a hell of a lot of maths just to point out that the mass has the same momentum at the start and end of the cycle, hence the the average force over the period in question has to be 0.
Daniel
1) The mass rotates through 360 degrees (2 Pi radians), and when it reaches it's original position it has the same angular velocity as when it started.
2) the forces are calculated from the point of view of the rotating mass. The force acting on the baseplate is equal and oppositte (good old Newton there).
The maths behind the y-axis force is similar and I'll leave as an excercise for the reader (I've spent way too much time typing equations into Word tonight).
Of course this really is a hell of a lot of maths just to point out that the mass has the same momentum at the start and end of the cycle, hence the the average force over the period in question has to be 0.
Daniel
Last edited by Deemar; 26th Jul 2007 at 14:31. Reason: Corrected the image link
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Meaningful answer
seeing this is an Interesting discussion. and everyone is refreshed on Centripetal and Centrifugal and torque forces . not to mention a Couple of Coupled moments.
Could some one help me on this one.
I am using my toy (DID I say toy ) Helicopter. RC.
here are the Specs.
Rotor Radius 27.5cm (oops. thats Rotor Radius, not diamiter my Boo
)
Rotor RPM 2600
Weight at tip 5g
Given these. I am trying to get a meaningful answer.
I am Using CF = M x Vsquared over Radius.
I get a Tip speed of near 270kph
and when I get an answer using
5 x 269.547squared over 275
I get 1321.01
what dose that mean ?? IE what is the pulling power of the 5g weight on the Axis of rotation ?
is that 1.321kg or 131g or is my use of metric wrong or >>>??? ...
Please help me get a Meaningful answer
Could some one help me on this one.
I am using my toy (DID I say toy ) Helicopter. RC.
here are the Specs.
Rotor Radius 27.5cm (oops. thats Rotor Radius, not diamiter my Boo
)Rotor RPM 2600
Weight at tip 5g
Given these. I am trying to get a meaningful answer.
I am Using CF = M x Vsquared over Radius.
I get a Tip speed of near 270kph
and when I get an answer using
5 x 269.547squared over 275
I get 1321.01
what dose that mean ?? IE what is the pulling power of the 5g weight on the Axis of rotation ?
is that 1.321kg or 131g or is my use of metric wrong or >>>??? ...
Please help me get a Meaningful answer
Last edited by BGRing; 10th Aug 2007 at 17:13.
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BGring, lots of ways to do this. My favourite is to stick to kg, m, and s.
Convert your 270kph into 75 m/s, your 5g into 0.005kg and your 27.5cm into 0.275m.
You find the force on the 5g weight is 102.3 N (=102.3 kg m/s^2). What does that mean? Well 1 N (Newton) is about the weight of 1 apple in Earth's gravity (easy way to think of it: Newton..apple), so 102.3N is the same force as 100 apples weigh.
If you want g then you don't want a force, you want the acceleration. It's probably already been discussed in this thread, but to keep the tip weight moving in a circle it's velocity must change (in direction, not necessarily in speed). So the acceleration responsible for changing the tip weight's velocity is just the v squared over r part of the equation.
This works out to 20,454 m/s^2 or about 2085 g's !
Easy to see why balancing is so important.
Convert your 270kph into 75 m/s, your 5g into 0.005kg and your 27.5cm into 0.275m.
You find the force on the 5g weight is 102.3 N (=102.3 kg m/s^2). What does that mean? Well 1 N (Newton) is about the weight of 1 apple in Earth's gravity (easy way to think of it: Newton..apple), so 102.3N is the same force as 100 apples weigh.
If you want g then you don't want a force, you want the acceleration. It's probably already been discussed in this thread, but to keep the tip weight moving in a circle it's velocity must change (in direction, not necessarily in speed). So the acceleration responsible for changing the tip weight's velocity is just the v squared over r part of the equation.
This works out to 20,454 m/s^2 or about 2085 g's !
Easy to see why balancing is so important.
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Post #1 Picture 2 top right diagram portion.
with the rotation velocity being Higher between points D to C than they are between Points A to B. it stands to reason that the CF is greater between the points D to C than A to B. thus . you are not getting any lift.
I remember when I was a Little Kid. Must have been 5. I ran around the house thinking I could lift myself off the ground. IE I Put one hand between my legs at front and one at back grabbed and pulled up. (
) luckily pain set in before I either ripped my arms out of their socket or did some Damage to the Little Crown jewels. I sat down that day and Scratched my head. Must have been all those Road runner cartoons I was watching back then. (1975).
Surmise.
