The total force hides the first limit总受力会掩盖先到极限的车轮

Keep the 2,000 kg car and 4,905 N of added downforce from Part 2. Now ask it to brake while turning. The total normal load remains 24.525 kN, yet one tyre in the example below carries 10.64 kN and another only 2.77 kN. A controller cannot spend those loads as if they belonged to one large tyre.沿用第二期的 2,000 kg 车辆与 4,905 N 额外下压力，现在让它一边制动、一边转弯。四轮法向载荷之和仍是 24.525 kN，但下面的算例中，一条轮胎承担 10.64 kN，另一条只有 2.77 kN。控制器无法把这些载荷当成属于同一条巨大轮胎的资源，任意支取。

The design task is to keep useful force available at each contact patch. That requires three checks: how the load moves, how tyre capacity changes with that load, and how much of the capacity is already being used to turn. Only then is it meaningful to decide where more downforce should act.设计要解决的是每个接触区还能提供多少有用的力。需要先检查三件事：载荷怎样转移，轮胎能力怎样随载荷改变，以及转弯已经用了多少余量。之后，才能判断新增下压力应该作用在哪里。

The geometry and tyre curve in this installment are explicit assumptions, not Tesla specifications or measured Roadster performance. The calculations are quasi-static force budgets: they omit body-motion transients, road irregularities, detailed slip behaviour and the mass and power cost of the downforce device already discussed in Part 2.本期的几何参数与轮胎曲线均为明确给定的假设，不是 Tesla 规格或 Roadster 实测性能。计算属于准静态受力预算：省略车身运动瞬态、路面起伏、详细滑移响应，以及第二期已经讨论的下压力装置质量与功耗代价。

Top view and shared-scale comparison of four normal loads. With 4.905 kN downforce, each starts at 6.131 kN. Braking shifts load to the front; adding cornering shifts load outside. Final front outer/front inner/rear outer/rear inner loads are 10.64/3.72/7.39/2.77 kN; the total stays 24.525 kN. 俯视图与同刻度曲线：固定4.905千牛下压力，四轮由各6.131千牛开始；制动使前轴增载，再叠加转弯使外侧增载。最终前外、前内、后外、后内为10.64、3.72、7.39、2.77千牛，合计始终为24.525千牛。
Figure 1 · Downforce adds load; braking and cornering redistribute it. The top view shows the final case; all tyre values are in kN. Solid and dashed lines identify outer and inner tyres. Lines compare quasi-static cases, not a time history; tyre shapes do not encode contact area.图 1 · 下压力增加总量，制动与转弯重新分配载荷。俯视图对应最终工况，轮载单位为千牛；实线表示外侧轮，虚线表示内侧轮。连线比较准静态工况，并非时间曲线；轮胎形状不编码接触面积。

First balance the four normal loads先把四轮载荷算平

Take a 2.8 m wheelbase L, equal 1.7 m tracks t and a centre-of-mass height h = 0.5 m, halfway between the axles. Use g = 9.81 m/s². Downforce D acts vertically on the vehicle centreline; a fraction β reaches the front axle and the rest the rear. For the load snapshots, β = 0.5. Positive longitudinal acceleration aₓ means driving forward; braking has a negative value.取轴距 L = 2.8 m、前后轮距均为 t = 1.7 m，质心位于两轴正中、离地 h = 0.5 m，重力加速度 g = 9.81 m/s²。下压力 D 沿车身中线竖直作用，其中 β 分配到前轴，其余分配到后轴；载荷快照取 β = 0.5。纵向加速度 aₓ 向前为正，制动时为负。

Normal load is the road reaction perpendicular to the surface at each tyre. Write the front and rear axle totals as Nf and Nr. Acceleration transfers this load from front to rear because the tyre forces act at road level while inertia acts through the higher centre of mass. The transferred amount Qₓ follows from pitch-moment balance. It is a redistribution: adding it at the rear subtracts the same amount at the front.法向载荷是路面对轮胎、垂直于路面的支撑力，前后轴的总量分别记作 Nf 与 Nr。加速时，轮胎作用力位于路面，惯性作用通过更高处的质心，俯仰力矩平衡要求载荷从前轴转移到后轴。转移量 Qₓ 是重新分配的载荷：后轴增加多少，前轴就减少多少。

