Two coordinate systems:
Cartesian = (fancy way of saying) the usual X/Y coordinates
Polar = Distance/Angle coordinate - or Magnitude/Direction
From Cartesian to Polar: Magnitude=Sqrt(X*X+Y*Y), Angle=ArcTan(Y/X)
From Polar to Cartesian: X=Magnitude*Cos(Angle), Y=Magnitude*Sin(Angle).
When you look at your screenshot, the property of that force is set to "Polar".
Fx is set to 42.426 (Lb?) as the force-magnitude (visualized as the size of the arrow)
Fy is set to 225 (°?) as the direction (visualized as the orientation of the arrow)
Setting the coordinates to "cartesian", and the horizontal-force Fx becomes (-30 Lb) or 30 Lb going left, and the vertical-force Fy becomes (-30 Lb) or 30 Lb going down.
(Don't know why it shows Fx/Fy without units for that, as [F]/Theta would be expected)
Add:
-- WM2D forgets to set the appropriate labels, they will be ok when swapping the option to cartesian and then back to polar --
WM2D
Moderator: scott
re: WM2D
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: WM2D
Could one of you good people please explain to me what the "moment of inertia"is on the properties table of a circle .
How does this number expressed in " lb-in^2 "affect the results of a sim in WM2D ?
Maybe this is one for you Marchello
Graham
How does this number expressed in " lb-in^2 "affect the results of a sim in WM2D ?
Maybe this is one for you Marchello
Graham
The moment of inertia for a thick ring is:
I = 0.5 * m * ( R1^2 + R2^2)
Where m=mass, R1=inner radius, R2=outer radius
For a thin ring R1=R2 --> I=m*(R^2)
For a disk R1=0 --> I=0.5*m*(R^2)
So a thin ring has more inertia than a disk.
What does that mean....
It affects questions like the one FCdriver posted:
For an ultra-light wheel (or a pendulum) that factor F is 1.
For a heavy wheel it goes towards zero.
As a ring has twice the inertia of a solid disk, the F of a ring is lower compared to a disk.
The actual factor-F co-depends on the dropped weight, so it's no easy tell what that actual factor-F is.
But what I actually can tell is that a weight on a weightless rod (a pendulum) the velocity (for that problem) is Vmax=5.68 m/s
When that weight is planted onto a disk its end-velocity (for that height drop) is smaller (because actually that pendulum has now a certain weight) is lower.
And when that weight is planted onto a flywheel (thus moment of inertia of a ring) the end velocity is even lower.
In theory an infinitely heavy wheel (without being concerned about its own gravitational field) should have a factor-F approaching zero, meaning that planted weight has little effect.
I hope it makes some sense.
Marchello E.
Sidenote 1: As a European I'm not used to Lbs - somehow I translate it as 'Lobsters')
Sidenote 2: What kind of vibrant window could possibly be found behind such dirty cloth?
I = 0.5 * m * ( R1^2 + R2^2)
Where m=mass, R1=inner radius, R2=outer radius
For a thin ring R1=R2 --> I=m*(R^2)
For a disk R1=0 --> I=0.5*m*(R^2)
So a thin ring has more inertia than a disk.
What does that mean....
It affects questions like the one FCdriver posted:
My answer: http://www.besslerwheel.com/forum/viewt ... 544#138544please tell me the Terminal velocity that a weight falling 65 inches, on a 84 inch radius ? Would it be better to have a heavier wheel, or a lighter wheel? Is it waste of energy, having the weight move a heavy wheel, or a benefit?
In the meantime I figured out the relationship with inertia as a factor (F) of that maximum velocity.A weight falling 65 inch should get a velocity of about 5.68 m/s [2*g*h=v^2].
(add: less when accounting for wheel inertia... I'll figure it out)
For an ultra-light wheel (or a pendulum) that factor F is 1.
For a heavy wheel it goes towards zero.
As a ring has twice the inertia of a solid disk, the F of a ring is lower compared to a disk.
The actual factor-F co-depends on the dropped weight, so it's no easy tell what that actual factor-F is.
But what I actually can tell is that a weight on a weightless rod (a pendulum) the velocity (for that problem) is Vmax=5.68 m/s
When that weight is planted onto a disk its end-velocity (for that height drop) is smaller (because actually that pendulum has now a certain weight) is lower.
And when that weight is planted onto a flywheel (thus moment of inertia of a ring) the end velocity is even lower.
In theory an infinitely heavy wheel (without being concerned about its own gravitational field) should have a factor-F approaching zero, meaning that planted weight has little effect.
I hope it makes some sense.
Marchello E.
Sidenote 1: As a European I'm not used to Lbs - somehow I translate it as 'Lobsters')
Sidenote 2: What kind of vibrant window could possibly be found behind such dirty cloth?
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: WM2D
OK I get it
A higher value of "moment of inertia "for something like a flywheel would require more time to reach max velocity than one with a lower value.
A logical fact and not something to fret or lose sleep over .
Thankyou again Marchello
I was born and raised in England and when they adopted the metric system I knew that I could never go back.
BTW how do you say "perpetual motion "in Holland?
