G force Calculation
Drifting
Joined: Oct 2003
Posts: 1,486
Likes: 10
From: Big Pine Key FL
I assumed that the acceleration was constant.
BTW gravitational constant 1G=32ft/sec*sec is an approximate value near the surface of the earth
We're talking about position as a function of time.
Acceleration is a unit of CHANGE of Velocity!
Constant velocity = ZERO acceleration
In order to change velocity some force has to be applied to change velocity
Speed is commonly used to talk about the magnitude (size/amount) of the instantanious velocity.
Average acceleration is infact as I stated: change in velocity over the change in time.
To know acceleration at 60' requires that we know the speed at 60' which I approximated. I didnt think about it to long so it may have been a bad approximation
OR if I knew the force that the tires were imparting against the ground I could get at it that way.
This might help... AND it even uses a Corvette as an example...
Approx. 0.5 G gets a 13 sec quarter
http://www.google.com/url?sa=t&ct=re...Ua-6Fr4dmcc82A
BTW 3 G's is a lot...
later
BTW gravitational constant 1G=32ft/sec*sec is an approximate value near the surface of the earth
We're talking about position as a function of time.
Acceleration is a unit of CHANGE of Velocity!
Constant velocity = ZERO acceleration
In order to change velocity some force has to be applied to change velocity
Speed is commonly used to talk about the magnitude (size/amount) of the instantanious velocity.
Average acceleration is infact as I stated: change in velocity over the change in time.
To know acceleration at 60' requires that we know the speed at 60' which I approximated. I didnt think about it to long so it may have been a bad approximation
OR if I knew the force that the tires were imparting against the ground I could get at it that way.
This might help... AND it even uses a Corvette as an example...
Approx. 0.5 G gets a 13 sec quarter
http://www.google.com/url?sa=t&ct=re...Ua-6Fr4dmcc82A
BTW 3 G's is a lot...
later
Last edited by 84rzv500r; Dec 21, 2006 at 09:29 AM.
Melting Slicks
Joined: Jan 2004
Posts: 2,672
Likes: 1
From: New Cumberland PA
Example:
Vf is 60mph
Vi is 60mph
t is 1 hour
(60-60)/1 =0
Your saying the average velocity is Zero, when we all know that it is 60.
That equation is for acceleration and in the above example the Acceleration is 0.
Racer
Joined: Apr 2005
Posts: 464
Likes: 0
From: Charleston South Carolina
My friend has the G-force device built into his head unit in the radio on his 6 cyl firebird with ugraded ecu and exhaust. He would pull .88 Gs max on takeoff. Now my 78 Chevy van can beat his firebird and can actually pull a 15.00 in the 1/4 and the 60 foot time was 2.02. So using the master converter program with the data from my van = .93 Gs. Cut my 60 foot time in half and you get the original posters time and you also get 1.86 Gs. THIS IS A REAL WORLD EXAMPLE AND NOT GUESS WORK. So what some of you guys are saying is that a device that is specifically designed to measure G-force is wrong?
Le Mans Master
Joined: Mar 2004
Posts: 7,470
Likes: 1,490
From: Little Rock AR
I'm looking for some help with an equation I've been asked to solve! I guess I need the help of a physics goru.
Ok, a friend of mine has a rear engine dragster that his 16 year old daughter drives. The car with driver weighs 1770 lbs. The 60 foot quauter mile time is 1.10 seconds. The 1/4 mile time is 7.965 seconds. The top speed at the end is 165mph.
Is this enough information to calculate the g's that the driver has to endure on the launch of this car. Any idea on what level of negative g's when the parachute is deployed from 165 mph!
Thanks so much if you can help me!
Indy
Ok, a friend of mine has a rear engine dragster that his 16 year old daughter drives. The car with driver weighs 1770 lbs. The 60 foot quauter mile time is 1.10 seconds. The 1/4 mile time is 7.965 seconds. The top speed at the end is 165mph.
Is this enough information to calculate the g's that the driver has to endure on the launch of this car. Any idea on what level of negative g's when the parachute is deployed from 165 mph!
Thanks so much if you can help me!
Indy
Launch G force - We can only make assumptions to come up with a best guess. The accelleration is not constant, it is variable. The G force is a moving target.
Parachute G force - how much drag does the parachute provide? We do know how much kinetic energy you have at he end of the run but a little parachute will make less G's than a big one.
Even then, the chute will hit pretty hard when it first opens. After that I think it is just the amount of air friction provided by the vehicle/chute which will vary with the square of the velocity.Not enough info.
-Mark.
Le Mans Master
Joined: Jul 2006
Posts: 8,117
Likes: 104
From: charlotte north carolina
use d=1/2 a*t*t
distance equals one half acceleration times time squared.you can use metric or english units.
the problem is that this car's acceleration is NOT going to be linear. there is no way that you can even come close to calculating the true acceleration due to tire slippage. i'm sure that after a split second of slip, when the tires grab there is a tremendous spike in acceleration that only on-board instruments could determine.
distance equals one half acceleration times time squared.you can use metric or english units.
the problem is that this car's acceleration is NOT going to be linear. there is no way that you can even come close to calculating the true acceleration due to tire slippage. i'm sure that after a split second of slip, when the tires grab there is a tremendous spike in acceleration that only on-board instruments could determine.
