Archie's Archery Depot

More Info On Mechanical Bow:

Cross Bows - Since Historical Mature

Have you ever learned about "Crossbows" - what are they? Crossbow is weapon, which comprises of mounted bow that is got on a origin that bucks rockets. This is also famed as bolts.

The Chinese previously used crossbows in the 5th century. This weapon cursorily became from one of the heavy components specially during warfare - the Warlike States period of time. Usually other bows rely on the force of the archer whereas in crossbow has the drive of its feature which is advanced with the help of mechanical activate.

Mechanical trigger is accommodating as at a time umteen buttons can be made even without arduous crossbowman. The crossbows became popular for hunting and royal entourages. At times passes there were few more advancements with the arm like it was kicked upstairs with multiple-firing crossbows, which were utilized for military campaigns.

The Chinese were experts with the getting of the crossbows. The betterment of crossbows was totally dependent on advanced bronze technology that allowed cooks up to ascent it with the accurate machined trigger-mechanisms.

With time the crossbow has been reinforced a lot and has been practiced widely during wars. It has been helpful in many ways during wartime as it is from one of the powerful weapons.

Evidence of usage of cross bows was found in the western civilization also. These weapons were popular in Previous and Greek and Romans in other times in Europe. At this time, crossbows were saw to be dominant and fatal artillery that was capable of penetrating any armor. It is said that crossbows found from Past china were capable to pierce numbers of layers of iron armor. In Quondam china the military strategy, chiefly was defence reaction and onslaught of walled cities of their conquest. Crossbow was thus an ideal arm for this strategy of them. Crossbow was treasured for it potentiality to riddle with severe force and give a fatal blow to their opposition.

In eleventh century during Chinese rule in china, cross bow is to Chinese as horse is to Khitan, the plus that always was discriminatory. In the field conflicts in china, their foot would have pike servicemen with shields, row of archers and a row of crossbow workforce. As the cavalry would approach, the crossbow men would shoot initially after pike hands and bowmen. Job of pike hands and archers was to give extend to crossbow servicemen, who were capable to have most of the damage in enemy lines.

As found by some researchers the crossbows from Chinese and Mediterranean companies were gathering bows glued horizontally to wooden tillers or stocks. An arrow or a bolt would displace down a groove or through a notch in a stool, when it was shot. Common device found in both of them was a stirrup at the front of a crossbow. A someone can brace the crossbow with his foot while drawing the string with both his hands (this can be done, when in sitting position) or with a hook attached to his belt.

Commonly both these crossbows employed trigger to release the cocked string. Roman employed a crossbow having a rotating nut held in place of a lever, while Chinese crossbows used a system of hooks and levers. Both of these crossbows were specialized by its firing mechanism. Historians believed that both these cultures were independent of each other. In short crossbow is an effective weapon system since Prehistoric times.

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Crossbows
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mechanical physics?

An archer puts a 0.2 kg arrow to the bowstring. An average force of 189.8 N is exerted to draw the string back 1.44 m. g=9.8 m/s/s and assume no friction loss. With what speed does the arrow leave the bow (in m/s) and how high does it go (m)?

There are certain ambiguities in the question: there's one piece of information that's missing, and it's unclear what assumptions we are allowed to make. That said, here's what I suggest.

Since this is a Physics question and not Archery, we will make the assumption that the bow, when extended, obeys Hooke's law: that means that the force of reaction of the bow is proportional to how much I try to stretch it---if I stretch twice as much, it's resisting twice as much, and so on. That law can be written mathematically as:

F = - k x

where x is the stretch displacement, F is the force of the bow for that stretch and k is a constant solely dependent on the bow. To determine the value of k, we are told that a stretch of 1.44 m produces a force of -189.9 N. (The minus sign is a reminder that the force from the bow is in the opposite direction of the stretch; it's a force of reaction, so to speak.) Therefore:

k = 189.9 / 1.44 = 132 kg/s/s

How much "work" is required to stretch that bow by a given displacement x, from a starting position where the bow is unstretched? You may remember that the "work" is the integral of the force along the path of displacement:

W = Integral_from_0_to_x ( F dx ) = 1/2 k x^2

That represents the potential energy in the stretched bow. If we assume that the string of the bow has no mass and the bow is perfectly elastic, then, on releasing the string, all that potential energy will be transfered as kinetic energy K to the arrow, which gives us the value of K:

K = W = 1/2 k x^2

Since by definition the kinetic energy is K = 1/2 m v^2, we can calculate the initial speed v of the arrow:

v = Sqrt(k/m) x

where m is the mass of the arrow, k the constant of the bow (calculated earlier) and x is the initial stretch. Finally let's plug in the values:

v = Sqrt(132/0.2) * 1.44 = 37 m/s

That's the answer to your first question!

To answer the second part of the question, we have to know in what direction the archer is aiming. Is he aiming straight upwards, or at a 45-degree angle with the ground, or at another random angle?

IF HE IS AIMING STRAIGHT UPWARDS, the arrow will reach the highest possible height. All its initial kinetic energy will become gravitational potential energy U. You may remember that the gravitational potential energy at a height h is:

U = m g h or h = U / (m g)

where g is the gravitational constant. The arrow reaches its highest point when U equals the initial kinetic energy K:

h = K / (m g) = 1/2 m v^2 / (m g) = 1/2 v^2 / g
= 70 meters in height.

ON THE OTHER HAND, if the archer wants the arrow to go as far as possible horizontally, he'd better aim at a 45-degree angle. In that situation, we must consider only the vertical component of the arrow's initial speed. The vertical component is:

v_y = v * sin(45 degrees) = v / Sqrt(2) = 26 m/s.

The kinetic energy contained in that vertical component will transform as potential energy as the arrow climbs higher and higher; the reasoning is identical to previously. In this new situation however, the height reached will be:

h = K / (m g) = 1/2 (v_y)^2 / g = 35 meters in height.

Studying the Boat Classifieds in February
Granted puddles are frozen, the cove glassy cold, tide so low the heron is probing mud-bars for wiggly things in the muck and only his ankles are wet. More sleet and snow are ever in the offing. Not propitious sailing weather.

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