Graham
A higher value of "moment of inertia "for something like a flywheel would require more time to reach max velocity than one with a lower value.
A logical fact and not something to fret or lose sleep over .
Thankyou again Marchello
I was born and raised in England and when they adopted the metric system I knew that I could never go back.
BTW how do you say "perpetual motion "in Holland?
Graham
Basically one could make the wheel react fast to motion shifts, or absorb annoying ones....
When a weight goes from top to bottom, that final velocity depends on Inertia.
Dutch: "Perpetuüm mobiel"
partitioned: Per-pe-tu-um mo-biel
sounds like:
(note the short/long indicated vowels are for my own convenience (dutch), and does not relate to the english sounding vowel )
Per - short "e", like the "her" in there, or better: Pear
pe - long "e", like pay but forget the focus on the"y"
tu - long "u", like the "e" as in new, without the "w"
um - short "u", like the "um" as in drum
Mo - long "o", like doh
biel - long "i", like keel
It combines to something like: Pear-pay-tew-uhm Moh-beel
oh, and a slight intonation on "pay" (you know Dutch huh!), and a smaller on "beel"... the rest is muffled along.
A bit weird to think about it this way, but good luck practising
When a weight goes from top to bottom, that final velocity depends on Inertia.
Dutch: "Perpetuüm mobiel"
partitioned: Per-pe-tu-um mo-biel
sounds like:
(note the short/long indicated vowels are for my own convenience (dutch), and does not relate to the english sounding vowel )
Per - short "e", like the "her" in there, or better: Pear
pe - long "e", like pay but forget the focus on the"y"
tu - long "u", like the "e" as in new, without the "w"
um - short "u", like the "um" as in drum
Mo - long "o", like doh
biel - long "i", like keel
It combines to something like: Pear-pay-tew-uhm Moh-beel
oh, and a slight intonation on "pay" (you know Dutch huh!), and a smaller on "beel"... the rest is muffled along.
A bit weird to think about it this way, but good luck practising
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
Re: re: WM2D
the circumference doubles, the ratio doubles, with gears its the number of teeth, 25 teeth to 50 teeth. The 25 tooth gear turn twiceMagnum wrote:Can anyone tell be how to setup a gear system to run 2:1. The larger wheel set at 1 and a smaller wheel set at twice the speed.
Ken T.
Forget your lust for the rich man's gold
All that you need is in your soul
And you can do this, oh baby, if you try
All that I want for you my son is to be satisfied
All that you need is in your soul
And you can do this, oh baby, if you try
All that I want for you my son is to be satisfied
The "how" is rather difficult, I don't think they are simulatable with WM2D.
The attachment, which I found somewhere on the net, shows some interesting indications of the "how".
Like a blue line (left) showing some trickling water-level, and on the right a Fellenius slope stability analysis.
Looking at these I think one can simply conclude: They do just fine.
The only wet feet we have here is from some heavy autumn rains...
Marchello E.
The attachment, which I found somewhere on the net, shows some interesting indications of the "how".
Like a blue line (left) showing some trickling water-level, and on the right a Fellenius slope stability analysis.
Looking at these I think one can simply conclude: They do just fine.
The only wet feet we have here is from some heavy autumn rains...
Marchello E.
- Attachments
-
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: WM2D
With global warming predicted to raise sea levels considerably in upcoming years dykes may become more popular around the globe.
It has been said that London England will be underwater in the future!!
Graham
It has been said that London England will be underwater in the future!!
Graham
re: WM2D
Hello Gentlemen
I am a newcomer to this forum and haven't even begun to read all of the many threads and topics. I have been contemplating the Bessler wheel for about 20 years and dabbled in a few construction attempts. I am not a physics guru or trained mechanical engineer, but have built many mechanical and electronic devices over the past 40 years as chief engineer for a small electronics company.
Not being trained in the nuances of writing scripts or doing complex computations on moving bodies, I am using a very simple approach with WM2D to vet wheel designs that would probably be useful especially to beginning users of WM2D. If this method is of interest, I can outline it here or start a new topic as I don't want to derail this thread.
What I am doing in the way of simple vetting of wheel designs using what I think is a unique approach, may have already been explored here, and I have not yet found it in a search.
Your input on which direction to go with this approach would be appreciated, but so far as a newcomer, I'm a bit disappointed at the lack of a single reply.
Kind regards
eeman
I am a newcomer to this forum and haven't even begun to read all of the many threads and topics. I have been contemplating the Bessler wheel for about 20 years and dabbled in a few construction attempts. I am not a physics guru or trained mechanical engineer, but have built many mechanical and electronic devices over the past 40 years as chief engineer for a small electronics company.
Not being trained in the nuances of writing scripts or doing complex computations on moving bodies, I am using a very simple approach with WM2D to vet wheel designs that would probably be useful especially to beginning users of WM2D. If this method is of interest, I can outline it here or start a new topic as I don't want to derail this thread.
What I am doing in the way of simple vetting of wheel designs using what I think is a unique approach, may have already been explored here, and I have not yet found it in a search.
Your input on which direction to go with this approach would be appreciated, but so far as a newcomer, I'm a bit disappointed at the lack of a single reply.
Kind regards
eeman