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**Synapse****Member**- Registered: 2005-08-08
- Posts: 6

Alright... we were kinda curious about how fast a halfling would fall.....

So it weighs 50 pounds with oh... say 20 pounds of equiptment... knives... some food and other stuff.... a halfling would carry... like spoons...

Plus it is nice to beable to say you know what the terminal velocity of a halfing is... thanks for the help!

-Synapse-

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Equipment would surely include a parachute, so I would increase that by about 10lb.

Now, we cannot determine the terminal velocity without knowing such things as drag coefficient and such. Could we suspend your subject in a wind tunnel before coming up with an answer?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Synapse****Member**- Registered: 2005-08-08
- Posts: 6

Yeah that sounds like a good idea... um lets say he is outstreached.... but I doubt that is too much drag. There are no parachutes in Farun.... well some one might have one..... some weirdo...

*Last edited by Synapse (2005-08-08 18:37:27)*

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

Terminal velocity? i think that involves differential equations

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**Synapse****Member**- Registered: 2005-08-08
- Posts: 6

Well what do you need to know? The gravity is the same as earths... and the drag for an outstrached halfling is not all that great... right? 70 some pounds. What is the TV for a human being? Or a really short person... something along the lines of a child withe equiptment.

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

This is from a physics forum:

One common model is that the resistance force is proportional to the speed.

Under that model, an object falling, under gravity has acceleration -g+ kv (k is the proportionality constant, v the speed. Since that is a function of v, it give the linear differential equation mdv/dt= -mg+ kv. The general solution to that is v(t)= Ce-kt/m-mg/k. For very large t, that exponential (with negative exponent) goes to 0 and the "terminal velocity" is -mg/k.Another common model is to set the resistance force proportional to the square of the speed. That means the net force is -g+ kv2 and v satisfies the differential equation mdv/dt= -g+ kv2. That's a non-linear differential equation but is separable and first order. We can integrate it by writing

dv/(kv2-g)/m= (-1/2√(g))(1/(√(k)v+√(g))dv/m+(1/2√(g))(√(k)v-√(g))dv/m= dt.

Integrating both sides, we get (1/2√(kg))ln((√(k)v-√(g))/(√(k)v+√(g))= mt+ C. For large t, the denominator on the left must go to 0: the terminal velocity is -√(g/k) which, you will notice, is independent of m. This model is typically used for very light objects falling through air or objects falling through water.

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

what is a halfling??

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**Synapse****Member**- Registered: 2005-08-08
- Posts: 6

Hm... as much as I know about math I know nothing about what kylekatarn just pointed out.

However I can tell you what a halfling is, think of a human being only 1/4th the size roughly... like a child that grows to be an adult in every way exept shorter life span and lacks the vertical height a human has. They are tricky little things. Read some TSR(Wizards books now) there are plenty of Halflings in those fantasy novels. Good books too.....

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Synapse wrote:

Hm... as much as I know about math I know nothing about what kylekatarn just pointed out.

kylekatarn just quoted from a physics site ... if you really need the formulas, someone could dissect it all for you. But what I got out of that is that it does not matter what the mass of the halfling is.

As he or she falls the velocity keeps increasing until the downwards acceleration caused by gravity equals the drag caused by the air. That is why skydivers can fall faster when they "arrow" themselves, and they can slow up when they "spreadeagle" themselves (and they really slow up when they let the parachute open!).

So, I am afraid we are back to that windtunnel for tests. We will also have to ask your halfling to try different positions to see how that affects the drag, too.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Synapse****Member**- Registered: 2005-08-08
- Posts: 6

So what you are saying is that there is no amount of math on paper that can calculate the TV of an outstreached halfing? Cause I am just looking for MPH.... I figured it would not be too bad to figure out but I guess it is way more to figure than I had imagined.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

As it **does** depend a lot on the shape of the halfling (arms outsretched or whatever) just go with the normal estimate of terminal velocity, which is about 55 m/s, or 200 km/h, or about 120 mph.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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