Hi all,
Its been many years since I have had to calculate accelerations, G force etc..
I was hoping someone here could help me out.
I want to calculate the G force exerted on a mass on a platform which moves up and down in a see-saw motion from a central axis. The platform moves a maximum distance of 13mm in either direction at 28 movements per second.
Should this be enough information to work it out?
It would be most appreciated if someone could answer this, and if possible show the working out.
It's undergoing harmonic motion with frequency 28Hz with amplitude 13mm = 1.3cm = 0.013m and assuming it starts at zero displacement (ie x=0 at t=0). Therefore it's position (ignoring the slight side to side motion from the see-saw) will be expressed as, taking x' = dx/dt etc,
x = 0.013sin(28*2pi*t)
To find this I've done the double integration of the harmonic equation x'' + wx = 0 for w = (2*pi*f)^2 with standard boundary conditions (I can walk you through this if needs be).
Force due to oscillation = F = ma = mx'' = m(-wx) = -m[(2*pi*28)^2]*0.013sin(28*2pi*t). This is extremised when |sin|=1.
So it'll be max/min when F = -/+0.013m[(2*pi*28)^2]. When F = -0.013m[(2*pi*28)^2] then it's being accelerated upwards, When F=-0.013m[(2*pi*28)^2] then it's being pulled downwards.
Now we add in gravity so the total force an object of mass m will feel is mg-m[(2*pi*28)^2]*0.013sin(28*2pi*t). The sign on the mg is because when the force is zero, you feel 1g (even though it's pulling you down). To get the G force found you divide this by mg (ie you divide by the weight of the object since 5g = 5 times usual weight etc) to get 1-(1/g)[(2*pi*28)^2]*0.013sin(28*2pi*t)
Thus the max G's will be 1+(0.013/g)[(2*pi*28)^2], experienced when the object begins to be pulled upwards and the min will be 1-(0.013/g)[(2*pi*28)^2]. This makes sense, because you feel heaviest in an elevator/lift when it accelerates upwards and lightest when it accelerates downwards.
All you need to do now is put in g=9.8 and use a calculator.
x = 0.013sin(28*2pi*t)
To find this I've done the double integration of the harmonic equation x'' + wx = 0 for w = (2*pi*f)^2 with standard boundary conditions (I can walk you through this if needs be).
Force due to oscillation = F = ma = mx'' = m(-wx) = -m[(2*pi*28)^2]*0.013sin(28*2pi*t). This is extremised when |sin|=1.
So it'll be max/min when F = -/+0.013m[(2*pi*28)^2]. When F = -0.013m[(2*pi*28)^2] then it's being accelerated upwards, When F=-0.013m[(2*pi*28)^2] then it's being pulled downwards.
Now we add in gravity so the total force an object of mass m will feel is mg-m[(2*pi*28)^2]*0.013sin(28*2pi*t). The sign on the mg is because when the force is zero, you feel 1g (even though it's pulling you down). To get the G force found you divide this by mg (ie you divide by the weight of the object since 5g = 5 times usual weight etc) to get 1-(1/g)[(2*pi*28)^2]*0.013sin(28*2pi*t)
Thus the max G's will be 1+(0.013/g)[(2*pi*28)^2], experienced when the object begins to be pulled upwards and the min will be 1-(0.013/g)[(2*pi*28)^2]. This makes sense, because you feel heaviest in an elevator/lift when it accelerates upwards and lightest when it accelerates downwards.
All you need to do now is put in g=9.8 and use a calculator.
Thankyou very much AlphaNumeric
A question,
By 28 movements per second, I mean zero > + peak > zero OR - peak > zero > + peak
If were talking of a sine wave it would be just 1/2 a cycle, does this mean I should use 14 in the equation rather than 28?
A question,
By 28 movements per second, I mean zero > + peak > zero OR - peak > zero > + peak
If were talking of a sine wave it would be just 1/2 a cycle, does this mean I should use 14 in the equation rather than 28?
Yes, you should put in 14 instead.
I found someone elses working on this a little different, can anyone tell me if it is wrong?
I can't see where he got his final formula, and it gives an answer way off the mark.
