開發紀錄:一台球型吊艙的完整力矩試算——把公式鏈套在真實外形上會得到什麼

這是 LocalPapa Notes 開發紀錄系列的第二十三篇。第二十一篇的入門手冊第三層走了一次完整的八步範例計算,數字跟馬達選型計算機的預設值逐位吻合。但那組預設值描述的是一個特定形態:力臂 50mm、迎風面積 0.025 m² 的裸露酬載——酬載直接暴露在氣流裡,離轉軸有一段距離。

大多數實際在用的雲台不長那樣。EO/IR 吊艙幾乎都做成球型,而球型化的整個重點就是把轉軸擺在接近壓力中心的位置,讓力臂小一個數量級。那麼換一台球型吊艙重算,結論會不會不一樣?

這篇就是那次重算。過程中還挖出一件手冊沒講、我自己也是算完才發現的事:那張拆解表裡的「摩擦 13.0%」根本不是那個設計的性質,是模型的恆等式。

力臂:球型吊艙與裸露酬載的差別 力臂:球型吊艙與裸露酬載的差別 圖示為 1:1 實際比例(mm) 轉軸 壓力中心 阻力 氣流 裸露酬載(手冊範例)力臂 50mm酬載離軸遠 → 力臂大T_wind = 0.2188 N·m 球型吊艙(本篇)力臂 12mm轉軸貼近壓力中心 → 力臂小T_wind = 0.0612 N·m
圖 1:兩種形態的力臂差異,依實際尺寸 1:1 繪製。阻力作用在壓力中心,力矩等於阻力乘上壓力中心到轉軸的距離——球型化把這段距離從 50mm 壓到 12mm。

這台是什麼

以一台中型三光球型吊艙為對象——EO 變焦、熱像、雷射測距三個感測器裝在同一顆球裡,掛在多旋翼或無人直升機下方。這是市面上很常見的一個級別。

重要:以下每一個輸入值都是我依這個級別推理出來的估算,不是任何廠商公布的規格。這篇的價值在於示範推導過程與敏感度,不在於這些數字本身。你手上有實際規格的話,換掉再算一次,方法完全一樣。

每個估值怎麼來的

先講清楚每個數字的來源,因為文章後半整個結論都掛在這上面。

  • 球體直徑 150 mm——塞得下 30x 變焦鏡組 + 熱像 + 雷測的最小尺寸。
  • 總重 1.8 kg,內框組件約 1.0 kg——同級吊艙的常見量級。
  • 迎風投影面積 A = 0.0177 m²——球體投影,π × 0.075² = 0.01767。三軸共用同一個外形。
  • 風阻係數 Cd = 0.60——光滑球體在這個雷諾數約 0.47,但第二十二篇整理的 AIAA SciTech 2021 那篇指出,真實砲塔含突出窗口與旋轉機構接縫,阻力係數比乾淨半球–圓柱高 40% 以上,所以往上調到 0.60。
  • 力臂 L_arm:Roll 5mm/Pitch 12mm/Yaw 8mm——這是球型吊艙跟裸露酬載最大的差別。球型設計讓三軸都通過接近球心的位置,只留下壓力中心偏移,我抓約 15% 半徑。三軸不同是因為受力方向不同:阻力沿飛行方向,對 Roll 軸(縱向)幾乎不產生力矩,對 Pitch 軸(橫向水平)則乘上垂直偏移,是三軸中最大的。
  • 轉動慣量 J:Roll 0.0012/Pitch 0.0020/Yaw 0.0038 kg·m²——用 J = m·k² 估,迴轉半徑分別抓 35/45/50 mm。Yaw 是最外層,扛著內兩層加球殼,所以最大。
  • 穩定頻寬 f_bw:Roll/Pitch 12 Hz、Yaw 10 Hz——這個級別的常見值。Yaw 通常較低。
  • 目標角偏移 θ_max = 0.1 mrad——30x 變焦對 LOS 殘留的要求。
  • 扭力常數 Kt = 0.10 N·m/A——這個尺寸的 frameless 直驅馬達。
  • 速度 v = 20 m/s、陣風係數 1.4——多旋翼或無人直升機的巡航上限。

先看模型:這一軸在跟什麼對抗

算式之所以長那樣,是因為它在描述一個很具體的物理情境。先把那個情境畫出來,後面每一項是哪裡來的就不用死記。

單軸力矩模型:Pitch 軸在跟什麼對抗 單軸力矩模型:Pitch 軸在跟什麼對抗 內框+酬載 氣流 28 m/s 壓力中心 · 12mm F_drag 5.10 N 轉軸(垂直紙面) T_wind T_inertia T_motor 風阻 T_wind氣流打在壓力中心,力乘上到轉軸的距離外界加在你身上(阻抗) 摩擦 T_friction軸承預壓與走線的拖曳,與風速無關外界加在你身上(阻抗) 慣性 T_inertia載體晃動時要反向加速,才能把 LOS 留在原處你要主動產生(需求) T_motor ≥ ( T_wind + T_inertia + T_friction ) × SF
圖 2:Pitch 軸的自由體圖。三項力矩性質不同——風阻是氣流打在壓力中心產生的穩態阻抗、摩擦是軸承與走線的機構阻抗(跟風速無關)、慣性不是阻力而是「載體晃動時要主動反向加速」的需求。馬達要同時應付這三者,再乘上安全係數。