It is the CF at the bottom when you are allowing it to Free (Though it is not because you are allowing it to free) that counteracts any upward momentum of the Frame.
I recall a little article about using Gyros to make Anty gravity devices. Never heard any more of it. (No it is not top secret. it is just not working)
Remember in the Road runner series or any other cartoon where they had a fan on a Sail boat. or the one where they would walk a plank and nail another one at the end and then walk that one and then nail another at the end of that and walk that one and then nail another one at the end of that etc etc... just doesn't work.
My last moment of delusion was when I thought that if we had a Mag lev rail launcher on the highest mountain in Ecuador. to launch a Lifting vehicle that could then launch a rocket into orbit, but then it was done with just a lifting vehicle. and Rocket 1.
OK. here is food for your thought jiff.
what if the whole wheel was rotating on another axis. that could re direct the CF of the down side between accel and Decel. I will Leave that one to you.
<DeleteOH. Please answer my other post some one. PLEASE
Delete>
<EDIT thanks matthew. I was typing this one and didnt see your reply. thanks , scratch that last EDIT>
with the rotation velocity being Higher between points D to C than they are between Points A to B. it stands to reason that the CF is greater between the points D to C than A to B. thus . you are not getting any lift.
I remember when I was a Little Kid. Must have been 5. I ran around the house thinking I could lift myself off the ground. IE I Put one hand between my legs at front and one at back grabbed and pulled up. (
) luckily pain set in before I either ripped my arms out of their socket or did some Damage to the Little Crown jewels. I sat down that day and Scratched my head. Must have been all those Road runner cartoons I was watching back then. (1975).Surmise.
It is the CF at the bottom when you are allowing it to Free (Though it is not because you are allowing it to free) that counteracts any upward momentum of the Frame.
I recall a little article about using Gyros to make Anty gravity devices. Never heard any more of it. (No it is not top secret. it is just not working)
Remember in the Road runner series or any other cartoon where they had a fan on a Sail boat. or the one where they would walk a plank and nail another one at the end and then walk that one and then nail another at the end of that and walk that one and then nail another one at the end of that etc etc... just doesn't work.
My last moment of delusion was when I thought that if we had a Mag lev rail launcher on the highest mountain in Ecuador. to launch a Lifting vehicle that could then launch a rocket into orbit, but then it was done with just a lifting vehicle. and Rocket 1.
OK. here is food for your thought jiff.
what if the whole wheel was rotating on another axis. that could re direct the CF of the down side between accel and Decel. I will Leave that one to you.
<DeleteOH. Please answer my other post some one. PLEASE
Delete><EDIT thanks matthew. I was typing this one and didnt see your reply. thanks , scratch that last EDIT>
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Purrfect Landing
Oh, why all the fuss, when my “anti-gravity/purr-petual motion” device is so much more successful, doesn’t need a power source, and could easily drive the tail rotor of any conventional helicopter.
This as yet unpatented device requires one cat and a generously buttered slice of fresh, warm toast. Simply staple the toast to the cat’s back, taking care to ensure the buttered side faces uppermost, opposing gravity, when the cat stands in the normal axis, ie, with its feet glued to the workbench where, afteral, they had to be, to achieve that earlier stapling assembly technique.
Having un-glued the animal’s feet, all that remains is to charge the device by inverting it through 180 degress. Simply drop the device from a height of at least three feet for the viewer to observe how the widely held physical laws of the cat “always” landing on its feet, and the one about toast “always” landing butter side down, now come into conflict. The initial equal and opposite forces are unbalanced by the cat’s natural instinct to twist and effect a four-point landing. This supremacy over the bread’s inability to think provides the initial rotation to start the device.
Now observe as momentum maintains the direction of rotation, while the natural switching on and off of the two separate physical laws powers the device indefinitely about a horizontal axis, or at least until all the howling results in a raid by those animal protection fanatics.
In the future I hope it may also be possible to harness the perpetual levitation evident in the device as a control mechanism to prevent ground contact during hover training.
Could someone draw me a force diagram for this, please?
This as yet unpatented device requires one cat and a generously buttered slice of fresh, warm toast. Simply staple the toast to the cat’s back, taking care to ensure the buttered side faces uppermost, opposing gravity, when the cat stands in the normal axis, ie, with its feet glued to the workbench where, afteral, they had to be, to achieve that earlier stapling assembly technique.
Having un-glued the animal’s feet, all that remains is to charge the device by inverting it through 180 degress. Simply drop the device from a height of at least three feet for the viewer to observe how the widely held physical laws of the cat “always” landing on its feet, and the one about toast “always” landing butter side down, now come into conflict. The initial equal and opposite forces are unbalanced by the cat’s natural instinct to twist and effect a four-point landing. This supremacy over the bread’s inability to think provides the initial rotation to start the device.
