What Does 0-60 In 2 Feels Like?
Discussion
Hello everyone, this question is for anyone that has sat in a car that went from 0-60 in 2 secs bracket. What does it feel like? A couple of months ago, I was in downtown DC waiting for the traffic light to turn green. I saw a low rider truck and a Nissan 350z. I guess they were going to gun it when it turns green. I try to keep up with them in my 89 Nissan Maxima. But the low rider disappeared within few seconds. The 350z was 10-20 feet behind during the standstill to 60mph or whatever speed it was doing. I think it's around 60 or so. But anyways, if that low rider truck did 0-60 in about 4 or 5 seconds, it looks pretty quick. So being the Ultima GTR can do 60s in the 3 seconds and 2 seconds bracket, it must be extremely quick. Does your head and body hurt when you hit the throttle all the way down from standstill to get a propre 60 launch? Anyone ever witnessed a GTR doing 60s in 2 secs in person? In the record run videos, the 0-60mph run in 2.7 seconds seemed slow. It doesn't look like 2 seconds rather than 4 or 5 seconds. I'm just curious. As the fastest car I have ever sat in was a 450hp 1994 Toyota Supra TT. So how much distance between a 4-5 seconds 0-60mph car from a 2-3 seconds car? Lets say the GTR 640 races against a 4-5 second car like the Elise, or a Dodge Charger SRT8. 4-5 car distance?
First its different for the passengers than for the driver, the passengers do get thrown around abit.
Secondly you don't just bang out a 2.7 sec 0-60, the tyres need to be hot and sticky, the clutch hot, tyre pressures must be correct and the dampers dialed in.
In my opinion its not the 0-60 that shocks people its the way it does not let off till passed 170mph then they are scared/impressed
Secondly you don't just bang out a 2.7 sec 0-60, the tyres need to be hot and sticky, the clutch hot, tyre pressures must be correct and the dampers dialed in.
In my opinion its not the 0-60 that shocks people its the way it does not let off till passed 170mph then they are scared/impressed
By the calculators I was able to find on the net, 0-60 in 2.7 seconds yields just under 3G of acceleration. That's a bit of a guess. So, if you weigh 200 lbs stangind still, you now weigh 600 under acceleration. However, those numbers are based on acceleration as a constant. The forces are probably greater in the torque band of the engine.
Top Fuel dragsters run the quarter mile at 336 mph and under 4.5 seconds! You do the math...
Top Fuel dragsters run the quarter mile at 336 mph and under 4.5 seconds! You do the math...
GreenV8S said:Huh? What? I have no idea what you just said, but it sounds better...probably right and I was way off.
60 mph is roughly 30 ms-1 or 3 G-seconds, so 3/2.7 gives an average acceleration of just over 1.1G.
I plugged some numbers into a webpage to get results. You know what they say: garbage in, garbage out.
Is it correct to use 32 feet per sec/sec as 1G for horizontal acceleration? I guess that makes sense. The first second, travel distance is 32 feet, or 21.82 mph. Second second, distance is 64 feet, or 43.64 mph. Third second, distance is 96 feet, or 65.45 mph. If so, 60 mph in just under 3 seconds would be around 1G. And, 130.91 mph in 6 seconds would still be 1G.
G Man said:
First its different for the passengers than for the driver, the passengers do get thrown around abit.
Secondly you don't just bang out a 2.7 sec 0-60, the tyres need to be hot and sticky, the clutch hot, tyre pressures must be correct and the dampers dialed in.
In my opinion its not the 0-60 that shocks people its the way it does not let off till passed 170mph then they are scared/impressed
Oh......
Thanks once again GMan for the run at VMAX. I've never experienced acceleration like it! Simply phenomenal. FWIW, I would expect it difficult to reach forward and touch the dash in at least the first 4 gears, under full acceleration.
builder said:
Is it correct to use 32 feet per sec/sec as 1G for horizontal acceleration?
32ft/sec/sec is 1g, whatever direction you are accelerating.
builder said:
The first second, travel distance is 32 feet, or 21.82 mph...