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
Thanks for the link! Interesting article, with some clear explanations. I like his simplified equation:
d(feet) = 16a (Gs) t (seconds) ^2
When I plug in 60ft and 1.1 seconds and solve for acceleration I get 3.1 Gs. Does that look right? (And that will keep me from challenging TheoUK and asking him to show his work!)
Sorry about my "avg velocity" mistake, and thanks for the clear example mandm1200 so I didn't have to think too hard to realize my error!!
For the 60ft time I think it is reasonable to assume constant acceleration over the 60ft, realizing that the answer represents an average, and sure, Gs will be higher over the first 30 ft than the 2nd 30ft.
Now to figure out what I was doing wrong to get 3.4 Gs, instead of 3.1 Gs. I think I was calculating a 1 sec 60ft time rather than 1.1 sec?
d(feet) = 16a (Gs) t (seconds) ^2
When I plug in 60ft and 1.1 seconds and solve for acceleration I get 3.1 Gs. Does that look right? (And that will keep me from challenging TheoUK and asking him to show his work!)
Sorry about my "avg velocity" mistake, and thanks for the clear example mandm1200 so I didn't have to think too hard to realize my error!!
For the 60ft time I think it is reasonable to assume constant acceleration over the 60ft, realizing that the answer represents an average, and sure, Gs will be higher over the first 30 ft than the 2nd 30ft.
Now to figure out what I was doing wrong to get 3.4 Gs, instead of 3.1 Gs. I think I was calculating a 1 sec 60ft time rather than 1.1 sec?
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
And as far as answering the actual question:
I think it is helpful to know that the driver will have to endure at the very least 3.1 gs (but in reality more) if the 60ft time is 1.1 secs.
Is this enough information to calculate the g's that the driver has to endure on the launch of this car
Pro
Joined: May 2003
Posts: 601
Likes: 0
From: Nottingham
When I plug in 60ft and 1.1 seconds and solve for acceleration I get 3.1 Gs. Does that look right? (And that will keep me from challenging TheoUK and asking him to show his work!)
I wondered if someone might ask me to show my workings (it's a bit like being an undergraduate again!) I didn't really have time to put them down in words yesterday.
Your answer from the equation there is right, again for the AVERAGE acceleration in that first 1.1 second increment. As lots of people have said, that of course isn't the maximum G that is likely to be experienced. But its a good order of magnitude answer for the general level of acceleration we're looking at.
As for workings, well, v = d/t, so d = v.t, or to put it another way, the AREA UNDER a graph of velocity against time will give you the distance travelled to that point.
So if you assume the velocity goes up linearly from zero to "V" (unknown) at 1.1 seconds (the 60' mark), then the total distance travelled is equal to the triangular area under that graph. So,
60 feet = 1/2 X V X t (area of triangle is half the base times the height)
Which you can solve for V, giving 109feet per sec.
That change in velocity, from zero to 109fps, occured in 1.1 seconds, so the acceleration was 109/1.1 = 99.17 f/s/s = 3.1Gs
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Drifting
Joined: Oct 2003
Posts: 1,486
Likes: 10
From: Big Pine Key FL
I wondered if someone might ask me to show my workings (it's a bit like being an undergraduate again!) I didn't really have time to put them down in words yesterday.
Your answer from the equation there is right, again for the AVERAGE acceleration in that first 1.1 second increment. As lots of people have said, that of course isn't the maximum G that is likely to be experienced. But its a good order of magnitude answer for the general level of acceleration we're looking at.
As for workings, well, v = d/t, so d = v.t, or to put it another way, the AREA UNDER a graph of velocity against time will give you the distance travelled to that point.
So if you assume the velocity goes up linearly from zero to "V" (unknown) at 1.1 seconds (the 60' mark), then the total distance travelled is equal to the triangular area under that graph. So,
60 feet = 1/2 X V X t (area of triangle is half the base times the height)
Which you can solve for V, giving 109feet per sec.
That change in velocity, from zero to 109fps, occured in 1.1 seconds, so the acceleration was 109/1.1 = 99.17 f/s/s = 3.1Gs
Team Owner
Joined: Apr 1999
Posts: 21,953
Likes: 1,445
From: Reno Nevada
2024 C3 of the Year Finalist- Modified
1.10 7.95@ 165 tells me a few things if this is not using a throttle stop to acquire these times.
The 60 foot is Okay, 1770 lbs is a heavy dragster. My 220 inch was 1380# with an injected small block. Good ET, but slow mph so I would guess that it's a big ci TQ motor like 500+ BBC. Speed is a function of HP per pound
I was at one of the last NHRA events and the 8.900 sec class were taking off slow using delay throttle stops and someone exceeded 200 mph.
I never liked packing a parachute so we never used ours other than testing. The other thing is you have to stop and dork with the chute on the return road. I just slowed for the turn off and then drove back to the pits. None of that hooking you up and towing.
