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Can a dragster really accelerate to 100 mph in a second? How many G is it pulling off the line?
These are commonly asked questions which I am going to try and answer using some school book maths and a bit of help from Ask Jeeves.
At the time of writing (April 2005), the fastest run outside of America took place on 6 September 2003 at 14:24 when Kim Reymond astounded the Santa Pod crowd with a magnificent 4.645 second 317.56 mph charge.
The incremental times and speeds on that run were as follows :
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Information courtesy of Andy Marrs of TSI Timers and Eurodragster.com
You would think with all that information that working out anything to do with the run would be a piece of cake. Wrong! The problem is that no dragster accelerates at a constant rate throughout the run. We all know that at the hit a top fueller leaves like a scalded cat, but you can clearly see from the figures in the table above that while Kim was still flying at the eighth mile, he had ‘only’ added another 56 mph to his speed at the end of the quarter. It is worth remembering that these cars are set up for elapsed times not terminal speeds, therefore the tune-up is all about getting the car moving as quickly as possible as soon as possible and then maintaining that speed.
This means that to work out the answers to the two questions posed at the top of this page you have to make assumptions which you know are not true. The question is - how much do they affect the results.
So, enough waffle, let’s have a go at answering the questions.
Can a dragster accelerate to 100 mph in a second?
Yes, err probably, almost certainly, yes . . . I think so.
What we need to do is fill in one of the statistics missing from the table above and that is the speed the car was doing at the 60 foot mark.
We know that Kim took 0.855 seconds to cover 60 feet so we can work out the average speed. Firstly, if we want the answer in miles per hour we have to convert the distance into miles and the time into hours both of which are pretty foreign concepts for top fuel dragsters.
If there are 1,320 feet in a quarter mile there must be 5,280 feet in a mile. If we divide 60 feet by 5,280 feet we find that our 60 foot dash is 0.011364 miles. This sounds about right because 60 feet is a very small fraction of a mile.
There are 60 minutes in an hour and 60 seconds in a minute which means there are 3,600 seconds in an hour. If we do the same as above and divide 0.855 seconds by 3,600 seconds, we find that Kim took all of 0.000238 hours to reach the 60 foot beam.
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47 mph for a top fuel dragster? That doesn’t sound very fast but remember this is the average speed. If you assume that the car was at a standstill at the start of the 60 feet, you would have to double that number to get its speed at the 60 foot mark which makes a much more respectable 95.69378 mph. That sounds a bit more like it.
The problem is that already assumptions are having to be made. As my old guv’nor used to say "assume makes an ass of u and me". We have assumed that the car was accelerating at a fixed rate which it was not, and that it was at a standstill at the start of the first 60 feet which it cannot have been or it could not have broken the start beam.
So far it looks as if the car might have been doing 95-odd miles per hour 60 feet into the run. That assumed speed occurred after 0.855 seconds so if we assume (again!) that the car continued to accelerate at the same rate for the next 0.145 seconds the speed after 1 second would be :
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Can this be right? The short answer is no. I won’t bore you with the calculations, but if you apply the same logic to the entire pass you come up with a terminal speed of 387 mph which is real Sammy Miller stuff and far faster than the 317 mph the car actually achieved. Interestingly, the position on the eighth mile is better with a calculated 286 mph which is not too far removed from the actual 261 mph.
It seems from this that as the distance gets shorter the calculation gets more accurate. Also, the fact that the car is actually moving at the beginning of the 60 feet (don’t ask me how fast!) would tend to raise the speed slightly at the 60 foot beam.
If you are still awake after all that, my conclusion is that on this run there is little doubt that Kim Reymond’s car was traveling at over 100 mph at 1 second into the run. Phew!
How many G is it pulling off the line?
Speed is measured in miles per hour or feet per second or kilometres per year or whatever you like. Acceleration is the rate at which speed is increasing and, in the case of feet per second, is measured in feet per second per second. 1G is said to be equivalent to an acceleration of approximately 32.14 feet per second per second at sea level.
So if we can find out how fast the car was accelerating we can soon work out the G-force being applied to Mr Reymond. The problem is we are trying to find the G-force at the hit but we only have information for the first 60 feet. More averages and assumptions coming up.
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We need to work this out in feet per second for reasons which will become obvious in a moment. 60 mph is 88 feet per second therefore 95.69378 mph (the calculated speed at 60 feet) is 140.3509 feet per second.
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As stated above, the force of G is equivalent to an acceleration of 32.14 feet per second per second. Therefore Kim Reymond experienced circa 5.11 G when he blasted off the line on the afternoon of 6 September 2003.
Because of the assumptions made above this figure cannot possibly be 100% accurate but I think it gives some idea of the G-force experienced by a driver of a top fuel dragster at the hit.
I shot this picture of Kim Reymond as he performed his burn out before unleashing that epic lap on the unsuspecting crowd at Santa Pod.
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And finally . . .
If the fueller was doing 95.69378 mph after 0.855 seconds that means its 0 to 60 time was fractionally over half a second (0.5361 seconds actually). Try laying that on some of your mates next time you are in the pub!
My thanks to Andy Marrs of TSI Timers for kindly vetting this article prior to publication.
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