No it isn't, because you haven't been doing 32ft/sec for a whole second. Starting from rest (and assuming constant acceleration) your average speed over the first second is 16ft/sec, not 32.
0-2 in to seconds!! not many road cars can do this mate!! if any!!
My Superkart/Gearbox kart does it in around 2 seconds tho! and It feels fast, very fast! :-)
Fastest time recorded by a SuperKart is 1.9 seconds! not bad really!
What you really want to feel is cornering force!! road car's get around 1g, kart gets around 3g!!
evil on the neck/arms/lungs.....
Then there's the brakes!!! :-)
Cheers
A
>> Edited by DABOSS on Tuesday 13th December 12:01
My Superkart/Gearbox kart does it in around 2 seconds tho! and It feels fast, very fast! :-)
Fastest time recorded by a SuperKart is 1.9 seconds! not bad really!
What you really want to feel is cornering force!! road car's get around 1g, kart gets around 3g!!
evil on the neck/arms/lungs.....
Then there's the brakes!!! :-)
Cheers
A
>> Edited by DABOSS on Tuesday 13th December 12:01
Andrew Noakes said:Of course. That's logical. It's only 32 feet/sec at the 1.0 second mark. The Ultima is quick, but not digital. It's making sense now.
No it isn't, because you haven't been doing 32ft/sec for a whole second. Starting from rest (and assuming constant acceleration) your average speed over the first second is 16ft/sec, not 32.
Then, 32 and 48 feet? So, 65 mph should be reached in a total of 96 feet. Here's an interesting G-Force calculator.
Nope, assuming constant acceleration, it would take about 132.6 feet to get to 60 MPH in 3.0 seconds, or 119.7 feet for the 2.7-second Ultima. 65 MPH would take an even longer distance (not 96', more than 120'!), although I'm not sure why you suddenly changed to 65 MPH in your last post...
This uses the equation D=1/2at^2 (D-istance, A-cceleration, and t-ime) if we start with a velocity of zero. And to find the “a” in that equation, we use the equation A = Vf/t (Vf = velocity at the end of the distance or time, or "final" velocity) -- again, this is only if the Vi (initial velocity) = 0 MPH -- which gives us an acceleration (a) of 0.915 g’s for 0-60 in 3.0 s. and 1.02 g’s for 0-60 in 2.7 s.; then plug this “a” into the above equation (1/2at^2) -- but remember to multiply the 0.915 and 1.02 by 32.2 to get the correct units(imperial/non-metric).
Think of it this way -- which is similar to your second-by-second method of determining this: An Ultima would be going 40 MPH (about 59 ft/s.) at 2 seconds and 60 MPH (about 88 ft/s.) at 3 seconds, for a constantly-accelerating Ultima that takes a full 3 seconds to get to 60 MPH. The average of 59 and 88 ft/s. is about 73.5 ft/s., so how would we go only 48 ft over that 1 second (avg. of 48 ft/s.) if we're accelerating constantly? This appears to be where you made your mistake (i.e., you miscalculated, and/or neglected that we’re starting at 59 ft/s., or 40 MPH; we start there even for the 0.915 g-force/3.0 seconds 0-60 run, let alone for the 1.00-g run you seem to have wanted to review -- so the Vi [initial velcity] is 59 ft./s. if we take only the 40-60 MPH (2.0 s. to 3.0 s.) portion of this run).
Now, if someone did 0-60 in 2.0 s., we'd reach 60 in about 88', just under the 96' you're theorizing; this is 1.37 g-forces, averaged.
>> Edited by jeff-texas on Friday 16th December 04:35
This uses the equation D=1/2at^2 (D-istance, A-cceleration, and t-ime) if we start with a velocity of zero. And to find the “a” in that equation, we use the equation A = Vf/t (Vf = velocity at the end of the distance or time, or "final" velocity) -- again, this is only if the Vi (initial velocity) = 0 MPH -- which gives us an acceleration (a) of 0.915 g’s for 0-60 in 3.0 s. and 1.02 g’s for 0-60 in 2.7 s.; then plug this “a” into the above equation (1/2at^2) -- but remember to multiply the 0.915 and 1.02 by 32.2 to get the correct units(imperial/non-metric).