During a turn, use the magnitude of lateral acceleration aᵧ. The amount transferred from the inner side to the outer side is Qᵧ. It is half the resulting outer-minus-inner load difference. Let λ = 0.60 be the front axle’s share of that transfer. It is an effective model parameter; real suspension geometry, roll stiffness and unsprung masses determine the split, as discussed in SAE 930763.转弯时取横向加速度 aᵧ 的大小。从内侧转移到外侧的载荷为 Qᵧ，外侧与内侧因此增加的载荷差是它的两倍。取 λ = 0.60，表示前轴承担这次横向载荷转移的 60%。这是模型中的等效参数；实际分配由悬架几何、侧倾刚度与非簧载质量等共同决定，参见 SAE 930763。

The plus sign belongs to the outer tyre and the minus sign to the inner tyre. At aₓ = −0.6g and aᵧ = 1g, 2.10 kN transfers to the front and 5.77 kN to the outside. Summing the four results still gives mg + D = 24.525 kN. A negative wheel load would mean the assumed four-contact state has ended; it cannot be repaired by simply clipping that value to zero.加号对应外侧轮，减号对应内侧轮。在 aₓ = −0.6g、aᵧ = 1g 时，2.10 kN 载荷转移到前轴，5.77 kN 转移到外侧；四轮相加仍为 mg + D = 24.525 kN。如果算出负载荷，就说明所假设的四轮接地状态已经失效，不能简单把负值截成零后继续沿用同一组平衡式。

Stiffer suspension can reduce body roll without eliminating this load transfer. In this model, the total lateral transfer is set by mass, acceleration, centre-of-mass height and track. Changing its front–rear distribution changes which axle pays the larger tyre-capacity penalty. Lowering the centre of mass addresses the total transfer itself.更硬的悬架可以减少车身侧倾，却不等于消除载荷转移。在这个模型中，横向转移总量由质量、加速度、质心高度与轮距决定；改变前后分配，则改变哪一轴承担更多轮胎能力损失。降低质心，才直接减少这里的转移总量。

More load buys less than proportional grip增加载荷，抓地力却不按比例增加

Part 2 used a constant friction coefficient. That makes the sum of four tyre capacities depend only on total normal load. A load-sensitive tyre breaks that simplification: increasing normal load raises peak force, but the effective coefficient can fall. Measured nonlinear tyre data in Kasprzak, Lewis and Milliken’s vehicle-optimisation study explicitly includes this dependence.第二期使用恒定附着系数，因此四条轮胎能力之和只取决于法向总载荷。轮胎的载荷敏感性会改变这个结果：法向载荷增大时，峰值力继续增加，但等效附着系数可能下降。Kasprzak、Lewis 与 Milliken 的整车优化研究使用的实测非线性轮胎数据，就包含这种依赖关系。

A tyre develops force through deformation and relative motion in its contact patch. Longitudinal slip describes a mismatch between wheel rolling speed and ground speed; slip angle describes the difference between where a wheel points and where it travels. The force builds with these motions and reaches a peak under particular conditions. The calculation below keeps only that peak capacity; it does not solve the slip needed to produce it.轮胎通过接触区的变形与相对运动建立力。纵向滑移描述车轮滚动速度与地面行进速度的差异；侧偏角描述车轮指向与实际行进方向的差异。受力随这些运动建立，并在特定条件下达到峰值。下面的计算只保留这个峰值能力，不求解产生它所需的滑移状态。

Use a deliberately simple peak-force law C(Fz), with reference load F₀ = 4,905 N, reference coefficient μ₀ = 1.5 and exponent p = 0.88. The exponent illustrates sublinear growth; it is not a fitted tyre measurement. Setting p = 1 recovers the constant-coefficient model. Temperature, pressure, camber, surface and slip conditions would require actual tyre data.这里用一个简化的峰值力函数 C(Fz)：参考载荷 F₀ = 4,905 N，参考系数 μ₀ = 1.5，指数 p = 0.88。这个指数用于呈现次线性增长，没有拟合某款轮胎的实测数据；p = 1 时就回到恒定附着系数模型。温度、胎压、外倾角、路面与滑移条件的影响，仍需真实轮胎数据确定。

