匿名使用者
匿名使用者 發問時間: 社會與文化語言 · 7 年前

翻譯20點 !!!!!!!!!!!20點 勿直接用看不懂文法

http://www.mip.berkeley.edu/physics/C+25+0.html

幫我大概敘述一下這個實驗在做神麼

他要證明或是說明的物理概念大概是神麼還有翻譯下文

The toy helicopter used in the experiment has two rotor

blades, a smaller stabilizer mounted above a larger rotor. Any

contribution of the stabilizer blade to the lift was ignored. The toy helicopter used for this experiment has a mass of 11 g. The

radius of the larger rotor blade is 6.5 cm. Take the density of

air to be 1.3 kg/m3 and the free-fall acceleration g to be 9.8

m/s2. The graph in Fig. 1 of the frequency squared versus the

mass lifted yields a slope that, when combined with the other

known values, enables one to experimentally determine the

value of l. The best fit to the data results in a value of l =

0.044. Since l is the ratio of inflow air speed to rotor tip speed,

it is possible to arrive at a speed for the air passing through

the plane of the rotor. At a frequency of 80 Hz, wR is 32.7 m/s.

That gives an air speed through the rotor plane of 1.4 m/s,

which seems a reasonable result.

It is possible with practice to fly the helicopter so that it

hovers with little motion. A stroboscopic measurement of the

rotational frequency of the flying helicopter while hovering

resulted in a frequency of 84 Hz. This measurement indicates

that the helicopter produces more lift when it is not tied to

a platform. That is interesting because it is well known that

a hovering helicopter produces better lift near the ground.

Some factor in being tethered to the platform makes the toy

helicopter seem less capable of lifting than it actually is. With

the aid of Eq. (1), a mass of 11 g for the freely hovering helicopter

and a frequency of 84 Hz for the rotor blade yields a

value of l = 0.052.

In spite of ignoring many complicated effects in the flight

of a helicopter, it is interesting that a simple physics model can

predict the relationship between the rotor frequency and the

lifting capability of a toy helicopter.

已更新項目:

如果還能幫我翻譯這個網頁說明一下會更好http://www.mip.berkeley.edu/physics/C+25+2.html

3 個解答

評分
  • 7 年前
    最佳解答

    在實驗中使用的玩具直升機有兩個轉子

    葉片,安裝一個較小的穩定劑以上的較大的轉子。任何

    被忽略的穩定葉片的升力的貢獻。在這個實驗中使用的玩具直升機有一個質量為11 g。 “

    較大的轉子葉片的半徑是6.5厘米。以密度

    空氣1.3 kg/m3和自由落體加速度g9.8

    m/s2之間。圖中的曲線圖。相對於1的頻率的平方

    質量抬起產量的斜面,當與其他組合

    已知的值,使一個實驗確定

    值為l。最適合的數據結果值L =

    0.044。由於l是流入空氣的速度的比率,在轉子前端速度,

    它是可能達到的速度,通過的空氣通過

    所述轉子的平面。在頻率為80 Hz,WR為32.7米/秒。

    這給出了一個空氣速度為1.4米/秒通過的轉子平面,

    這似乎是一個合理的結果。

    這是可能與實際飛行的直升機,以便它

    徘徊很少運動。頻閃測量

    飛行直升機的旋轉頻率,而盤旋

    導致的頻率為84赫茲。這種測量表明

    這架直升機產生更多的升力,當它不依賴於

    一個平台。這是有趣的,因為這是眾所周知的

    一個盤旋的直升機產生更好的電梯附近的地面上。

    某些因素被拴到平台使得扭蛋

    直升機似乎能夠提升比它實際上是。同

    式的援助。 (1),11克的質量為自由懸停直升

    和用於轉子葉片的84赫茲的頻率產生一個

    值L =0.052。

    儘管忽略了很多複雜的影響,在飛行

    直升機,有趣的是,一個簡單的物理模型可以

    預測轉子的頻率之間的關係和

    提升能力的玩具直升機。

  • 匿名使用者
    5 年前

    到下面的網址看看吧

    ▶▶http://qoozoo201409150.pixnet.net/blog

  • 7 年前

    你這題等於一般的四題,打字也會累的!

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