Now observe as momentum maintains the direction of rotation, while the natural switching on and off of the two separate physical laws powers the device indefinitely about a horizontal axis, or at least until all the howling results in a raid by those animal protection fanatics.
In the future I hope it may also be possible to harness the perpetual levitation evident in the device as a control mechanism to prevent ground contact during hover training.
Could someone draw me a force diagram for this, please?
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Diameter / Radius
BGRing, bit picky I know, but I think that you may have used diameter and not radius to calculate tip speed.
Matthew Parsons' calculations and explanation are spot on, but a factor of 2 out because of this. If you need to use the result then take this into account
Matthew Parsons' calculations and explanation are spot on, but a factor of 2 out because of this. If you need to use the result then take this into account
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One blade is 25.3cm Blade bolt to Blade tip (Not going Diagonal, IE same distance from LE at tip). and the Bolt to Hub axis is about 2.3cm So my calculations are wrong because I used wrong length to tip.
Just so I can Check what I think I understand so far. and to Highlight a Question or two.
this is How I did It. (Now using a Rotor RPM of 2400)
2 x Pie x 27.6 = Circumference of 173.415cm
173.415 x 2400 = Distance in a Min of 416198.194cm
But to stick with Matthews Example, I will divide 416198.194cm/m by 60 to get cm/s and then Divide by 100 to get the working number of 69.366 m/s
Same weight at the tip of 5Grams. Or .005kg and a Radius of 0.276m
So.
Newton's is 87.167 N (= 87.167 kg m/s^2)
is that g as is Gravity? or grams.. ? I am guessing it is G-force as you have not implied any mass amount.
With my new Example, I get the figure of 17433.485 m/s^2 .
Now, how did you get the next Number? (2085)
So if this is gravitation or G-force or ?. how do I use this number with my 5grams to get a pulling weight in Grams on the Blade bolt ?
is it just times. IE previous example , 2085 x 5 = 10.425kg?
Thanks in advance.
(to make it easy . I Bolded and Redden my questions. <Correct me where wrong please >)
Just so I can Check what I think I understand so far. and to Highlight a Question or two.
this is How I did It. (Now using a Rotor RPM of 2400)
2 x Pie x 27.6 = Circumference of 173.415cm
173.415 x 2400 = Distance in a Min of 416198.194cm
Previously I would times this by 60 to get per hour. then Divide by 100 to get m/h and divide by 1000 to get kph (249.718kph)
Same weight at the tip of 5Grams. Or .005kg and a Radius of 0.276m
So.
Newton's is 87.167 N (= 87.167 kg m/s^2)
If you want g then you don't want a force, you want the acceleration. It's probably already been discussed in this thread, but to keep the tip weight moving in a circle it's velocity must change (in direction, not necessarily in speed). So the acceleration responsible for changing the tip weight's velocity is just the v squared over r part of the equation.
This works out to 20,454 m/s^2 or about 2085 g's !
Now, how did you get the next Number? (2085)
So if this is gravitation or G-force or ?. how do I use this number with my 5grams to get a pulling weight in Grams on the Blade bolt ?
is it just times. IE previous example , 2085 x 5 = 10.425kg?
Thanks in advance.
(to make it easy . I Bolded and Redden my questions. <Correct me where wrong please
>)
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Sorry, your original post had rotor diameter listed as 27.5cm
If it is rotor radius that is 27.5 cm I retract my previous post.
Using your latest post I agree with your calculations that give the CF as approximately 87.17 N
(I prefer to work with angular rate which is ~ 251 rad/s
CF accn = angular rate ^2 * radius
CF accn ~ 17434 m/s^2)
If it is rotor radius that is 27.5 cm I retract my previous post.
Using your latest post I agree with your calculations that give the CF as approximately 87.17 N
(I prefer to work with angular rate which is ~ 251 rad/s
CF accn = angular rate ^2 * radius
CF accn ~ 17434 m/s^2)
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waspy, I should have mentioned that. Thanks. The tip speed, rotational speed and rotor length only worked if it was a radius, as it was used later in BGring's post.
BGring. The g used later on is as in G-force, which isn't a force, which unfortunately leads to more misunderstanding. 'g' is a unit of acceleration based on the gravitational pull of the earth. With non-spherical shape of the planet, the mass distribution, and the varying rotational speeds based mostly on latitude, the value of g is not constant. However, it changes so little that conventionally we use 1 g = 9.81 m/s^2. So dividing your answer of 17,433.485 by 9.81 you get 1777 g (not grams).
