G force Calculation
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Pro
Joined: Oct 2000
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G force Calculation
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
Team Owner
Joined: Aug 2006
Posts: 24,125
Likes: 15
From: Columbia Missouri
Get and install one of the new Kenwood Excelon head units. They all have something called a G-Force analyzer. You set it, and take off. It'll measure all directions, and show you as you do it or in graph form when you're done. It's really funny when kids put them in their Civics and drive around a mall trying to get good numbers. Hah. I tried it in my Durango and actually did good. Pulled a lot on launch.
Racer
Joined: Apr 2005
Posts: 464
Likes: 0
From: Charleston South Carolina
2 Gs would be 64 fps if 1 G is 32 fps. If the car can get to 60 ft in 1.1 seconds then it's pulling about 1.8 Gs up to the 60 ft mark. The G force will change as the horsepower is being overcome by drag and result in a lower G-force by the time it gets to the 1/4 mile mark. The most G-force will be felt in the beginning of the run. Or just get a G-Timer and plug it into a 12 volt source. I think they are relatively cheap.
Last edited by strokervette; Dec 20, 2006 at 10:47 PM. Reason: Added info
Team Owner
Joined: Jul 1999
Posts: 26,545
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From: Huntersville NC
1G, or the acceleration of one G, is 32feet per second per second. That means after one second you are traveling at a speed of 32 ft per second. It doesn't mean you have traveled 32 ft in that first second. In fact, you must have traveled less than 32 feet. It's calculus, which I am not good at.
But I do know that there are 5280 ft in a mile, and if after one second you are going 32ft per second, you will go 1920 ft in 60 seconds and 115,200 ft in one hour which is only 21.8 miles, so you after 1 second you are only going 21.8 miles per hour and I don't know how far you have gond. Now if you keep accelerating at 1 G, after 2 seconds you will be going 64 ft per second or 43.6 mph. So after 7 seconds at 1G you are going 153 mph. Thing is, I don't think a dragster accelerates uniformly. I think highest Gs are off the line, then it tails off a little.
But I do know that there are 5280 ft in a mile, and if after one second you are going 32ft per second, you will go 1920 ft in 60 seconds and 115,200 ft in one hour which is only 21.8 miles, so you after 1 second you are only going 21.8 miles per hour and I don't know how far you have gond. Now if you keep accelerating at 1 G, after 2 seconds you will be going 64 ft per second or 43.6 mph. So after 7 seconds at 1G you are going 153 mph. Thing is, I don't think a dragster accelerates uniformly. I think highest Gs are off the line, then it tails off a little.
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
If you are going 64 ft per second, then you will cover 64 ft .... in one second!
But he starts at zero, so he has to achieve a bit more than 2Gs acceleration to cover that first 60ft in just over 1 sec! If he is accelerating at 2Gs then he will be accelerating at 64ft per second, but he doesnt reach that speed of 64ft per second until the entire second has elapsed. Somebody use calculus pls!!
So... according to my calculations you have to pull nearly 4Gs to get your 60ft down to 1 second.
In meters, 10m/s = 1G, after 1 second you only cover 5 meters.
40m/s = 4Gs, after 1 second you cover 20m, which is more than 60ft... so probly around 3.5 g's. Sounds crazy!
Last edited by PRNDL; Dec 20, 2006 at 11:11 PM.
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Racer
Joined: Apr 2005
Posts: 464
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From: Charleston South Carolina
G force does start at a determined speed which in this case it is 0mph. It's like if they were already going 15 mph when they crossed the start and and lets say it took the 1 second to get to 60 ft still at 15 mph, then they are still not achieving any Gs. G force is acceleration force. So it don't matter where you start at, 0 mph or 100 mph. Being that we are talking about 0 mph at point A and it takes them 1.1 sec to get to point B (60 ft) then the average G-force is about 1.8 Gs according to one of the previous posts. They might not be getting traction at the first 10 feet but then hook up at the last 25 feet, so the G-force is changing the whole time and we can only estimate the average unless you have a device like Durangoboy or me have mentioned where it records the whole ride to 60 feet or even the 1/4 mile. My friend had one in his head unit in the radio and when he would take off from a stop it would peak out after like a second or sothen slowly drop off. So the max G-force would be somewhere in the 0-60 foot time for sure. Or if you have the engine dynoed you could see what rpm the engine makes max torque and see where on the run in first gear does it come in. I would think that would be a good indicator where your max G-force is.
Last edited by strokervette; Dec 20, 2006 at 11:27 PM.
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
1.8 Gs won't get you there. 2 Gs will get you to 64 ft/second in one second, and you will have averaged 32ft per second during that time, so you will only have covered 32 ft in that first second. 2Gs will barely get you halfway there! But I will leave it at that. I think all the math whizes are in OT.
Drifting
Joined: Oct 2003
Posts: 1,486
Likes: 10
From: Big Pine Key FL
1g is a unit of acceleration 32 ft/second PER SECOND
I'm gonna skip tha calculus...
Aave = (Vf-Vi)/t
Where:
Aave is the average Acceleration
Vf is the final velocity
Vi is the initial velocity
and t is the time
Soooooo....
again the acceleration of a dragster is NOT constant as it changes velocity rather quickly off the line and then tapers off
in the posters example.....
Vf = 165mi/hr * 5280ft/1mi * 1hr/3600sec => 242 ft/sec
Vi = 0 mi/hr => 0 ft/sec
t=7.9 sec
Aave=(242-0)ft/sec
-------------------
7.9
= 242/7.9 = 30.6 ft/sec/sec and 1g is 32 ft/sec/sec
remember it was the average acceleration over the entire 7.9 seconds.
so this is where the calculus comes in as you can calculate the average acceleration of smaller chunks of time...
