Physics 101 is a great course ... D = ½gt²

However, since we don't aim directly at everything regardless of the distance, you need to use the ballistic formulas, corrected for drag, etc. which are major influences on Trad arrow flight.

Although I have a Masters in Physics, I don't think of these things.

My arrows drop the amount I allow for them, based on experience (called hueristics in fancy physics talk).

FWIW, at 30 yds I figure the apogee for 165 fps is about 15" above the sight-target line ... providing I don't put more or less pressure on my third finger, don't put more or less heel into the grip, get the same draw length, and the same 'clean' release, etc. and etc. and etc. ...

Quotewhat is the optimum angle to get maximum distance at a  flight shoot. I would guess that to be about 45 degrees. Now... Anyone know the answer to that?

Yeah, it's about 37º when drag is taken into account.

gjarcher, where have you been? Ha! I don't see how the "1/2" apllies to this. What I remember is distance = 32fps x time squared. This of course ignoring atmosphere.

After I posted the last time I saw something about flight shooting that stated that about 45 degrees was optimum depending on wind conditions. I wonder what O L Adcock would say? He's won most of the records.

Bud B, hanging with Harry will get you in trouble. When are you coming back to shoot on Tuesday night? We'll run up a pole at a target outside and put up some markers to determine where Harry's aiming.

"Hi ho Silver Away...Up I Say"      

Ok, so to throw a wrench in the cogs of this conversation.  An arrow has to actually have hit a specific height above the ground before it can fall.  So despite the excellent example of trajectory drawing shown above I do not think we have resolved the question at hand.

I think we need some clarity on where the arrow is falling or dropping from.  If we are measuring from the bow hand then drop will only be the distance below the bow hand and will not include the rise above the hand.

If on the other hand we want to calculate total drop at a set yardarge from the bow hand we have to measure from the top of the arrows arc to the point of arrow impact at that distance.  

For this conversation I believe we are looking for the distance the arrow drops based on impact points at 20 yards, 30 yards, and 40 yards.  So I would say we need to run the calculations with the zero elevation being a dead on impact point of 20 yards.  That leaves the arrow drop at 20 yards as zero.  Then it is a simple physics equation to calculate the drop at 30 and 40 yards with the matching arrow weight and velocity.  

If someone has a ballistics calculator feel free to run the numbers, if not I will look up the proper trajectory equations and run them.

my head is hurting reading all this!    ... i may need to take a calculator in the stand next time

Bjorn,

Here are the definitive numbers. I used Friends calculator link to run the numbers.  With a zero elevation of 20 yards the arrow has not dropped at all.  It rose 6.7 inches above the bow hand then fell back to the same height.

At 30 yards the arrow has fallen 20.6 inches below the zero elevation.  At 40 yards the arrow has fallen 55.9 inches below the zero elevation. 55.9 / 12 = 4.65 feet of drop below the zero elevation.  And that my friends is the laws of physics applied to arrow ballistics.  

Mild R/D longbow
Draw Weight 55 lb
584 grain arrow
Assumed velocity of 165 fps


Zero Distance = 20 yards

These numbers reflect the distance for every yard from the hand. Positive numbers are above the bow hand elevation and negative numbers are below the bow hand elevation.



1   1.3    
2   2.4    
3   3.4    
4   4.2    
5   5.0  
6   5.6    
7   6.0  
8   6.4    
9   6.6  
10   6.7    
11   6.6    
12   6.4  
13   6.1    
14   5.6  
15   5.0    
16   4.3  
17   3.4    
18   2.4    
19   1.3    
20   0.0    
21   -1.4  
22   -3.0    
23   -4.7    
24   -6.5    
25   -8.5    
26   -10.7  
27   -12.9    
28   -15.4    
29   -17.9    
30   -20.6    
31   -23.5    
32   -26.5    
33   -29.7    
34   -33.0    
35   -36.4    
36   -40.0    
37   -43.8    
38   -47.7  
39   -51.7    
40   -55.9

QuoteOriginally posted by Ragnarok Forge:
[QB]  It rose 6.7 inches above the bow hand then then fell back to the same height.


Did the arrow actually rise ? Or was it launched with an upward angle, to coincide with a chosen zero distance and crossed the line of sight?

I don't understand what the bow hand has got to do with the matter at hand...other than raise or lower the arrow ??? but then again physics is not one of my stronger areas...but I have played the ballistics game a lot of years:0

Rod,

Did the arrow actually rise ? Or was it launched with an upward angle, to coincide with a chosen zero distance and crossed the line of sight?

The arrow was fired at a raised trajectory as you mention.  That raised trajectory is what accounts for the arrows rise above the bow hand  

To calculate trajectory for any projectile you must have a known starting elevation or a known ending elevation.  In the case of archery the starting elevation is always the bow hand and the ending elevation is the X in your target at a set yardage.  Setting the zero elevation at 20 yards gives us a fixed ending elevation.  This allows the calculator to reverse the math and figure the bow hand elevation as part of the equation. So for a thirty yard shot to hit the target you have to raise the bow hand a set amount that will change the arrow trajectory enough to account for the 20.6 inches of drop.

I've no idea, i never shoot over 15 ...

good luck !

Clay...I'm slow    I understand the bow hand relevance now....your using the bow hands relationship to the line of sight, to make corrections, instead of the arrow point.

I'm glad I shoot instinctive,this has giving me a headache trying to understand where to aim.

is it hunting season yet?
lol

knowing the physics doesn't hurt, as long as you don't try to apply it to your shooting...