Greysides

06-19-2010, 02:11 PM

From the FITA Field Archery Guidelines (http://archery.ie/field.pdf):

In Field Archery an important part of the shooting technique consists of making accurate range

estimations on the unmarked courses. In order to compete with the best archers this knowledge

can not be entirely dependent on your intuition or on your terrain evaluation, as these methods

are far too inaccurate and you will end up losing too many important points. A field archer will

have to find his own way to appreciate the distances, and he will have to practice this as part

of his shooting form. The most accurate methods are based on geometrical concepts as shown below.

Most of the methods, if not all, are based on the application of the Thales Theorem, by which we

can find the wanted distance if we know the distance from the dominant eye to a measuring

item (i.e. sight ring, scope, arrow rest, etc) placed on the bow, called d, which width is

called a, and the width of the projection of that item on the target as you see it, or which can be

calculated (the size of what you see on the target), called A. The relationship between these

elements will give you the distance to the target, called D, by simply applying the relationship :

a / d = A / D

Using as a measuring item any permitted part of the equipment, as for instance a sight component,

the arrow rest, etc.

In order to make it easier to understand, and to avoid the need to be applying any mathematics

on the field course, and in order to get the distance to the target as simply and quickly as possible

use the following principle; If the width of the measuring item (i.e. sight ring etc) happens

to be exactly one hundredth of the distance from the retina of your eye to the measuring item (i.e.

the sight ring is 8mm and the distance from the eye to it is 80cm) or if you can adjust your equipment

to make that relationship , then the relationship will be :

D = A * 100

Which if taking D in meters, and A in centimeters, will become : D (metres) = A (cm)

So that the range (in metres), D, results from the simple calculation of the measuring part's projection

width (in centimeters), A, on the target.

The knowledge of A is based entirely on the assumption that we know the target size. For

instance if the target on the figure is an 80cm diameter face, A would cover half of the face,

plus one division and a half, that is : 40+8+4 = 52cm, and we would conclude that the distance

to the target is 52 meters. If on the other hand it was a 60cm target face, then the calculation

would be : 30+6+3 = 39cm, and the distance 39 meters based on the above mentioned relationship.

This simple and immediate relationship is not always possible, and then the archer needs to

find his own. However, most archers do not apply any maths when doing the measuring,

they simply compare by experience based on the described principle.

http://img59.imageshack.us/img59/7664/targetcomparison.jpg

For instance, when practicing you will shoot from various distances on the various target

face sizes. By practice you will find how much of your sighting device, or anything else, you can

see in front of you, is covering the target face or buttress.

Hopefully you can follow that. Essentially I used the Thales Theorum part of it.

To cut a long story short, I hold up the bows sight window against one edge of the target and see how many rings are obscured (taking it as 10 rings per target).

With a 20 cm face each ring covered adds 1.2m to the distance from the target.At 12m the width of the face is completely hidden.

Unmarked FITA distance with this face 5-10m.

With a 40 cm face each ring covered is a 2.4m to the distance from the target. At 24m the face is completely covered.

Unmarked FITA distance for this face is 10-20m.

With a 60 cm face each ring covered is a 3.6m to the distance from the target.

Unmarked distance range for this face is 15-30m.

The distance to width relationship is linear not exponential as I had thought- rubbishing previous attempts at distance comparisons to pats of my button which gave that impression.

By finding the distance your sight window (or other measuring part) covers the entire face and dividing by 10 to get the distance per ring you should be quickly able to suit the method to yourself for each size target.

This method compares well with my practical results of playing about with targets and distances.

It would be interesting to hear anybodies experience of this. It shouldn't be hard to try it out.

In Field Archery an important part of the shooting technique consists of making accurate range

estimations on the unmarked courses. In order to compete with the best archers this knowledge

can not be entirely dependent on your intuition or on your terrain evaluation, as these methods

are far too inaccurate and you will end up losing too many important points. A field archer will

have to find his own way to appreciate the distances, and he will have to practice this as part

of his shooting form. The most accurate methods are based on geometrical concepts as shown below.

Most of the methods, if not all, are based on the application of the Thales Theorem, by which we

can find the wanted distance if we know the distance from the dominant eye to a measuring

item (i.e. sight ring, scope, arrow rest, etc) placed on the bow, called d, which width is

called a, and the width of the projection of that item on the target as you see it, or which can be

calculated (the size of what you see on the target), called A. The relationship between these

elements will give you the distance to the target, called D, by simply applying the relationship :

a / d = A / D

Using as a measuring item any permitted part of the equipment, as for instance a sight component,

the arrow rest, etc.

In order to make it easier to understand, and to avoid the need to be applying any mathematics

on the field course, and in order to get the distance to the target as simply and quickly as possible

use the following principle; If the width of the measuring item (i.e. sight ring etc) happens

to be exactly one hundredth of the distance from the retina of your eye to the measuring item (i.e.

the sight ring is 8mm and the distance from the eye to it is 80cm) or if you can adjust your equipment

to make that relationship , then the relationship will be :

D = A * 100

Which if taking D in meters, and A in centimeters, will become : D (metres) = A (cm)

So that the range (in metres), D, results from the simple calculation of the measuring part's projection

width (in centimeters), A, on the target.

The knowledge of A is based entirely on the assumption that we know the target size. For

instance if the target on the figure is an 80cm diameter face, A would cover half of the face,

plus one division and a half, that is : 40+8+4 = 52cm, and we would conclude that the distance

to the target is 52 meters. If on the other hand it was a 60cm target face, then the calculation

would be : 30+6+3 = 39cm, and the distance 39 meters based on the above mentioned relationship.

This simple and immediate relationship is not always possible, and then the archer needs to

find his own. However, most archers do not apply any maths when doing the measuring,

they simply compare by experience based on the described principle.

http://img59.imageshack.us/img59/7664/targetcomparison.jpg

For instance, when practicing you will shoot from various distances on the various target

face sizes. By practice you will find how much of your sighting device, or anything else, you can

see in front of you, is covering the target face or buttress.

Hopefully you can follow that. Essentially I used the Thales Theorum part of it.

To cut a long story short, I hold up the bows sight window against one edge of the target and see how many rings are obscured (taking it as 10 rings per target).

With a 20 cm face each ring covered adds 1.2m to the distance from the target.At 12m the width of the face is completely hidden.

Unmarked FITA distance with this face 5-10m.

With a 40 cm face each ring covered is a 2.4m to the distance from the target. At 24m the face is completely covered.

Unmarked FITA distance for this face is 10-20m.

With a 60 cm face each ring covered is a 3.6m to the distance from the target.

Unmarked distance range for this face is 15-30m.

The distance to width relationship is linear not exponential as I had thought- rubbishing previous attempts at distance comparisons to pats of my button which gave that impression.

By finding the distance your sight window (or other measuring part) covers the entire face and dividing by 10 to get the distance per ring you should be quickly able to suit the method to yourself for each size target.

This method compares well with my practical results of playing about with targets and distances.

It would be interesting to hear anybodies experience of this. It shouldn't be hard to try it out.