注意虛線那一項的性質不一樣:風阻與摩擦是外界加在你身上的,慣性項是你自己要主動產生的。這也是為什麼前兩者算的是「頂住」、後者算的是「加速度儲備」。

三軸完整計算

共用步驟。設計風速與阻力對三軸都一樣(同一個外形):

v_design = 20 × 1.4 = 28 m/s

F_drag = ½ × 1.225 × 28² × 0.60 × 0.0177 = 5.0997 N

Pitch 軸(瓶頸軸)逐步

  • T_wind = 5.0997 × 0.012 = 0.06120 N·m
  • α = (2π × 12)² × 0.0001 = 0.56850 rad/s²
  • T_inertia = 0.0020 × 0.56850 = 0.00114 N·m
  • T_friction = (0.06120 + 0.00114) × 15% = 0.00935 N·m
  • T_required = (0.06120 + 0.00114 + 0.00935) × 2.5 = 0.17921 N·m
  • I_cont = 0.06120 / 0.10 = 0.612 A
  • I_peak = 0.17921 / 0.10 = 1.792 A

把上面七步畫成一張圖,會比逐行讀清楚——哪個輸入餵給哪一步、三條路徑在哪裡合流:

公式鏈:從輸入到馬達電流(Pitch 軸實際數字) 公式鏈:從輸入到馬達電流(Pitch 軸實際數字) 環境輸入ρ v 1.4 × 1.4 設計風速 v_design28 m/s ½ρv²·Cd·A 氣動阻力 F_drag5.0997 N × L_arm 風阻力矩 T_wind0.06120 動態輸入f_bw θ_max (2πf)²·θ 峰值角加速度 α0.5685 × J 慣性力矩 T_inertia0.00114 15% 摩擦力矩 T_friction0.00935 三項相加0.07169 × SF 2.5 總需求 T_required0.17921 ÷ Kt 0.10 峰值電流 I_peak1.792 A 粗框是這條鏈的三個產出;每個箭頭上是該步套用的運算
圖 3:整條公式鏈與 Pitch 軸的實際數字。三條路徑(風阻/慣性/摩擦)各自算到自己的力矩,合流之後才乘安全係數、再換成電流。粗框的是這條鏈的三個產出。

三軸結果(單位 N·m):

  • RollT_wind 0.02550T_inertia 0.00068T_friction 0.00393T_required 0.07527I_cont 0.255 AI_peak 0.753 A
  • PitchT_wind 0.06120T_inertia 0.00114T_friction 0.00935T_required 0.17921I_cont 0.612 AI_peak 1.792 A
  • YawT_wind 0.04080T_inertia 0.00150T_friction 0.00634T_required 0.12161I_cont 0.408 AI_peak 1.216 A

ΣI_peak = 3.761 A(三軸同時吃峰值)。Pitch 是瓶頸軸,因為它的力臂最大。


發現一:那張拆解表裡的「摩擦 13.0%」是恆等式

手冊第三層的範例算出三項佔比是「風阻 85.1%、摩擦 13.0%、慣性 1.8%」。我這台球型吊艙的形態完全不同——力臂只有四分之一、面積更小、速度更高——結果算出來:

  • Roll:風阻 84.70%、慣性 2.26%、摩擦 13.04%
  • Pitch:風阻 85.37%、慣性 1.59%、摩擦 13.04%
  • Yaw:風阻 83.87%、慣性 3.08%、摩擦 13.04%
三項佔比:摩擦那一欄是恆等式 三項佔比:摩擦那一欄是恆等式 風阻 慣性 摩擦 手冊範例(裸露酬載) 85.11% 1.84% 13.04% Roll 軸 84.70% 2.26% 13.04% Pitch 軸 85.37% 1.59% 13.04% Yaw 軸 83.87% 3.08% 13.04% 四種形態的摩擦欄都是 13.04%——0.15 ÷ 1.15,與設計無關
圖 4:四種形態的三項佔比。虛線是摩擦段的左緣——四列完全對齊,因為那一段的寬度根本不受設計影響。

摩擦三個軸都是 13.04%,一位不差。手冊範例也是 13.04%。這不是巧合:

T_friction / T_總和 = 0.15(T_wind + T_inertia) / [(T_wind + T_inertia) × 1.15] = 0.15 / 1.15 = 13.04%