Because this curve bends downward, starting from equal tyre loads and transferring some load to the outer tyre loses more capacity at the inner tyre than it gains at the outer one. The axle keeps the same total normal load, but its peak-force sum falls. That is why the outer tyre cannot fully compensate for the inner tyre’s unloading.这条曲线向下弯曲。从原本均匀的两轮向外侧转移载荷、加剧载荷不均时，内侧轮损失的能力大于外侧轮接收同样载荷所增加的能力。车轴的法向总载荷没有变，峰值力之和却下降了。这就是外侧轮无法完全补回内侧轮卸载损失的原因。

A concave assumed tyre force law shows why unequal normal loads reduce the sum of two tyre capacities at fixed axle load. 假设的轮胎峰值力曲线向下弯曲；同一车轴总载荷下，两轮载荷不均会降低两条轮胎的峰值力之和。
Figure 2 · The curve is an illustrative law, not tyre-test data. The linear p = 1 comparison removes load sensitivity and therefore removes the redistribution penalty.图 2 · 曲线是示例规律，不是轮胎台架数据。线性的 p = 1 对照没有载荷敏感性，因此也没有重新分配载荷造成的能力损失。

Even before transfer, added downforce has diminishing returns in this example. Increasing all four loads by 25% increases their summed capacity by , about 21.7%. The balanced-load sum rises from 29.43 to 35.82 kN; the braking-and-turning load snapshot gives 35.32 kN. Neither sum proves that the car can use all of it in a desired direction while also satisfying yaw balance.即使没有载荷转移，这个算例中的下压力也存在边际收益递减。四轮载荷都增加 25%，峰值力之和增加的是 ，约 21.7%。均匀载荷时，总能力从 29.43 增至 35.82 kN；前面的制动转弯载荷快照得到 35.32 kN。这两个总和都没有证明：在满足横摆力矩平衡的同时，车辆能够把全部能力用在所需方向上。

Choose a balance with room left over选择仍有余量的下压力分配

Keep D = 4,905 N and vary only its front fraction β. First isolate steady cornering: no longitudinal acceleration, a centred mass and no imposed yaw moment. Under the small-steering-angle approximation, neglecting tyre-to-body force-projection differences, each axle must supply half the required lateral force. Spare rear capacity cannot simply fill a front deficit while preserving this moment balance.保持 D = 4,905 N，只改变分给前轴的比例 β。先单独考察稳态转弯：没有纵向加速度、质心居中、没有额外横摆力矩。采用小转角近似，忽略轮胎坐标到车身坐标的受力投影差异，前后轴各承担一半横向力。后轴的剩余能力，不能在保持力矩平衡的同时直接填补前轴缺口。

For each candidate lateral acceleration, recompute the four loads and the two axle capacities. Both axles must meet demand, and all four loads must remain positive. Increasing acceleration raises demand while also increasing the transfer that reduces tyre capacity. Solving these coupled constraints gives a force-budget upper bound, not a measured cornering limit.对每个候选横向加速度，都重新计算四轮载荷和两轴能力。前后轴都要满足需求，四轮载荷也都要为正。加速度提高，一方面增加需求，另一方面又通过载荷转移削弱轮胎能力。联立这些约束，得到的是受力预算上界。

A front fraction of 30% gives about 1.58g; 50% gives 1.70g. The unconstrained peak is near 56% front and 1.74g, where the two axle bounds meet. Compared with 50:50, that is only about a 2.2% gain. At the peak, the inner-front load is just 254 N. Optimising this number alone leaves little room for unloading disturbances.前轴分到 30% 下压力时，上界约为 1.58g；分到 50% 时，约为 1.70g。不另设轮载余量的峰值出现在前轴约占 56%、加速度约为 1.74g 的位置，两轴上限在这里相遇。相对前后各半，仅提高约 2.2%，此时内侧前轮只剩 254 N 载荷。只追求这个峰值，几乎没有为卸载扰动留出空间。

Make the reserve explicit. As an illustrative screening rule, require every wheel to retain at least 500 N. This is a chosen comparison threshold, not a safety standard. At a specified target of 1.65g, accept only balances that satisfy both axle-force demands and that minimum load; the highlighted interval in Figure 3 is the resulting model-allowed range. For p = 0.88, it runs from about 53.3% to 70.1% front downforce. The low end reaches the inner-front load reserve; the high end reaches the rear-axle force bound.把余量写成约束。这里示范性地要求每个车轮至少保留 500 N 载荷；500 N 仅用于比较，不是安全标准。给定 1.65g 目标后，只有同时满足两轴受力需求与最小轮载的分配比例才予以保留，图 3 高亮的区间就是这个模型允许的范围。p = 0.88 时，前轴比例约为 53.3% 至 70.1%；下界由内侧前轮的轮载余量决定，上界由后轴受力能力决定。