You can use g by itself, but to get the force you can either go back to the CF equation you started with or you can multiply g by the weight of the object. To get the weight of a 5g object, you multiply it by the acceleration due to gravity (9.81 m/s^2) and you get 0.005kg x 9.81 m/s^2 = 0.049N, or 1/20 of an apple ~one small bite.
Matthew.
BGring. The g used later on is as in G-force, which isn't a force, which unfortunately leads to more misunderstanding. 'g' is a unit of acceleration based on the gravitational pull of the earth. With non-spherical shape of the planet, the mass distribution, and the varying rotational speeds based mostly on latitude, the value of g is not constant. However, it changes so little that conventionally we use 1 g = 9.81 m/s^2. So dividing your answer of 17,433.485 by 9.81 you get 1777 g (not grams).
You can use g by itself, but to get the force you can either go back to the CF equation you started with or you can multiply g by the weight of the object. To get the weight of a 5g object, you multiply it by the acceleration due to gravity (9.81 m/s^2) and you get 0.005kg x 9.81 m/s^2 = 0.049N, or 1/20 of an apple ~one small bite.
Matthew.
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Thanks waspy, I changed that first post to Radius. you were correct and I thank you for Pointing it out
looking at you method of calculation. I understand what you are doing. you are getting the 251 by dividing the m/s by the radius. seems to come to the same answer (Similar).
Mathew. Thanks again.
here is what I think I have. is the 9.81 an exact # or an Approximation.
8.885kg
is the 9.81 because we are being lightened a little due to earth rotation?
I think dantruck has something there. Though, I think it may be easier to use two pieces of buttered bread or two cats (this would give a truly = and Opposite Force)
looking at you method of calculation. I understand what you are doing. you are getting the 251 by dividing the m/s by the radius. seems to come to the same answer (Similar).
Mathew. Thanks again.
here is what I think I have. is the 9.81 an exact # or an Approximation.
8.885kg
is the 9.81 because we are being lightened a little due to earth rotation?
I think dantruck has something there. Though, I think it may be easier to use two pieces of buttered bread or two cats (this would give a truly = and Opposite Force)
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Interesting idea BGRing!
You won't be forgotten when I beat Jiff to the Nobel prize.
The 'Cat 2' version would double the howl problem, however. Meanwhile your 'two toast' version offers significant weight saving potential, I'm sure. Though we'd still need a trigger to initiate rotation. Do you think a thin film of "I Can't Believe It's Not Butter" on one of the slices would induce sufficient potential difference to achieve engine start?
Meanwhile...I am perturbed that my ideas are so popular they are already being ripped off, complete with force diagram, here...
http://en.wikipedia.org/wiki/Buttered_cat_paradox
Watch out Jiff. There's no containing brilliance in this world. And the truth will always out.
Dan
You won't be forgotten when I beat Jiff to the Nobel prize.
The 'Cat 2' version would double the howl problem, however. Meanwhile your 'two toast' version offers significant weight saving potential, I'm sure. Though we'd still need a trigger to initiate rotation. Do you think a thin film of "I Can't Believe It's Not Butter" on one of the slices would induce sufficient potential difference to achieve engine start?
Meanwhile...I am perturbed that my ideas are so popular they are already being ripped off, complete with force diagram, here...
http://en.wikipedia.org/wiki/Buttered_cat_paradox
Watch out Jiff. There's no containing brilliance in this world. And the truth will always out.
Dan
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BGring, g=9.81 is an approximation. It changes depending on where you are, assumed to be close to the surface, and is based on a reference frame relative to the planet, so the rotation does decrease it.
Dantruck, if you tie a catnip ball to the cat's tail I think that would initiate rotation.
Dantruck, if you tie a catnip ball to the cat's tail I think that would initiate rotation.
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Thanks Mathew P'...good Canadian thinking there. Perhaps you'd be kind enough to build a true northern hemisphere test unit for me?
Here in southern Spain we're just too close to the equator to rule out Coriolis, plus I'm concerned about the effect of the Hadley cell under which we sit at this latitude.
Are you also a 'Barenaked Ladies' fan, perchance?
Dan
Here in southern Spain we're just too close to the equator to rule out Coriolis, plus I'm concerned about the effect of the Hadley cell under which we sit at this latitude.
Are you also a 'Barenaked Ladies' fan, perchance?
Dan
On 26 July 2007, it was written:
Quote:
Originally Posted by Jiff
My force diagram will be posted here within two days.
Jiff?
Quote:
Originally Posted by Jiff
My force diagram will be posted here within two days.
Jiff?
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