best way to describe this -

[tris-]

it appears ive had some kind of mental block. take this column of numbers -

0.01
1.01
4.02
9.05
16.10
25.15
36.21

then minus 1 from 4, to get 3 (ofc), then 4 from 9, to give 5 etc, keep doing that down the table. it gives the number pattern 3 5 7 9 11.

problem is i cannot for the life of me think how to describe this pattern

anyone have an idea?

 

[Solo]

for an odd sequence 1,3,5,7,9,11 its 2n-1, where n=1,2,3,4,5 ? Is that what you mean.

 

[Ingafgrinn Macabre]

2n+1 for n in N*

since the serie starts at 3

 

[tris-]

what is meant by N* ?

i proibably should of put the complete table in at the start -

1 0.01
10 1
20 4 (4-1) =3
30 9 (9-4)=5
40 16 (16-9)=7
50 25 (25-16)=9
60 36 (36-25)=11

with the left column being increased by me, with an increase of 10. the second column is the result of a calculation and the third being what i propose to be a relationship between the numbers.

 

[scarloc]

N is the set of integers, in this case.

 

[psyco]

are you looking for a specific name? looks like it could be fibanachee(sp?) variant

 

[tris-]

somehow i found a simpler relation (left column/9.97)^2 shows the rate of increase.

found by sheer luck, no idea how to explain it :\

 

[Ingafgrinn Macabre]

N* = {1, 2, 3, 4, 5, ...}
so yeah, set of integers

just a question, where and how did you get those numbers? and is that the complete list? Because if they're measurements or something, couldn't it just be a deviation of n^2?

 

[tris-]

its the difference of stopping distances between static and kinetic friction. ive opted out of using 3 5 7 9 11 thing, and gone for the second formula which is simpler, and works. the only thing i dont know is what 9.97 has to do with anything :\

 

[tris-]

It looks to be just noise in the data set, obviously you wont get PERFECT measurements for a number of reasons (environmental conditions and so on). If I were you, I'd go with n^2 and leave the other part out.

true. but it doesnt explain why dividing one number by a seemingly random other number, then squaring it, gives the answer

its basically, if you take the velocity of your car, divide it by 9.97 and square it, you get the difference in breaking distance between your wheels locking and not locking.

 

[scarloc]

It looks to be just noise in the data set, obviously you wont get PERFECT measurements for a number of reasons (environmental conditions and so on). If I were you, I'd go with (n^2)/10 and leave the decimal out, but report it as noise or whatever.

 

[Ezteq]

my cats name is mittens.......

 

[Ingafgrinn Macabre]

true. but it doesnt explain why dividing one number by a seemingly random other number, then squaring it, gives the answer

its basically, if you take the velocity of your car, divide it by 9.97 and square it, you get the difference in breaking distance between your wheels locking and not locking.

don't have all too much time at the moment, but static and dynamic frictionconstants and their formula's aren't difficult, will detail it a bit when I get back

 

[psyco]

my cats name is mittens.......

im so tempted

 

[tris-]

don't have all too much time at the moment, but static and dynamic frictionconstants and their formula's aren't difficult, will detail it a bit when I get back

yeh their formula arnt difficult. but what im asking is how come if you divide a velocity, by a seemiingly random number (9.97) and then square the result, you get the rate of increase in stopping distance between the two types of friction

 

[Ingafgrinn Macabre]

Well, here we go.

Let's start with a basic physics formula displaying the relationship between accelleration, startingspeed, endspeed, starting distance and enddistance:

[01]


Our endspeed is 0 (break to full stop), and our startingdistance is 0 (only need the distance from the point of breaking)
The frictionconstant is nothing more or less than the factor to which it can convert downforce to lateral forces, so for instance a frictionconstant of 0.7 at a wheel which has a load of 1000N can at max exert a force of 700N lateral when breaking or accelerating.
Since however we'd prefer an accelerationfactor instead of a plain force, we change it a bit. Instead of using the downforce of said 1000N, we'll use g.
I won't bore you with the formula's to prove that this works, but it does.
So, our accelerationforce a can be reformed as follows:

[02]


With these three constraints we can rewrite [01] to this:

[03]

Removing the both negatives against eachother results to:

[04]

 

[Ingafgrinn Macabre]

We're looking for the travelled distance and not the acceleration, therefor we'll reform [04] as followed:

[05]


There are two frictionfactors. The static frictionfactor is always (well, I atleast haven't seen a different situation) higher than the kinetic friction factor. It has to do with the microscopic structure of the contact surfaces, intentation etc. anyway, you might have these two factors:

[06]


They actually don't matter much, since we'll not be solving it numerically, but it does perhaps make it a tad easier to follow.
We can now set up two formulas for the static and kinetic breakingdistance:

[07] __ and __

[08]

the breakingdistance x_tk will always be longer than the breakingdistance x_ts.