Think of it this way -- which is similar to your second-by-second method of determining this: An Ultima would be going 40 MPH (about 59 ft/s.) at 2 seconds and 60 MPH (about 88 ft/s.) at 3 seconds, for a constantly-accelerating Ultima that takes a full 3 seconds to get to 60 MPH. The average of 59 and 88 ft/s. is about 73.5 ft/s., so how would we go only 48 ft over that 1 second (avg. of 48 ft/s.) if we're accelerating constantly? This appears to be where you made your mistake (i.e., you miscalculated, and/or neglected that we’re starting at 59 ft/s., or 40 MPH; we start there even for the 0.915 g-force/3.0 seconds 0-60 run, let alone for the 1.00-g run you seem to have wanted to review -- so the Vi [initial velcity] is 59 ft./s. if we take only the 40-60 MPH (2.0 s. to 3.0 s.) portion of this run).
Now, if someone did 0-60 in 2.0 s., we'd reach 60 in about 88', just under the 96' you're theorizing; this is 1.37 g-forces, averaged.
builder said:
No it isn't, because you haven't been doing 32ft/sec for a whole second. Starting from rest (and assuming constant acceleration) your average speed over the first second is 16ft/sec, not 32.
>> Edited by jeff-texas on Friday 16th December 04:35
jeff-texas said:
This uses the equation D=1/2at^2 (D-istance, A-cceleration, and t-ime) if we start with a velocity of zero. And to find the “a” in that equation, we use the equation A = Vf/t (Vf = velocity at the end of the distance or time, or "final" velocity) -- again, this is only if the Vi (initial velocity) = 0 MPH -- which gives us an acceleration (a) of 0.915 g’s for 0-60 in 3.0 s. and 1.02 g’s for 0-60 in 2.7 s.; then plug this “a” into the above equation (1/2at^2) -- but remember to multiply the 0.915 and 1.02 by 32.2 to get the correct units(imperial/non-metric).
Ooopps, sorry guys! Must have logged on to the Open University forum by mistake
Fascinating stuff though...
Haha, are you trying to say I'm a... what's that British word, a "swot"?
Well, I made the first paragraph easier to read for laymen, and that (the number of feet, the results) are what builder was interested in calculating -- probably without my wordy explanation, but then, how's he going to calculate it next time, when he wants to see what drag-racers would call the "60-foot time," for example? The general name if anyone ever needs to google for it in the future, is "kinematics equations".
Well, I made the first paragraph easier to read for laymen, and that (the number of feet, the results) are what builder was interested in calculating -- probably without my wordy explanation, but then, how's he going to calculate it next time, when he wants to see what drag-racers would call the "60-foot time," for example? The general name if anyone ever needs to google for it in the future, is "kinematics equations".gtr-gaz said:
This uses the equation D=1/2at^2 (D-istance, A-cceleration, and t-ime) if we start with a velocity of zero. And to find the “a” in that equation, we use the equation A = Vf/t (Vf = velocity at the end of the distance or time, or "final" velocity) -- again, this is only if the Vi (initial velocity) = 0 MPH -- which gives us an acceleration (a) of 0.915 g’s for 0-60 in 3.0 s. and 1.02 g’s for 0-60 in 2.7 s.; then plug this “a” into the above equation (1/2at^2) -- but remember to multiply the 0.915 and 1.02 by 32.2 to get the correct units(imperial/non-metric).
Ooopps, sorry guys! Must have logged on to the Open University forum by mistake
Fascinating stuff though...
gtr-gaz said:
Fascinating stuff
Even more so when you start considering real-world conditions where acceleration varies throughout rather than being a constant.
The actual acceleration and distance figures for the Ultima 0-100mph-0 record run are available from Ultima's website.
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