T_windT_inertia 在分子分母同時被約掉了。 只要摩擦比例填 15%,這一格永遠是 13.04%,跟你的雲台長什麼樣、飛多快、多重完全無關。

這件事的意義是:那張拆解表其實只有兩個自由度,不是三個。 你唯一在讀的資訊是「風阻與慣性的相對比例」,摩擦那一欄不帶任何設計資訊。如果你曾經看著那張表想「還好,摩擦只佔 13%」——那個 13% 不是量出來的,是你自己填的 15% 除以 1.15。

發現二:所以比例摩擦模型在球型吊艙上會低估

比例模型算出 Pitch 軸的摩擦只有 0.00935 N·m。但球型吊艙配的是精密軸承加集電環,這一類機構的靜摩擦力矩實際量級在 0.02 N·m 上下——摩擦是機構決定的絕對量,跟今天有沒有風、飛多快沒有關係

改用絕對值 0.02 N·m 重算:

  • RollT_required0.07527 變成 0.11545×1.5),I_peak 1.155 A
  • Pitch:從 0.17921 變成 0.20583(×1.1),I_peak 2.058 A
  • Yaw:從 0.12161 變成 0.15574(×1.3),I_peak 1.557 A

ΣI_peak 從 3.761 A 上升到 4.770 A

Roll 軸差最多,而且原因很直接:它的 T_wind 最小(力臂只有 5mm),比例模型跟著縮水,算出來的摩擦只有 0.00393 N·m——低估了五倍。

規則可以記成一句話:T_wind 越小,那個 15% 越不能用。氣動主導的設計(大力臂、高速)用比例模型還行;球型化、低速、室內的設計,摩擦必須當成絕對量另外估或量。

發現三:L_arm 主導一切,而它最難估

把 Pitch 軸的力臂從 0 掃到 50mm(摩擦固定用絕對值 0.02):

  • L_arm = 0mmT_wind 0.0000T_required 0.0528I_peak 0.53 A、風阻佔比 0.0%
  • L_arm = 2mmT_wind 0.0102T_required 0.0783I_peak 0.78 A、風阻佔比 32.5%
  • L_arm = 3.9mmT_wind 0.0200、風阻恰好等於絕對摩擦——這是兩者的交叉點0.02 ÷ 5.0997 = 3.92mm
  • L_arm = 5mmT_wind 0.0255T_required 0.1166I_peak 1.17 A、風阻佔比 54.7%
  • L_arm = 12mmT_wind 0.0612T_required 0.2058I_peak 2.06 A、風阻佔比 74.3%(本篇基準)
  • L_arm = 20mmT_wind 0.1020T_required 0.3078I_peak 3.08 A、風阻佔比 82.8%
  • L_arm = 30mmT_wind 0.1530T_required 0.4353I_peak 4.35 A、風阻佔比 87.9%
  • L_arm = 50mmT_wind 0.2550T_required 0.6903I_peak 6.90 A、風阻佔比 92.3%(手冊範例的裸露酬載)
力臂掃描:T_required 隨 L_arm 變化(Pitch 軸) 力臂掃描:T_required 隨 L_arm 變化(Pitch 軸) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 10 20 30 40 50 力臂 L_arm(mm) 力矩(N·m) 交叉點 3.9 mm 12mm T_required 0.6903 N·m T_wind 0.2550 N·m T_friction(絕對值) 0.0200 N·m
圖 5:力臂掃描。橘線(風阻)與綠虛線(絕對摩擦)在 3.9mm 交叉——在那之左,摩擦才是主導項。本篇的基準 12mm 已經在風阻主導區。

從 0 到 30mm,T_required8.2 倍球型吊艙的整個機構價值就在把這個數字壓小——壓到 3.9mm 以下,摩擦就取代風阻成為主導項了。

麻煩的是:L_arm 的物理意義是壓力中心到轉軸的距離,而在有流動分離的鈍體上,壓力中心會隨迎角移動。這不是量尺量得出來的,是這條公式鏈裡最不確定、也最有影響力的一項。這正是第二十二篇第四類的結論:有了 CFD 之後,正確做法是直接取鉸鏈力矩取代整條 Cd × A × L_arm 分解,而不是回頭校正 Cd

其餘參數的敏感度排序

以 Pitch 軸、L_arm = 12mm、絕對摩擦為基準,逐項單獨變動:

  • v 20 → 25 m/sT_required +41.8%
  • 摩擦 0.02 → 0.04 N·m:+24.3%
  • Cd 0.60 → 0.75:+18.6%
  • v 20 → 15 m/s:−32.5%
  • Cd 0.60 → 0.45:−18.6%
  • θ_max 放寬 10 倍(0.1 → 1 mrad):+12.4%
  • f_bw 12 → 20 Hz:+2.5%
  • J 加倍(0.0020 → 0.0040)+1.4%
敏感度:單項變動對 T_required 的影響 敏感度:單項變動對 T_required 的影響 -50% -25% 0 +25% +50% 飛行速度 v(15↔25 m/s) -32.5% +41.8% 風阻係數 Cd(0.45↔0.75) -18.6% +18.6% 摩擦(0.01↔0.04 N·m) -12.1% +24.3% θ_max(0.1→1 mrad) +12.4% 頻寬 f_bw(12→20 Hz) +2.5% 轉動慣量 J(加倍) +1.4% T_required 變化(%) J 加倍只差 1.4%——量測力氣該花在 v 與摩擦上
圖 6:把上面的清單改用 tornado 呈現,每一列是同一個參數的低端與高端。列的長短就是「這一項值不值得花力氣量準」——由上而下遞減。