 

[Ingafgrinn Macabre]

To get the difference in breakingdistance between the two we'll substract x_ts from x_tk:

[09]

Step 1: equalise quotients
step 2: combine quotients
step 3: seperate v_0 from the rest


You can see that with the totally reformed quotient only the starting speed is still a variable.

[10]

This [10] is the formula to which your numbers should adhere.

 

[Ingafgrinn Macabre]

Well, let's see if they do.

Cos I'm lazy I renamed your left column to lcol.
First let's break up your formula to something that's easier to handle:

[11]


Second, let's try and get x_t_Diff to conform to the format of the above formula:

[12]

Step 1: First I devide the starting velocity with a constant so I don't actually need that velocity anymore, but a non-equal but equivalent denominator (like your lcol)
Step 2: Seperating the constant from the denominator.
Step 3: Throwing everything apart from the variable denominator below the quotientline

This results in the following formula:

[13]

It seems to me that this fits nicely to your formula in [11].
It wouldn't surprise me if every constant in that part below the quotientline would end up being that 99.something if you choose the constant C so that:

[14]

 

[Tasslehoff]

Inga makes me cry

 

[psyco]

Inga makes me cry

i think everyone is crying

 

[Ingafgrinn Macabre]

hmmm, my choice of wording up there could be confusing. I just found out that denominator is the english word for the lower part of a fraction, so perhaps it would've been better to call it a factor or whatever, but I guess my intention what I tried to say was clear

and Tassle, it wasn't that difficult, although it did take a bit of time due to getting those formula's accurately displayed and readable

 

[Ingafgrinn Macabre]

You know Tris, It's nice to get some feedback, something that makes you know that you haven't just spend a large portion of your day helping other people for jack shit :/
Just a little sign that your work is appreciated is often enough.

Since I've posted the work here above you've posted 18 replies on other threads, a lot of 'em pretty useless, yet you can't find the time to post a simple thanks. Goodie, really makes a man feel like half a day's work well spent :/

 

[tris-]

sorry, i assumed it would be implyed that im thankful for inputs.

took some time to understand what youve done as youve actually used another formula to what ive used. i didnt use the suvat equations, but something i derived from the formula for k.e. and another formula the tutor supplied

over the last two days ive also found that multiplying the result from that formula by 4.2243 will give you the rate of increase on a wet road.

 

[Ingafgrinn Macabre]

sorry, i assumed it would be implyed that im thankful for inputs.

I'm afraid that doesn't really show when there's absolutely no response :/

took some time to understand what youve done as youve actually used another formula to what ive used. i didnt use the suvat equations, but something i derived from the formula for k.e. and another formula the tutor supplied

The formula [01] I used is a variation on (and combination of) the following standard phisics formulas:

[15]

If you're gonna do more with motion phisics you'll probably gonna need these more often.

It is totally possible to do it thru kinetic energies and such, but it's much more work and so if you want to do it accurately that way.
Also, it's a bit beating round the bush with kinetic energy, since this matter is a simple friction matter you'll then need to know how much energy gets dissipated to heat and all that stuff.

over the last two days ive also found that multiplying the result from that formula by 4.2243 will give you the rate of increase on a wet road.

All that happens on a wet road is that the friction constants get lowered, so the formula's you use stay thesame but the \mu_{s} and \mu_{k} get lowered. This in turn changes the resulting value in formula [10] (and indeed [13]

 

[tris-]

I'm afraid that doesn't really show when there's absolutely no response :/

sorry about that then

thankfully this is the last i ever have to do with motion physics

 

[Serbitar]

this is quite worrying. I did maths & further maths at a level and got an A and a B. then i did two first years of degree level maths (failing miserably, mainly due to the perils of alcohol) and this thread makes about as much sense to me as this...

 

[Amanita]

What do you mean? That singing fish is seriously cool





ok, so I lie.

 

[psyco]

this is quite worrying. I did maths & further maths at a level and got an A and a B. then i did two first years of degree level maths (failing miserably, mainly due to the perils of alcohol) and this thread makes about as much sense to me as this...

i love that fish i was going to get the whole taxidemy series

 

[Tasslehoff]

and Tassle, it wasn't that difficult, although it did take a bit of time due to getting those formula's accurately displayed and readable

See, that's the problem, I just don't get it!

Hope I won't need to make maths at that level, if I ever get the need though, I know who to call