Can anyone see where this guy is coming from?
I can't see where he got his final formula, and it gives an answer way off the mark.
QUOTE
To calculate "G" acceleration: G = ((F*6.28)^2 * A/2)/9.8 where F is the frequency in Hz and A is the amplitude in meters. If you enter the amplitude in millimeters (or mm), the value of the expression is approximated with a 1% accuracy by
G = F/10 * F/10 * 0.2 * mm
For example at 30 Hz and 2 mm you obtain
G= 3 * 3 * 0.2 * 2 = 3.6 g
G = F/10 * F/10 * 0.2 * mm
For example at 30 Hz and 2 mm you obtain
G= 3 * 3 * 0.2 * 2 = 3.6 g
Can anyone see where this guy is coming from?
The first formula looks equivalent to the one AlphaNumeric gives, except the 1G for gravity is missing, and pi is approximated by 3.14 (A is the peak to peak amplitude).
The second also, for less precise values of g and pi.
I get roughly (+/- 1%) the same result with all 3 formulae if the extra 1G is ignored.
The second also, for less precise values of g and pi.
I get roughly (+/- 1%) the same result with all 3 formulae if the extra 1G is ignored.
Ok, you are talking with someone who has not touched a calculator in several years, so please go easy.
Based on Alphanumeric's equation
I get 42.0574 G's @ 28Hz, 13mm amplitude
Based on
I get 42.0574 G's @ 28Hz, 13mm amplitude
Based on
G = ((F*6.28)^2 * A/2)/9.8
I get 20.5080 G's @ 28Hz. 13mm amplitude
or based on
G = F/10 * F/10 * 1.3 * cm
I get 10.192 G's @ 28Hz 13mm amplitude
So could someone please point out (in laymans terms) what I am doing wrong here, and how, as Wibble points out, these two formulas give similar results.
Thanks in advance.
The A here is the peak to peak amplitude, 26mm (0.026m)
The A here is the peak to peak amplitude, 26mm (0.026m)
G = F/10 * F/10 * 0.2 * mm
This one is a bit vague, mm is being used as a variable for "A in millimeters", i.e. 0.026. The 0.2 should stay as 0.2.
Do you mean - peak to - peak amplitude? The amplitude of one full cycle rather than just peak to peak. Because in my example the peak to peak amplitude is 13mm, not 26mm.
There is still a problem in one of these formulas
Do you mean - peak to - peak amplitude? The amplitude of one full cycle rather than just peak to peak. Because in my example the peak to peak amplitude is 13mm, not 26mm.
There is still a problem in one of these formulas
Thus the max G's will be 1+(A/g)[(2*pi*F)^2]
To show you what I mean, lets use the example given here, where F=30 and A=2mm
To show you what I mean, lets use the example given here, where F=30 and A=2mm
To calculate "G" acceleration: G = ((F*6.28)^2 * A/2)/9.8 where F is the frequency in Hz and A is the amplitude in meters. If you enter the amplitude in millimeters (or mm), the value of the expression is approximated with a 1% accuracy by
G = F/10 * F/10 * 0.2 * mm
For example at 30 Hz and 2 mm you obtain
G= 3 * 3 * 0.2 * 2 = 3.6 g
Answer 3.6 G's
Now put these values into Alphanumeric's equation
G= 1 + (0.002/9.8)[(2*pi*F)^2]
Answer 1 + 7.25 G's
Obviously the first answer is around half of the second answer, so which way is correct?
Ahh, thats where I screwed up.
Thanks again Wibble
Based on Alphanumeric's equation
QUOTE
Thus the max G's will be 1+(0.013/g)[(2*pi*28)^2]
I get 42.0574 G's @ 28Hz, 13mm amplitude
Based on
QUOTE (->
| QUOTE |
| Thus the max G's will be 1+(0.013/g)[(2*pi*28)^2] |
I get 42.0574 G's @ 28Hz, 13mm amplitude
Based on
G = ((F*6.28)^2 * A/2)/9.8
I get 20.5080 G's @ 28Hz. 13mm amplitude
or based on
G = F/10 * F/10 * 1.3 * cm
I get 10.192 G's @ 28Hz 13mm amplitude
So could someone please point out (in laymans terms) what I am doing wrong here, and how, as Wibble points out, these two formulas give similar results.