J 加倍只差 1.4%——跟手冊範例得到同一個結論(那邊是 2.1%)。別花力氣去精算 J,那個時間拿去確認飛行速度上限跟量摩擦,回報大得多。

選型:這三軸該配什麼馬達

Pitch 軸需求 T_required 0.206 N·m(絕對摩擦版)、T_cont 0.061 N·m。注意 T_required 已經含了 SF 2.5,所以下面的「餘裕」是在 2.5 倍之上再多出來的部分,不是全部的安全係數。

計算機型錄裡有完整電機常數的三顆比對:

  • CubeMars GL30 KV290Kt 0.038I_peak 7.4AI_cont 2.13A):峰值扭矩 0.28 N·m ✅、連續 0.08 N·m ✅,餘裕 ×1.4/×1.3
  • CubeMars GL60 II KV28Kt 0.385I_peak 2.75AI_cont 1.56A):峰值 1.06 N·m ✅、連續 0.60 N·m ✅,餘裕 ×5.1/×9.8
  • CubeMars GL60 KV25Kt 0.444I_peak 4AI_cont 1.35A):峰值 1.78 N·m ✅、連續 0.60 N·m ✅,餘裕 ×8.6/×9.8

三顆都過關,但 GL30 只剩 ×1.4 餘裕,以估算值來說太緊——我的 L_arm 只要估低一倍,它就不夠了。而 L_arm 恰好是前面證明過最不確定的那一項。在你有 CFD 或實測鉸鏈力矩之前,這種餘裕不該用掉。

GL60 II 是比較合理的選擇,代價是重量。這也是選型的本質:餘裕買的是「估錯了還撐得住」,而你對哪一項最沒把握,就決定了你需要多少餘裕。

一個要提醒的方法論問題

計算機把 θ_max 定義為「允許殘留的最大晃動角度」,但公式 α = (2π·f_bw)² × θ_max 的行為是規格訂越嚴、需要的扭矩越少——上面敏感度表裡「θ_max 放寬 10 倍反而 +12.4%」就是這個現象。

這跟直覺相反,因為它其實不是抗擾動的物理推導,而是在估「頻寬邊緣要保留多少角加速度權限」的啟發式。把那個式子畫出來就看得懂它在算什麼:

α 從哪來:把「頻寬內壓住 θ_max」翻成角加速度 α 從哪來:把「頻寬內壓住 θ_max」翻成角加速度 f_bw 12 Hz f_bw 24 Hz(加倍) 角度 θ(t) θ_max 0.1 mrad 角加速度 α(t) 2.2740(四倍) α 峰值 0.5685 rad/s² 時間(12 Hz 的一個週期) 同樣的角度,頻率加倍 → α 變四倍。這就是 α = (2π·f)² × θ_max 的平方關係。
圖 7:α 的由來。假設雲台要在頻寬邊緣做一個振幅 θ_max 的簡諧修正動作,對時間微分兩次就得到角加速度的峰值。上下兩條 α 的 θ_max 完全相同,只有頻率差一倍——幅值卻差四倍,這就是式子裡那個平方。

手冊已經標明計算機用的是簡化簡諧振動估算而非完整動力學模型;實際設計要推 T_inertia,應該用待抵消的擾動幅值(載具在 f_bw 附近的姿態擾動有多大),而不是殘留目標。

這篇照工具的公式算,是為了讓你打開計算機對照時數字對得上。但這一項要往上做,得換一套推導。

這篇的估算邊界

再說一次,因為這關係到你能怎麼用這篇:

  • 所有輸入值都是估算,來源是「這個級別的吊艙大概長這樣」的工程推理,不是任何廠商的公布規格。
  • 有價值的是敏感度與相對關係L_arm 差 8.2 倍、J 只差 1.4%、摩擦 13.04% 是恆等式),這幾條不隨輸入值改變。
  • 絕對數值請不要直接拿去選型。 換上你自己的尺寸、重量、速度重算一次,計算過程完全一樣。

怎麼跟我協作

想把這套算在你自己的雲台上,手上準備這幾項會快很多:

  • 球體(或酬載)的外徑與迎風投影面積
  • 轉軸到壓力中心的偏移量,或更好的——CFD/風洞給的鉸鏈力矩
  • 逐軸的轉動慣量,或內外框各自的質量與大致尺寸
  • 載具的最大飛行速度
  • 軸承配置、預壓,以及走線是集電環還是直接跨軸(決定摩擦是不是能沿用比例模型)

缺哪一項就說缺,我不會替你猜一個填進去。

上一篇:力矩分析的論文文獻回顧。想試算馬達規格?雲台馬達選型計算機。想追蹤系列後續?把 LocalPapa Notes 加入書籤吧。

Dev Log: A Complete Torque Budget for a Ball Gimbal — What Happens When You Run the Formula Chain on a Real Shape

This is the twenty-third post in the LocalPapa Notes dev-log series. Layer 3 of the handbook in post twenty-one worked through a complete eight-step example whose numbers match the motor sizing calculator's defaults digit for digit. But those defaults describe one particular form factor: a payload with a 50mm moment arm and 0.025 m² frontal area, hanging exposed in the airstream, some distance from the rotation axis.

Most gimbals in actual use don't look like that. EO/IR pods are almost always built as balls, and the entire point of that shape is to put the rotation axes near the center of pressure so the moment arm shrinks by an order of magnitude. So does the conclusion change if you rerun the numbers for a ball gimbal?

This post is that rerun. Along the way it turns up something the handbook doesn't mention, and that I only noticed after doing the arithmetic: the "13.0% friction" in that breakdown table isn't a property of that design at all — it's an identity of the model.

Moment arm: ball gimbal vs exposed payload Moment arm: ball gimbal vs exposed payload Drawn 1:1 to actual scale (mm) Rotation axis Center of pressure Drag Airflow Exposed payload (handbook example)arm 50mmPayload far from axis → long armT_wind = 0.2188 N·m Ball gimbal (this post)arm 12mmAxis near center of pressure → short armT_wind = 0.0612 N·m
Figure 1: the moment arm in both form factors, drawn 1:1 to actual size. Drag acts at the center of pressure, and torque is that force times its distance to the rotation axis — going to a ball shape cuts that distance from 50mm to 12mm.

What this is

The subject is a mid-size tri-sensor ball gimbal — EO zoom, thermal, and laser rangefinder in a single ball, slung under a multirotor or unmanned helicopter. A very common class on the market.

Important: every input value below is my estimate for this class of device, not a published specification from any manufacturer. The value of this post is in the derivation and the sensitivities, not in these particular numbers. If you have real specs, substitute them and rerun — the method is identical.

Where each estimate comes from

Worth being explicit up front, because every conclusion in the second half rests on these.

  • Ball diameter 150 mm — the minimum that fits a 30x zoom lens group plus thermal plus LRF.
  • Total mass 1.8 kg, inner assembly about 1.0 kg — typical for the class.
  • Frontal area A = 0.0177 m² — ball projection, π × 0.075² = 0.01767. Shared across all three axes.
  • Drag coefficient Cd = 0.60 — a smooth sphere runs about 0.47 at this Reynolds number, but the AIAA SciTech 2021 paper covered in post twenty-two found realistic turrets with protruding windows and rotational-mechanism seams exceed a clean hemisphere-cylinder by 40%+, so it's raised to 0.60.
  • Moment arm L_arm: Roll 5mm / Pitch 12mm / Yaw 8mmthis is the big difference from an exposed payload. The ball layout puts all three axes near the sphere's center, leaving only the center-of-pressure offset, which I put at roughly 15% of the radius. The three differ because the force direction does: drag acts along the flight direction, so it produces almost no moment about the Roll (longitudinal) axis, while about the Pitch (lateral horizontal) axis it multiplies by the vertical offset — the largest of the three.
  • Inertia J: Roll 0.0012 / Pitch 0.0020 / Yaw 0.0038 kg·m² — from J = m·k² with radii of gyration of 35 / 45 / 50 mm. Yaw is outermost, carrying both inner stages plus the shell, so it's the largest.
  • Bandwidth f_bw: 12 Hz roll/pitch, 10 Hz yaw — typical for the class; yaw is usually lower.
  • Target angular deviation θ_max = 0.1 mrad — what 30x zoom demands of residual LOS.
  • Torque constant Kt = 0.10 N·m/A — a frameless direct-drive motor of this size.
  • Speed v = 20 m/s, gust factor 1.4 — cruise ceiling for a multirotor or unmanned helicopter.

First, the model: what this axis is working against

The formula looks the way it does because it describes a very specific physical situation. Draw that situation once and you no longer have to memorise where each term comes from.

