How to solve a mathtype problem...
How to solve a mathtype problem...
There are things I can do in math, but there are a great many which are simply beyond me. What I'm really interested in here is understanding the howto behind figuring out how to come up with a formula for something.
In driving, there is a concept I've heard called the "twosecond rule". It's something I've used most of my driving life, and the physical doing of it is not a big deal, but now I want to know the actual distances being created.
Here's how this works: when you are behind another vehicle, the minimum safe distance is generally touted as two seconds. That is, if the car in front of you stops or slows down suddenly, two seconds represents a comfortable margin for you being able to see it happening and respond accordingly. Basically, at whatever speed you are going, there should always be two seconds' worth of space between you and the car in front of you. That means, obviously, that if you and the car in front are driving at 1 mph, that would be a certain distance. If you're driving at 10 mph, that distance would be longer, and at 45 mph it would be even greater, and so on and so forth.
Usually, you pick an object on the road, whether it's a bit of street marking or an object off the side of the road, and the time between when the car in front of you clears it and the time you encounter it should be two seconds.
What I want to understand is how to come up with the formula to be able to calculate what two seconds' worth of distance is, in whatever measurement system you like (Imperial or Metric) at any given rate of speed.
In driving, there is a concept I've heard called the "twosecond rule". It's something I've used most of my driving life, and the physical doing of it is not a big deal, but now I want to know the actual distances being created.
Here's how this works: when you are behind another vehicle, the minimum safe distance is generally touted as two seconds. That is, if the car in front of you stops or slows down suddenly, two seconds represents a comfortable margin for you being able to see it happening and respond accordingly. Basically, at whatever speed you are going, there should always be two seconds' worth of space between you and the car in front of you. That means, obviously, that if you and the car in front are driving at 1 mph, that would be a certain distance. If you're driving at 10 mph, that distance would be longer, and at 45 mph it would be even greater, and so on and so forth.
Usually, you pick an object on the road, whether it's a bit of street marking or an object off the side of the road, and the time between when the car in front of you clears it and the time you encounter it should be two seconds.
What I want to understand is how to come up with the formula to be able to calculate what two seconds' worth of distance is, in whatever measurement system you like (Imperial or Metric) at any given rate of speed.
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Re: How to solve a mathtype problem...
My quick math:
Mph * 5280 = Fph (feet per hour)
Fph / 3600 = Fps (feet per second)
Fps * 2 = Safe following distance
Mph * 5280 = Fph (feet per hour)
Fph / 3600 = Fps (feet per second)
Fps * 2 = Safe following distance
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Re: How to solve a mathtype problem...
"There is, ultimately, only one truth  cogito, ergo sum  everything else is an assumption."  Me, my swansong.
 catweazel
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Re: How to solve a mathtype problem...
That's a little innacurate as it doesn't account for creating a natural asymptotic stability of traffic
"There is, ultimately, only one truth  cogito, ergo sum  everything else is an assumption."  Me, my swansong.
Re: How to solve a mathtype problem...
Yup. catweazel nailed it! Please ignore my previous assumptions.
Re: How to solve a mathtype problem...
catweasel is fast with the fingersI'll give him that.
multiply mph x 3 to get feetplenty close and easy to calculate driving within 2 seconds of oblivion
multiply mph x 3 to get feetplenty close and easy to calculate driving within 2 seconds of oblivion
Re: How to solve a mathtype problem...
Yes, mph x 3 = feet is pretty close (actual = 2.933). One good way to come up with formulas like this is by using the cancellation of units to know whether to multiply or divide. For this problem, start with the formula distance = velocity x time, or d = vt. We know the velocity in miles/hour and the time in seconds, so we have to get the solution in feet. We need to multiply or divide by some numbers (in square brackets below) to get the proper units:
d = (V miles/hour) x (2 seconds) x [ (5280 feet/mile) x (1 hour/3600 seconds) ] = 2.933 feet x V where V is in mph.
In this equation, we can see that miles, hours, and seconds cancel out, leaving only V and feet. Dimensional analysis has guided us to the correct formula.
I can't resist mentioning how ideal the sliderule (colloquially called a "slipstick" in the USA) is for the solution of problems like this. With a single setting of the slide, you can simply move the cursor to a given velocity on one scale, and instantly read off the corresponding distance on another scale. For example, move the slide to pull the center index of the B scale to 2.93 on the A scale, then you can slide the cursor hairline to any desired velocity on the B scale and read the distance in feet under the hairline on the A scale.
@catweazel 
d = (V miles/hour) x (2 seconds) x [ (5280 feet/mile) x (1 hour/3600 seconds) ] = 2.933 feet x V where V is in mph.
In this equation, we can see that miles, hours, and seconds cancel out, leaving only V and feet. Dimensional analysis has guided us to the correct formula.
I can't resist mentioning how ideal the sliderule (colloquially called a "slipstick" in the USA) is for the solution of problems like this. With a single setting of the slide, you can simply move the cursor to a given velocity on one scale, and instantly read off the corresponding distance on another scale. For example, move the slide to pull the center index of the B scale to 2.93 on the A scale, then you can slide the cursor hairline to any desired velocity on the B scale and read the distance in feet under the hairline on the A scale.
@catweazel 
In theory, theory and practice are the same. In practice, they ain't.