Thanks in advance.
QUOTE
G = ((F*6.28)^2 * A/2)/9.8
The A here is the peak to peak amplitude, 26mm (0.026m)
QUOTE (->
| QUOTE |
| G = ((F*6.28)^2 * A/2)/9.8 |
The A here is the peak to peak amplitude, 26mm (0.026m)
G = F/10 * F/10 * 0.2 * mm
This one is a bit vague, mm is being used as a variable for "A in millimeters", i.e. 0.026. The 0.2 should stay as 0.2.
QUOTE
The A here is the peak to peak amplitude, 26mm (0.026m)
Do you mean - peak to - peak amplitude? The amplitude of one full cycle rather than just peak to peak. Because in my example the peak to peak amplitude is 13mm, not 26mm.
There is still a problem in one of these formulas
QUOTE (->
| QUOTE |
| The A here is the peak to peak amplitude, 26mm (0.026m) |
Do you mean - peak to - peak amplitude? The amplitude of one full cycle rather than just peak to peak. Because in my example the peak to peak amplitude is 13mm, not 26mm.
There is still a problem in one of these formulas
Thus the max G's will be 1+(A/g)[(2*pi*F)^2]
QUOTE
G = ((F*6.28)^2 * A/2)/9.8
To show you what I mean, lets use the example given here, where F=30 and A=2mm
QUOTE (->
| QUOTE |
| G = ((F*6.28)^2 * A/2)/9.8 |
To show you what I mean, lets use the example given here, where F=30 and A=2mm
To calculate "G" acceleration: G = ((F*6.28)^2 * A/2)/9.8 where F is the frequency in Hz and A is the amplitude in meters. If you enter the amplitude in millimeters (or mm), the value of the expression is approximated with a 1% accuracy by
G = F/10 * F/10 * 0.2 * mm
For example at 30 Hz and 2 mm you obtain
G= 3 * 3 * 0.2 * 2 = 3.6 g
Answer 3.6 G's
Now put these values into Alphanumeric's equation
QUOTE
Thus the max G's will be 1+(A/g)[(2*pi*F)^2]
G= 1 + (0.002/9.8)[(2*pi*F)^2]
Answer 1 + 7.25 G's
Obviously the first answer is around half of the second answer, so which way is correct?
Both are correct, but A in one is not the same as A in the other.
AlphaNumeric uses zero to peak amplitude. (And interprets the intial question as saying that 13mm is the zero to peak amplitude, as do I).
The other uses A = peak to peak amplitude ( = 2 * zero to peak amplitude, hence the A/2 term).
You could consider the second to be slightly incorrect, as zero to peak is the normal usage for amplitude.
Peak to peak means +peak to -peak.
AlphaNumeric uses zero to peak amplitude. (And interprets the intial question as saying that 13mm is the zero to peak amplitude, as do I).
The other uses A = peak to peak amplitude ( = 2 * zero to peak amplitude, hence the A/2 term).
You could consider the second to be slightly incorrect, as zero to peak is the normal usage for amplitude.
Peak to peak means +peak to -peak.
QUOTE
AlphaNumeric uses zero to peak amplitude.
Ahh, thats where I screwed up.
Thanks again Wibble
Sorry to bring up and old thread, but I have found the information provided by AlphaNumeric to be extremely helpful in something I am trying to accomplish and I needed to get into contact with you and the board would not allow me to PM or e-mail you.
AlphaNumeric, I would like to use your explanation you provided in a publication outlining the force that is exerted on your body when using an oscillating whole body vibration exercise machine. Would you be interested in giving me permission to use it? I would like to use appropriate citation and give you credit for your explanation. I can be contacted at by e-mail at patrick@zaazstudios.com
AlphaNumeric, I would like to use your explanation you provided in a publication outlining the force that is exerted on your body when using an oscillating whole body vibration exercise machine. Would you be interested in giving me permission to use it? I would like to use appropriate citation and give you credit for your explanation. I can be contacted at by e-mail at patrick@zaazstudios.com
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