Single-axis torque model: what the pitch axis is working against Single-axis torque model: what the pitch axis is working against Inner frame + payload Airflow 28 m/s Center of pressure · 12mm F_drag 5.10 N Rotation axis (into page) T_wind T_inertia T_motor Wind T_windAirflow acts at the center of pressure, force timesits distance to the axisImposed on you (resistance) Friction T_frictionBearing preload and cable drag — independent ofairspeedImposed on you (resistance) Inertia T_inertiaCounter-accelerate when the airframe moves, to holdthe LOS stillYou must produce it (demand) T_motor ≥ ( T_wind + T_inertia + T_friction ) × SF
Figure 2: free-body diagram of the pitch axis. The three torques differ in kind — wind is steady resistance from airflow acting at the center of pressure, friction is mechanical resistance from bearings and cabling (independent of airspeed), and inertia is not resistance at all but the demand to actively counter-accelerate when the airframe moves. The motor has to cover all three, then a safety factor on top.

Note that the dashed term is different in kind: wind and friction are imposed on you from outside, while the inertia term is something you have to actively produce. That is why the first two are about holding position and the third is about reserving acceleration.

The full three-axis calculation

Shared steps. Design speed and drag are identical across axes (same external shape):

v_design = 20 × 1.4 = 28 m/s

F_drag = ½ × 1.225 × 28² × 0.60 × 0.0177 = 5.0997 N

Pitch axis (the bottleneck), step by step:

  • T_wind = 5.0997 × 0.012 = 0.06120 N·m
  • α = (2π × 12)² × 0.0001 = 0.56850 rad/s²
  • T_inertia = 0.0020 × 0.56850 = 0.00114 N·m
  • T_friction = (0.06120 + 0.00114) × 15% = 0.00935 N·m
  • T_required = (0.06120 + 0.00114 + 0.00935) × 2.5 = 0.17921 N·m
  • I_cont = 0.06120 / 0.10 = 0.612 A
  • I_peak = 0.17921 / 0.10 = 1.792 A

Those seven steps read more clearly as a picture — which input feeds which step, and where the three paths merge:

The formula chain: inputs to motor current (actual pitch-axis numbers) The formula chain: inputs to motor current (actual pitch-axis numbers) Environmentρ v 1.4 × 1.4 Design speed v_design28 m/s ½ρv²·Cd·A Drag force F_drag5.0997 N × L_arm Wind torque T_wind0.06120 Dynamicsf_bw θ_max (2πf)²·θ Peak ang. accel. α0.5685 × J Inertia torque T_inertia0.00114 15% Friction torque T_friction0.00935 Sum of three0.07169 × SF 2.5 Total T_required0.17921 ÷ Kt 0.10 Peak current I_peak1.792 A Heavy borders mark the chain's three outputs; each arrow carries the operation applied
Figure 3: the full formula chain with the pitch axis's actual numbers. Three paths (wind, inertia, friction) each produce their own torque; only after they merge does the safety factor apply, and only then the conversion to current. Heavy borders mark the chain's three outputs.

All three axes (N·m):

  • Roll: T_wind 0.02550, T_inertia 0.00068, T_friction 0.00393, T_required 0.07527, I_cont 0.255 A, I_peak 0.753 A
  • Pitch: T_wind 0.06120, T_inertia 0.00114, T_friction 0.00935, T_required 0.17921, I_cont 0.612 A, I_peak 1.792 A
  • Yaw: T_wind 0.04080, T_inertia 0.00150, T_friction 0.00634, T_required 0.12161, I_cont 0.408 A, I_peak 1.216 A

ΣI_peak = 3.761 A with all three at peak simultaneously. Pitch is the bottleneck, because its moment arm is the largest.


Finding 1: the "13.0% friction" in the breakdown table is an identity

The handbook's Layer 3 example reports the three shares as "wind 85.1%, friction 13.0%, inertia 1.8%." My ball gimbal has a completely different form factor — a quarter of the moment arm, smaller area, higher speed — and it comes out as:

  • Roll: wind 84.70%, inertia 2.26%, friction 13.04%
  • Pitch: wind 85.37%, inertia 1.59%, friction 13.04%
  • Yaw: wind 83.87%, inertia 3.08%, friction 13.04%
The three shares: the friction column is an identity The three shares: the friction column is an identity Wind Inertia Friction Handbook (exposed payload) 85.11% 1.84% 13.04% Roll axis 84.70% 2.26% 13.04% Pitch axis 85.37% 1.59% 13.04% Yaw axis 83.87% 3.08% 13.04% All four read 13.04% — that is 0.15 ÷ 1.15, independent of the design
Figure 4: the three shares across four form factors. The dashed line marks the left edge of the friction segment — all four align exactly, because that segment's width doesn't depend on the design at all.

All three axes land on 13.04%, to the digit. So does the handbook's example. That's not a coincidence:

T_friction / T_total = 0.15(T_wind + T_inertia) / [(T_wind + T_inertia) × 1.15] = 0.15 / 1.15 = 13.04%

T_wind and T_inertia cancel between numerator and denominator. As long as the friction ratio is set to 15%, that cell reads 13.04% forever — regardless of what your gimbal looks like, how fast it flies, or what it weighs.