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Re: How to solve a mathtype problem...
Two seconds is not enough. Three second rule is more appropriate. it is not only for driving but in aviation also. Three seconds meauserd from when situation actually changed, through recognition, your decition and finally your action.
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Re: How to solve a mathtype problem...
I've been watching Walter Lewin's 801 and 802 lectures, he's helped me understand some of this math.
https://www.youtube.com/watch?v=q9IWoQ1 ... hlFQAs8e
https://www.youtube.com/watch?v=q9IWoQ1 ... hlFQAs8e
Re: How to solve a mathtype problem...
Well, duh! Obviously!
[And now to figure out how in the heck to work that formula... ]

What I was really aiming for, folks, in all seriousness, is understanding how to come up with a formula.
While I'm certain my math skills have improved as I've completed all the various stages of cognitive development, I still as an adult have some pretty serious limits, specifically in the area of math. I'm quite capable in other areas of analytical thinking and abstract reasoning.
A few years ago, when I was actively taking classes in college, I got up as far as College Algebra. About a fifth of the curriculum, towards the end of the semester, was on logarithms. While I was just. barely. able. to make it through asymptotes, which is what immediately preceded it, I could not do anything with logarithms. I burned up I can't tell you how much time with professor office hours, math lab tutors, and even collegerecommended online resources, and it didn't help in the slightest. They would do stuff, and I just could not repeat it nor apply it on my own.
That is a cool series of videos. Thank you for posting them, even if they don't address my own needs. Knowledge should always be free, and if your posting of them here helps another human being, then it was well worth it.stormryder wrote: ⤴Sun Feb 25, 2018 10:40 amI've been watching Walter Lewin's 801 and 802 lectures, he's helped me understand some of this math.
https://www.youtube.com/watch?v=q9IWoQ1 ... hlFQAs8e
My issue with them is primarily he's not trying to explain how one would figure out how to define the values, nor how to figure out the construction of the formulas he is writing next to the graphs. There's a lot of assumed prior knowledge (and quite likely a few levels of math class between College Algebra and what he's teaching) so it's also way outside my area of knowledge, understanding, and competence.
Last edited by Portreve on Sun Feb 25, 2018 11:16 am, edited 1 time in total.
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“They said my computer wasn't strong enough for Windows 10, so I had to get one with more strongerness.”
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Re: How to solve a mathtype problem...
The two seconds thing is wrong. The MAIN THING to remember about traffic and stopping:What I want to understand is how to come up with the formula to be able to calculate what two seconds' worth of distance is, in whatever measurement system you like (Imperial or Metric) at any given rate of speed.
stopping distance ~= square of the velocity.
So if you can stop in 30 feet at 30mph, it'll take 120 feet at 60mph.
And since the deceleration is (more or less) constant, if you can stop in 2 seconds at 30mph, it'll take 4 seconds at 60mph.
Please edit your original post title to include [SOLVED] if/when it is solved!
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Re: How to solve a mathtype problem...
There's two issues with this. The first is the mass of the vehicle. The greater the mass, the greater the inertia. Therefore, whatever your speed, the stopping distance is impacted by mass. It's also affected by road surface type (how much friction the surface can produce), and condition of tires (how much friction they can produce).Flemur wrote: ⤴Sun Feb 25, 2018 11:05 amThe two seconds thing is wrong. The MAIN THING to remember about traffic and stopping:What I want to understand is how to come up with the formula to be able to calculate what two seconds' worth of distance is, in whatever measurement system you like (Imperial or Metric) at any given rate of speed.
stopping distance ~= square of the velocity.
So if you can stop in 30 feet at 30mph, it'll take 120 feet at 60mph.
And since the deceleration is (more or less) constant, if you can stop in 2 seconds at 30mph, it'll take 4 seconds at 60mph.
The other issue is that this has nothing to do with the two second rule. The rule is related to reaction time.
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Re: How to solve a mathtype problem...
Actually it's not, at least not much, because the stopping force (friction) increases with the mass at the same rate as the inertia (both linear). Which is why a motorcycle stops in about the same distance as a car (actually a car can often stop faster because less skill is involved). And the size of the tires doesn't matter much either, although the composition does.
Please edit your original post title to include [SOLVED] if/when it is solved!
Your data and OS are backed up....right?
Your data and OS are backed up....right?
Re: How to solve a mathtype problem...
Power (torque or other factors) of a motorcycle is, relatively speaking, far greater than that of a car, simply based on the relative mass of the motorcycle relative a car, or a truck, or a semi, etc. As far as skill of driving a motorcycle, I can only base opinion on having ridden a bicycle, and having had conversations with motorcycle riders, and I would tend to agree with you on that point.Flemur wrote: ⤴Sun Feb 25, 2018 11:43 amActually it's not, at least not much, because the stopping force (friction) increases with the mass at the same rate as the inertia (both linear). Which is why a motorcycle stops in about the same distance as a car (actually a car can often stop faster because less skill is involved). And the size of the tires doesn't matter much either, although the composition does.
I didn't say anything about tire size. I was talking about tire surface conditions. What condition is the rubber in? How worn down is the rubber? I would include what the composition of the rubber is, but I assume something like that is relatively standard in the automotive industry.
Zen Buddhism: The Journey Is The Reward
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Remember to mark your fixed problem [SOLVED].
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— Overheard random customer
Re: How to solve a mathtype problem...
I bought a 2017 Subaru Crosstrek and it does the math for me  It warns me when I get too close to the vehicle in front and will jam on the breaks to avoid a crash if I fail to put on the breaks in time.  problem solved. But I always have gone with the one car length for each 10mph as a rule of thumb.

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Re: How to solve a mathtype problem...
Your vehicle has a specific braking distance named 60  0 stopping distance. No math should be required.
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