As for "G" force. Get a G-tech meter for $150 and record the runs. no guesing
The 60 foot is Okay, 1770 lbs is a heavy dragster. My 220 inch was 1380# with an injected small block. Good ET, but slow mph so I would guess that it's a big ci TQ motor like 500+ BBC. Speed is a function of HP per pound
I was at one of the last NHRA events and the 8.900 sec class were taking off slow using delay throttle stops and someone exceeded 200 mph.
I never liked packing a parachute so we never used ours other than testing. The other thing is you have to stop and dork with the chute on the return road. I just slowed for the turn off and then drove back to the pits. None of that hooking you up and towing.
As for "G" force. Get a G-tech meter for $150 and record the runs. no guesing
Burning Brakes
Joined: Jun 2004
Posts: 1,012
Likes: 5
From: Granbury, TX
I wondered if someone might ask me to show my workings (it's a bit like being an undergraduate again!) I didn't really have time to put them down in words yesterday.
Your answer from the equation there is right, again for the AVERAGE acceleration in that first 1.1 second increment. As lots of people have said, that of course isn't the maximum G that is likely to be experienced. But its a good order of magnitude answer for the general level of acceleration we're looking at.
As for workings, well, v = d/t, so d = v.t, or to put it another way, the AREA UNDER a graph of velocity against time will give you the distance travelled to that point.
So if you assume the velocity goes up linearly from zero to "V" (unknown) at 1.1 seconds (the 60' mark), then the total distance travelled is equal to the triangular area under that graph. So,
60 feet = 1/2 X V X t (area of triangle is half the base times the height)
Which you can solve for V, giving 109feet per sec.
That change in velocity, from zero to 109fps, occured in 1.1 seconds, so the acceleration was 109/1.1 = 99.17 f/s/s = 3.1Gs
This sums it up quite nicely. If you have the change in velocity and time, you may calculate acceleration. Amaze your friends!
For example:
My 60 ft time was 5.6 sec, 61 MPH
1 feet per second = 0.681818182 mph
61 mph = 89.46 fps
89.46 fps/ 5.6 sec = 15.97 fps/ sec
15.97 fps/ sec/ 32.2 fps/sec = .5 g
1/8 mi time was 8.75 at 77 mph
77 mph = 113.23 fps per above
(113.23 fps/ 8.75 sec)/ 32.2 = .4 G
Which proves.........................
I need a faster car!
Instructor
Joined: Jun 2006
Posts: 123
Likes: 0
From: Westminster MD
Velocity is the distance traveled divided by the time it took to travel, which is:
V = D/t
Acceleration is the change in velocity divided by the time of the change, which is:
A = V/t
This can be expressed as:
A = V/t = (D/t)/t = D/(t*t) = D/t^2
In words: acceleration is the change in distance divided by time squared.
So the simple way to calculate this is:
A = D/t^2 = 60/(1.1^2) = 60/(1.1*.1.1) = 60/1.21 = 49.6 ft/sec^2
since 1G = 32.2 ft/sec^2
the dragster's acceleration in Gs is: 49.6/32.2 = 1.54 G
Of course the above assumes the start is at 0 MPH and the acceleration is linear for the first 1.1 seconds.
(Yes I confess I'm an engineer.)
V = D/t
Acceleration is the change in velocity divided by the time of the change, which is:
A = V/t
This can be expressed as:
A = V/t = (D/t)/t = D/(t*t) = D/t^2
In words: acceleration is the change in distance divided by time squared.
So the simple way to calculate this is:
A = D/t^2 = 60/(1.1^2) = 60/(1.1*.1.1) = 60/1.21 = 49.6 ft/sec^2
since 1G = 32.2 ft/sec^2
the dragster's acceleration in Gs is: 49.6/32.2 = 1.54 G
Of course the above assumes the start is at 0 MPH and the acceleration is linear for the first 1.1 seconds.
(Yes I confess I'm an engineer.)
Melting Slicks
Joined: Jan 2004
Posts: 2,672
Likes: 1
From: New Cumberland PA
The Velocity that was used to calculate the acceleration was erronous. The figure used was average velocity over the full durataion of the time period. In order to calculate acceleration, the velocity at the 60' mark is needed and not the average velocity during the full 60'. When calculating acceleration the velocity at certain points must be known or calcualted.
Edit: (for clarification)
V=d/t This equation is only used and only useful with the assumption the acceleration is 0.
Last edited by mandm1200; Dec 22, 2006 at 11:33 PM.
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
So the simple way to calculate this is:
A = D/t^2 = 60/(1.1^2) = 60/(1.1*.1.1) = 60/1.21 = 49.6 ft/sec^2
since 1G = 32.2 ft/sec^2
the dragster's acceleration in Gs is: 49.6/32.2 = 1.54 G
Of course the above assumes the start is at 0 MPH and the acceleration is linear for the first 1.1 seconds.
(Yes I confess I'm an engineer.)
I think mandm1200 has the answer: You need to use the speed at the 60 ft mark, not the average speed... or "velocity" if you prefer.
