Dropping to flat

Anonymous

weeksy59 wrote:

yeehaamcgee wrote:

So go on.... explain how forward speed affects the downwards acceleration from gravity, and the loss of that downwards speed when you hit the ground.

If speed and velocity wasn't a factor then bullets would just fall on the ground as soon as they leave the barrel of a gun.

The speed/velocity decreases the ammount of effect grvity has, therefore making it appear to defy the gravitational laws.

bullets DO fall to the ground as soon as they leave the barrel of a gun. The only effect of their forwards motion is to ensure that at the time they reach the floor, they're a very long way away.
The speed of the bullet has no effect on gravity. It should also be pointed out that a bullet creates no lift.
So, if you fire a bullet out of a gun, and also drop one from exactly the same height simultaneously, they will both contact the ground at precisely the same time.

supersonic

Yeehaa is correct. If a bullet is fired perfectly horizontally, with no velocity in the x direction it will fall to the ground in the same time as if it was just dropped.

http://www.antonine-education.co.uk/Phy ... proj_5.gif

weeksy59

yeehaamcgee wrote:

So, if you fire a bullet out of a gun, and also drop one from exactly the same height simultaneously, they will both contact the ground at precisely the same time.

Nah, i'm not having that at all.

A bullet droped out of my hand will impact the floor within a second or so.

You're trying to tell me a bullet fired straight out will impact within a second ?

Based upon that logic, how does a sniper manage to keep a bullet in the air for 10 seconds for a long range shot ? by firing at a 45 deg angle and hoping it drops in the right place (yes i accept they'll aim slightly high). ?

weeksy59

supersonic wrote:

Yeehaa is correct. If a bullet is fired perfectly horizontally, with no velocity in the x direction it will fall to the ground in the same time as if it was just dropped.

I don't get it.


Maybe i'm a thicko but i can't see how that's possible.

ride_whenever

only from a blunderbus, rifling prevents the dropping.

myopic

yeehaamcgee wrote:

myopic wrote:

yeehaamcgee wrote:

forward motion makes no difference when dropping to flat. However, it does affect drops to a slope, because your forward speed, once in contact with the slope, allows you to continue traveling downwards as well as along.

(i.e. being in motion on a slope means you're still losing altitude whereas being in motion on the flat gives no change in altitude)

I'm sure this isn't right! If I was going to do a drop to flat (unless it was just off a kerb) - I would be more confident if I had a run up and some momentum. The higher it was, the more I would feel this was of benefit.

So go on.... explain how forward speed affects the downwards acceleration from gravity, and the loss of that downwards speed when you hit the ground.

If you are moving forward, with wheels rotating, then some of the downward force is translated into forward motion. I think of it like the difference between skimming a stone across water (or anyhitng if it is going fast enough), compared to dropping it in.

Or another way of looking at it is the faster you are going, once you are airborne, the closer your achieved trajectory is to the flat surface you will land on. Which is analogous to matching the transition off a jump. Obviously, with a flat landing the trajectory changes to become less convergent until with a high enough drop off you will ultimately only achieve vertical travel. But if you land before then you will be close enough not to feel it the same. Your argument would only hold out if you got to the drop and then went straight down

supersonic

weeksy59 wrote:

supersonic wrote:

Yeehaa is correct. If a bullet is fired perfectly horizontally, with no velocity in the x direction it will fall to the ground in the same time as if it was just dropped.

I don't get it.


Maybe i'm a thicko but i can't see how that's possible.

Have a look on this page:

http://hyperphysics.phy-astr.gsu.edu/hbase/grav.html

Air resistance may have an effect (for the reasons RW point out).

myopic

I agree with the bullet analogy but I still think that tfe force felt from a drop to flat will be less if there is enough forward momentum so that the trajectory is closer to flat than would be the case from only vertical movement. In the extreme case 90 degrees to landing, but with enough speed this might only 10-20 degrees different which must make a difference to the force that is felt!

Anonymous

myopic wrote:

If you are moving forward, with wheels rotating, then some of the downward force is translated into forward motion. I think of it like the difference between skimming a stone across water (or anyhitng if it is going fast enough), compared to dropping it in.