What this means: the breakdown table has two degrees of freedom, not three. The only information you're actually reading is the ratio of wind to inertia; the friction column carries no design information at all. If you've ever looked at that table and thought "fine, friction is only 13%" — that 13% wasn't measured, it's the 15% you typed in, divided by 1.15.

Finding 2: so the proportional friction model underestimates on a ball gimbal

The proportional model puts Pitch axis friction at 0.00935 N·m. But a ball gimbal runs precision bearings plus a slip ring, and static friction torque for that kind of mechanism is realistically around 0.02 N·mfriction is an absolute quantity set by the mechanism, and has nothing to do with whether there's wind today or how fast you're flying.

Rerunning with an absolute 0.02 N·m:

  • Roll: T_required goes from 0.07527 to 0.11545 (×1.5), I_peak 1.155 A
  • Pitch: from 0.17921 to 0.20583 (×1.1), I_peak 2.058 A
  • Yaw: from 0.12161 to 0.15574 (×1.3), I_peak 1.557 A

ΣI_peak rises from 3.761 A to 4.770 A.

Roll shifts the most, for a direct reason: it has the smallest T_wind (only a 5mm arm), so the proportional model shrinks along with it and returns 0.00393 N·m — a fivefold underestimate.

The rule fits in one sentence: the smaller T_wind is, the less that 15% can be trusted. Aerodynamics-dominated designs (long arm, high speed) can live with the proportional model; ball-shaped, low-speed, or indoor designs need friction estimated or measured as an absolute quantity.

Finding 3: L_arm dominates everything, and it's the hardest to estimate

Sweeping the Pitch axis moment arm from 0 to 50mm (friction held at the absolute 0.02):

  • L_arm = 0mm: T_wind 0.0000, T_required 0.0528, I_peak 0.53 A, wind share 0.0%
  • L_arm = 2mm: T_wind 0.0102, T_required 0.0783, I_peak 0.78 A, wind share 32.5%
  • L_arm = 3.9mm: T_wind 0.0200 — wind exactly equals absolute friction; this is the crossover (0.02 ÷ 5.0997 = 3.92mm)
  • L_arm = 5mm: T_wind 0.0255, T_required 0.1166, I_peak 1.17 A, wind share 54.7%
  • L_arm = 12mm: T_wind 0.0612, T_required 0.2058, I_peak 2.06 A, wind share 74.3% (this post's baseline)
  • L_arm = 20mm: T_wind 0.1020, T_required 0.3078, I_peak 3.08 A, wind share 82.8%
  • L_arm = 30mm: T_wind 0.1530, T_required 0.4353, I_peak 4.35 A, wind share 87.9%
  • L_arm = 50mm: T_wind 0.2550, T_required 0.6903, I_peak 6.90 A, wind share 92.3% (the handbook's exposed payload)
Moment-arm sweep: T_required vs L_arm (pitch axis) Moment-arm sweep: T_required vs L_arm (pitch axis) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 10 20 30 40 50 Moment arm L_arm (mm) Torque (N·m) crossover 3.9 mm 12mm T_required 0.6903 N·m T_wind 0.2550 N·m T_friction (absolute) 0.0200 N·m
Figure 5: the moment-arm sweep. The orange line (wind) crosses the dashed green line (absolute friction) at 3.9mm — to the left of that, friction is the dominant term. This post's 12mm baseline already sits well inside the wind-dominated region.

From 0 to 30mm, T_required spans 8.2×. The entire mechanical value of a ball gimbal is in driving that number down — get it under 3.9mm and friction takes over from aerodynamics as the dominant term.

The awkward part: L_arm physically means the distance from the center of pressure to the rotation axis, and on a bluff body with separated flow, the center of pressure moves with angle of attack. A ruler can't measure it. It is simultaneously the most uncertain and the most influential term in the chain — which is exactly the conclusion of cluster 4 in post twenty-two: once you have CFD, take the hinge moment directly and replace the whole Cd × A × L_arm decomposition rather than calibrating Cd.

Sensitivity ranking for everything else

Baseline is the Pitch axis at L_arm = 12mm with absolute friction, varying one term at a time:

  • v 20 → 25 m/s: T_required +41.8%
  • friction 0.02 → 0.04 N·m: +24.3%
  • Cd 0.60 → 0.75: +18.6%
  • v 20 → 15 m/s: −32.5%
  • Cd 0.60 → 0.45: −18.6%
  • θ_max relaxed 10× (0.1 → 1 mrad): +12.4%
  • f_bw 12 → 20 Hz: +2.5%
  • J doubled (0.0020 → 0.0040): +1.4%
Sensitivity: effect of varying one term on T_required Sensitivity: effect of varying one term on T_required -50% -25% 0 +25% +50% Speed v (15↔25 m/s) -32.5% +41.8% Drag coeff. Cd (0.45↔0.75) -18.6% +18.6% Friction (0.01↔0.04 N·m) -12.1% +24.3% θ_max (0.1→1 mrad) +12.4% Bandwidth f_bw (12→20 Hz) +2.5% Inertia J (doubled) +1.4% Change in T_required (%) Doubling J moves it 1.4% — measurement effort belongs on v and friction
Figure 6: the same list as a tornado, one row per parameter spanning its low and high end. Row length is exactly "how much is this term worth measuring precisely" — descending from the top.