Back to the average.... We can do the 60ft acceleration the same way
The rest is left to the reader....
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
But you didn't answer his question! I think your calculations show that less than 1G acceleration will get you to 7.9 @ 165 mph. I did say earlier that, "if you keep accelerating at 1 G, after 2 seconds you will be going 64 ft per second or 43.6 mph. So after 7 seconds at 1G you are going 153 mph."
But the original post asks, "Is this enough information to calculate the g's that the driver has to endure on the launch of this car."
I think the 1.1 sec 60ft time is a better way to answer that question. This is where strokervette and I disagree. I think it has to be around 3.5 Gs. Can you check the math?
But the original post asks, "Is this enough information to calculate the g's that the driver has to endure on the launch of this car."
I think the 1.1 sec 60ft time is a better way to answer that question. This is where strokervette and I disagree. I think it has to be around 3.5 Gs. Can you check the math?
Drifting
Joined: Oct 2003
Posts: 1,486
Likes: 10
From: Big Pine Key FL
It's an approximation....
Vf=60/1.1=54.5 ft/sec
Vi=0
t=1.1
54.5-0
--------- = Aave
1.1 sec
=49.5 ft/sec/sec
49.5(ft/sec/sec) 1G
------------------------------ = 1.5G ave as the ft/sec/sec cancel out and your left with G's
32 (ft/sec/sec)
Vf=60/1.1=54.5 ft/sec
Vi=0
t=1.1
54.5-0
--------- = Aave
1.1 sec
=49.5 ft/sec/sec
49.5(ft/sec/sec) 1G
------------------------------ = 1.5G ave as the ft/sec/sec cancel out and your left with G's
32 (ft/sec/sec)
Racer
Joined: Apr 2005
Posts: 464
Likes: 0
From: Charleston South Carolina
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
Last edited by strokervette; Dec 21, 2006 at 02:06 AM. Reason: Forgot link
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
So it looks like my educated guess was correct
Again, accelerating at 60 feet per second per second does not mean you cover 60 ft in one second. It means that after one second your speed has increased from zero to 60 ft per second. You have been accelerating for one second. Your average speed is 30 ft/second over that entire second. So even tho you are now traveling at a speed of 60 ft/second you have only gone 30 ft. Your 1.8 Gs only gets you halfway there.
With all the fancy calculations, I think 84rzv500r is making the same error. I see the average speed is 54.5 ft/second over that first 60 ft, but not sure why you would divide a second time by 1.1. I may have to seek enlightenment elsewhere. I will gladly admit I am wrong if I am wrong.
If you are going to average 54.5 ft/second over the first 60 ft, then you will be traveling at 109 ft per second at the 60 ft mark. 109 ft/sec = 3.4 Gs
Last edited by PRNDL; Dec 21, 2006 at 03:03 AM.
Racer
Joined: Apr 2005
Posts: 464
Likes: 0
From: Charleston South Carolina
We are not looking to find out how much distance he covers in one second, we know that it's ABOUT 60 feet. But what kind of G-force is being applied. Inconsequentially, it also does not matter how much the item weighs or how much drag there is because we already have a model to go by,so we can rule that out also. Gravitation is a unit by itself. According to a calculator/converter anything would feel 1.86 Gs to accomplish 60 feet in 1 second. It's that simple !!!
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
It should be obvious that if you start at a speed of zero and you are accelerating at a rate of 60 ft per second per second, that will not get you 60 feet in the first second.
Team Owner
Joined: Jul 1999
Posts: 26,545
Likes: 46
From: Huntersville NC
(Vf-Vi)/t gives your average velocity, not your average acceleration.
Let's work backwards:
1. Do we agree that 3.4 Gs is 109 ft/sec^2 ? (109 ft per second squared)
2. So at that rate of acceleration, after one second you are going 109 ft per second, right?
3. Your average velocity is (109-0)/2 = 54.5 feet per second. Average velocity.
That means in one second you have gone 54.5 ft.... 1.1 secs you go 60 ft.
Why am I still banging my head on the wall?
Last edited by PRNDL; Dec 21, 2006 at 04:44 AM.
Pro
Joined: May 2003
Posts: 601
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From: Nottingham
Given that 60 foot time, the AVERAGE acceleration over the first sixty feet is 3.08 G's.
That assumes that you accelerate at a constant rate from the start to the sixty foot line. Sticking with that assumption, you'd be doing a little over 70mph at the 60' line! Swift!
In the real world though, your friend's daughter will pull MORE than 3 g's at the very start, and slightly less by the time she's made sixty feet. Remember 3.08 G's is just the average over the first 1.1 seconds. You can only really get a more accurate answer than "over 3 G's" if you had the time for the 30 foot, or 20, or 10 foot mark etc. The shorter the time/distance interval the more accurate your calcs could be.
Just using a bit of intuition, I'd say that to AVERAGE 3.1 G's over 60 feet, she'd have to pull maybe up nearer 4G off the starting line, which would reduce to something below 3G by the time she's made it to the 60 foot mark.
That assumes that you accelerate at a constant rate from the start to the sixty foot line. Sticking with that assumption, you'd be doing a little over 70mph at the 60' line! Swift!
In the real world though, your friend's daughter will pull MORE than 3 g's at the very start, and slightly less by the time she's made sixty feet. Remember 3.08 G's is just the average over the first 1.1 seconds. You can only really get a more accurate answer than "over 3 G's" if you had the time for the 30 foot, or 20, or 10 foot mark etc. The shorter the time/distance interval the more accurate your calcs could be.
Just using a bit of intuition, I'd say that to AVERAGE 3.1 G's over 60 feet, she'd have to pull maybe up nearer 4G off the starting line, which would reduce to something below 3G by the time she's made it to the 60 foot mark.