This is a common misconception.
Your trajectory consists of two elements or vectors, when dropping, horizontal (x), and vertical (y).
Those two elements added together give you the direction you're headed in, x+y, but neither will affect the other, they are exclusive (unless you have some lift-creating surface like a wing). therefore, x+y will never result in x being less than x, or y being less than y.
It doesn't matter how fast you're going forwards, the downwards vector x is only affected by gravity, and you will hit the floor with the same force whether you're moving forwards or not.
Likewise, your downwards acceleration or movement will not affect your forward speed.

myopic

:? OK - I don't know enough (any!) physics to argue with this. Just going by what I feel when riding. But doesn't the rotational force from the wheels help to translate some of one force into another? A question - not an argument!

supersonic

There is another factor worth discussing.

If you leave a drop off (flat to flat) at speed, both wheels will leave the ledge at almost the same time. The front will hit the floor first just before the rear - it did start to drop first.

If you ride off slowly, the front leaves first, and starts to drop - it will will land first by a margin.

To counter this you pull front end up and if judged correctly both wheels touchdown at the same time (or whatever you prefer).

However it does seem that as now the front wheel started from higher, it hits the deck harder! Maybe this is why many prefer a rear wheel landing first to flat.

Something to mull over, I need to think about it more (Obviously both wheels are connected by a frame, and we have the frame rotation too)

Basically it is whether we treat th bike as a point source or not.

Anonymous

ride_whenever wrote:

only from a blunderbus, rifling prevents the dropping.

Nope. rifling does not affect the dropping. It keeps the bullet spinning in one direction, and therefore stabilizing it (preventing it from turning end over end) in the air.
If a sniper puts a bullet in the air for 10 seconds, then his/her firing point must be 10 seconds worth of freefall above the final target.
But honestly, after 10 seconds, a high-speed sniper round would be very, very, very far away indeed, and I doubt that even the military have the neccesary telescopic sights to line a target up at that range.
I'm not sure what the muzzle velocity of a typical sniper rifle is, but I know that a 50-cal round can leave the barrel at up to 2,000mph. That's over half a mile per second, so in ten seconds, that 50-cal round would have traveled horizontaly around 5 miles, or if you include air resistance slowing it down, I'd still estimate it to be about 4 miles. Of course, this can only happen if the rifle is very very high up to begin with, as it would have impacted the floor.

Basically what I'm saying is that the 10-second sniper bullet is likely an exageration.

myopic

I'm going to stop thinking about it: my brain has started bleeding and running out of my ears......

weeksy59

yeehaamcgee wrote:

Basically what I'm saying is that the 10-second sniper bullet is likely an exageration.

It was just a random figure picked to prove/disprove/make a point.

Which you guys then say is wrong...

However... i dunno... i'm with brain bleeding boy above and not opening it any more

supersonic

To go back to the gun, you need to be 5m up for it to hit the ground in 1s.

If we say the gun has a muzzle velocity of 250m/s, and we are 5m up and in a vacuum, it will hit the floor after 250m. I think that is easier to see. Would you aim say 5m (or whatever) higher at a traget 250m away?! I would probably start higher than that!

Anonymous

myopic wrote:

:? OK - I don't know enough (any!) physics to argue with this. Just going by what I feel when riding. But doesn't the rotational force from the wheels help to translate some of one force into another?!

Nope, the rotational force of the wheels will not translate any of one force vector into the other.

A question - not an argument

That's fine, I didn't take your tone to be argumentative at all!
you weren't the one who was being an ass because you didn't know physics, but thought you did.

TommyEss

supersonic wrote:

There is another factor worth discussing.

If you leave a drop off (flat to flat) at speed, both wheels will leave the ledge at almost the same time. The front will hit the floor first just before the rear - it did start to drop first.

If you ride off slowly, the front leaves first, and starts to drop - it will will land first by a margin.

To counter this you pull front end up and if judged correctly both wheels touchdown at the same time (or whatever you prefer).

However it does seem that as now the front wheel started from higher, it hits the deck harder! Maybe this is why many prefer a rear wheel landing first to flat.

Something to mull over, I need to think about it more (Obviously both wheels are connected by a frame, and we have the frame rotation too)

Basically it is whether we treat th bike as a point source or not.

Suspect the perceived bigger hit when you've lifted the front is more to do with your weight distribution on the bike, and your centre of mass being rotated about the centre point of the bike a bit further (since you effectively rotate it back a bit when you lift the front up)

Like lifting a Newton's Cradle ball a bit higher - it was further back, so it will go further forwards (negating wind resistance etc!)

Anonymous

supersonic wrote:

To go back to the gun, you need to be 5m up for it to hit the ground in 1s.

If we say the gun has a muzzle velocity of 250m/s, and we are 5m up and in a vacuum, it will hit the floor after 250m. I think that is easier to see. Would you aim say 5m (or whatever) higher at a traget 250m away?! I would probably start higher than that!