Doubling J moves the answer 1.4% — the same conclusion the handbook example reached (2.1% there). Don't spend effort computing J precisely. That time is far better spent nailing down the airframe's speed ceiling and measuring friction.

Sizing: what motor these axes need

Pitch requires T_required 0.206 N·m (absolute-friction version) and T_cont 0.061 N·m. Note T_required already includes SF 2.5, so the "headroom" below is what's left on top of that 2.5×, not the total safety factor.

Comparing the three catalog entries with complete motor constants in the calculator:

  • CubeMars GL30 KV290 (Kt 0.038, I_peak 7.4A, I_cont 2.13A): peak torque 0.28 N·m ✅, continuous 0.08 N·m ✅, headroom ×1.4 / ×1.3
  • CubeMars GL60 II KV28 (Kt 0.385, I_peak 2.75A, I_cont 1.56A): peak 1.06 N·m ✅, continuous 0.60 N·m ✅, headroom ×5.1 / ×9.8
  • CubeMars GL60 KV25 (Kt 0.444, I_peak 4A, I_cont 1.35A): peak 1.78 N·m ✅, continuous 0.60 N·m ✅, headroom ×8.6 / ×9.8

All three pass, but GL30 has only ×1.4 left, which is too tight for estimated inputs — if my L_arm is low by a factor of two, it no longer fits. And L_arm is precisely the term shown above to be the most uncertain. Until you have CFD or a measured hinge moment, that headroom shouldn't be spent.

GL60 II is the more defensible pick, at a weight cost. That's the essence of sizing: headroom buys you "still works when the estimate was wrong," and whichever term you're least sure of is what sets how much you need.

One methodological caveat

The calculator defines θ_max as "the maximum residual deviation allowed," yet the formula α = (2π·f_bw)² × θ_max behaves such that a tighter spec demands less torque — which is what produces the "relaxing θ_max 10× costs +12.4%" line in the sensitivity table above.

That's counterintuitive because it isn't a disturbance-rejection derivation; it's a heuristic for how much angular acceleration authority to reserve at the bandwidth edge. Drawing the expression makes clear what it actually computes:

Where α comes from: turning "hold θ_max within the bandwidth" into acceleration Where α comes from: turning "hold θ_max within the bandwidth" into acceleration f_bw 12 Hz f_bw 24 Hz (doubled) Angle θ(t) θ_max 0.1 mrad Angular acceleration α(t) 2.2740 (4x) α peak 0.5685 rad/s² Time (one 12 Hz period) Same angle, double the frequency, four times the α. That is the square in α = (2π·f)² × θ_max.
Figure 7: where α comes from. Assume the gimbal performs a simple-harmonic correction of amplitude θ_max at the bandwidth edge; differentiate twice with respect to time and you get the peak angular acceleration. The two α traces have identical θ_max and differ only by a factor of two in frequency — yet their amplitudes differ by four. That is the square in the formula.

The handbook already flags that the calculator uses a simplified simple-harmonic estimate rather than a full dynamic model. For real design work, T_inertia should be driven by the amplitude of the disturbance you must cancel (how much attitude disturbance the airframe produces near f_bw), not by the residual target.

This post follows the tool's formula so the numbers match when you open the calculator alongside it. But pushing this term further requires a different derivation.

The limits of these estimates

Worth restating, because it governs how you can use this:

  • Every input is an estimate, derived from "this class of pod is roughly like this" engineering reasoning — not from any manufacturer's published specification.
  • What's durable is the sensitivities and relative relationships (L_arm spanning 8.2×, J moving only 1.4%, friction pinned at 13.04% by identity). Those don't change when the inputs do.
  • Do not take the absolute numbers into a sizing decision. Substitute your own dimensions, mass, and speed and rerun — the process is exactly the same.

How to work through this with me

To apply this to your own gimbal, having these ready makes it much faster:

  • Outer diameter and frontal projected area of the ball (or payload)
  • The offset from the rotation axis to the center of pressure — or better, a hinge moment from CFD or a wind tunnel
  • Per-axis inertia, or the mass and rough dimensions of the inner and outer frames
  • The airframe's maximum flight speed
  • Bearing arrangement, preload, and whether cables route through a slip ring or cross the axis directly (this decides whether the proportional friction model still applies)

If something's missing, say it's missing — I won't guess a value and fill it in for you.

Previous post: A Literature Review of Torque Analysis. Want to size a motor? Gimbal Motor Sizing Calculator. Want to follow the series? Bookmark LocalPapa Notes.

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