I'm not sure about 5m in one second, but it's about right. Too busy to do the maths right now (well, suposed to be!)
But yeah, as you say, you'd aim higher. I've never held a military sniper scope, but I've seen various diagrams and such that have notches horizontally and vertically to allow the marksman to compensate for wind, and distance.
Maybe someone from the military could clarify that?

supersonic

Yeah, was an approximation. Of course neglecting air resistance makes a big difference - 250m/s is quite high, many air pistols are a lot less, and will be a lot less when get to the target lol.

ride_whenever

again military grade rifles are rifled, but you do end up compensating, especially when shooting at a target a mile away.

Although that is mostly for the wind, as you zero your rifle to account of the drop at range. Coriolis effect also comes in at large ranges.

supersonic

Coriolis, really, wow!

I'll have to use that as a bike excuse when I land a dodgy jump lol.

Daz555

http://en.wikipedia.org/wiki/Rob_Furlong

Apparently this guy had to aim 256 feet above his target in order to get a the longest ever sniper kill at over 1.5 miles - 256 ft comes from the assumption that the bullet was in the air for as many as 4 secs. :shock:

http://www.builtonfacts.com/2008/05/21/ ... f-sniping/

supersonic

Incredible!

dave_hill

yoohoo999 wrote:

OK. Here is exactly what I'm talking about!

Here is me doing a small drop (about 3-4ft) at a relatively quick pace, to flat landing. I was trying to clear the path.

Temple Newsham by any chance?

yoohoo999

given that the gravitational acceleration is around 9.8 m/s2, if you did a drop from about 1.5m, you would be hitting the ground at around 5.4 m/s

V= squareroot(2gh) (g is gravitation accereration, h is height)
=squareroot (2 * 9.8 * 1.5)
=squareroot(29.4)
=5.422 m/s

Therefore, a rider + bike weighing 90kg would hit the ground with 1312 joules of kinetic energy. using KE = 1/2 (M * (V * V))

As I understand it, the bike would hit the ground with the same kinetic energy regardless of whether the rider dropped to flat or hit it at 100mph, provided that his y axis speed (ie the drop speed) was the same.

Therefore, if the kinetic energy is identical, surely all the difference in speed, or indeed technique in landing does is dissipate that energy by a different (and hopefully more efficient) manner. Down to the action of the suspension, rider position, rider movement etc.

Or am I missing something?

yoohoo999

dave_hill wrote:

yoohoo999 wrote:

OK. Here is exactly what I'm talking about!

Here is me doing a small drop (about 3-4ft) at a relatively quick pace, to flat landing. I was trying to clear the path.

Temple Newsham by any chance?

yup, pretty building eh? We have great fun doing the ampitheatre thing round the other side, it's so much fun having a race to see who can jump down to the bottom first! I think it's 6 or 7 drops in very quick succession

TommyEss

yoohoo999 wrote:

given that the gravitational acceleration is around 9.8 m/s2, if you did a drop from about 1.5m, you would be hitting the ground at around 5.4 m/s

V= squareroot(2gh) (g is gravitation accereration, h is height)
=squareroot (2 * 9.8 * 1.5)
=squareroot(29.4)
=5.422 m/s

Therefore, a rider + bike weighing 90kg would hit the ground with 1312 joules of kinetic energy. using KE = 1/2 (M * (V * V))

As I understand it, the bike would hit the ground with the same kinetic energy regardless of whether the rider dropped to flat or hit it at 100mph, provided that his y axis speed (ie the drop speed) was the same.

Therefore, if the kinetic energy is identical, surely all the difference in speed, or indeed technique in landing does is dissipate that energy by a different (and hopefully more efficient) manner. Down to the action of the suspension, rider position, rider movement etc.

Or am I missing something?

Think that's right - essentially - at height of 1.5m it has gravitational potential energy. In the drop that energy is converted into kinetic energy, from which you can calculate the speed at which it lands.

When you make the jump though, as discussed previously, the trajectory of your jump is a vector comprised of the vertical drop and the speed you are travelling - this gives you the velocity, which is a vector and not a scalar quantity (did I get that bit right - A level physics was a wee while ago)

dave_hill

yoohoo999 wrote:

Or am I missing something?

Friends?

yoohoo999

ha ha! touche!

i'm a corporate lawyer believe it or not (we normally don't touch numbers), even physicists make us lot look cool 8)

TommyEss

yoohoo999 wrote:

i'm a corporate lawyer

Boo... Hiss... And other assorted pantomime noises!