Allowed front-downforce intervals at a 1.65g pure-cornering target with a 500 N minimum wheel load, for three assumed tyre exponents. 在1.65g纯转弯目标与500牛最小轮载约束下，三种假设轮胎指数对应的前轴下压力允许比例区间。
Figure 3 · Each band satisfies the same 1.65g target and 500 N minimum load under one assumed tyre exponent. The shared interval passes all three scenarios. Axle peak forces are assumed jointly usable; compatible slip angles and transient stability remain outside this calculation.图 3 · 每条色带对应一个轮胎指数，同时满足 1.65g 目标和 500 N 最小轮载；共同区间通过了这三种情景的筛选。计算假定每轴能够共同利用两条轮胎的峰值，尚未求解相容侧偏角或瞬态稳定性。

Changing the assumed tyre exponent changes that interval too. The p = 0.84, 0.88 and 0.92 cases are sensitivity scenarios, not a confidence interval from measurements. Their common accepted interval is about 53.3% to 66.8%, under the same target and reserve. A setting should not be called robust just because it sits near the peak of one curve. β is also a design choice for the pressure resultant; a real single-chamber floor may not be able to move it independently on command.改变假设的轮胎指数，允许区间也会移动。图中的 p = 0.84、0.88、0.92 是敏感性对照，不是从测量得到的置信区间；在同一目标和余量要求下，三种情景共同允许的前轴比例约为 53.3% 至 66.8%。一个设置靠近某条曲线的峰值，不足以说明它对参数变化仍有余量。β 还是压力合力位置的设计选择，真实的单腔底板未必能够按指令独立移动它。

Compare the changes, not only the peak把不同改动放在一起比较

Hold total downforce fixed and change one input at a time. Moving β to its peak gives about 1.740g with 254 N on the lightest wheel. Keeping β = 0.5 and lowering the assumed centre of mass from 0.50 to 0.45 m gives about 1.732g with 734 N. Keeping the original height and changing the front transfer share λ from 0.6 to 0.5 gives about 1.747g with 1,090 N.保持总下压力不变，每次只改一个输入。把 β 调到峰值位置，得到约 1.740g，最轻车轮载荷约 254 N。保持 β = 0.5，把假设质心高度从 0.50 m 降到 0.45 m，得到约 1.732g、734 N。保持原来的高度，把前轴载荷转移份额 λ 从 0.6 改为 0.5，则得到约 1.747g、1,090 N。

Four design cases compare lateral-force upper bound and minimum wheel load: baseline, downforce rebalance, lower centre of mass and equal front/rear transfer share. 基线、下压力前移、降低质心和均分载荷转移四种方案，分别比较横向受力上界与该上界处的最小轮载。
Figure 4 · Each row changes one input from the baseline. The minimum load is evaluated at that row’s own force upper bound, so it measures proximity to unloading there. It is not the reserve at one common acceleration, and the hardware costs are not included.图 4 · 每行只改变基线的一个输入。最小轮载在该行自身的受力上界处计算，用于比较到达各自上界时离卸载还有多远；它不是同一加速度下的余量，图中也没有计入硬件代价。

The useful conclusion is a shortlist for investigation. Lowering the centre of mass or changing transfer distribution can preserve similar force capacity with more wheel-load reserve in these assumptions. Neither change is free: packaging may prevent the height reduction, while changing λ requires geometry and stiffness choices whose transient effects are absent here. Measure those constraints before selecting hardware; 56% is not a universal target.这组计算用于筛选值得继续核查的方向。在当前假设下，降低质心或改变载荷转移分配，能够在保持相近受力能力时留下更多轮载余量。但降低质心可能受空间布置限制，改变 λ 也需要调整几何与刚度，其瞬态影响尚未计算。应先核查这些约束，再选择硬件；56% 并不是通用目标。

Finish the braking-and-turning budget把开头的弯中制动算完

One contact patch transmits both longitudinal and lateral force. With peak capacity C, a friction circle approximates the shared budget. Kapania’s Stanford thesis introduces this coupling. For example, a tyre with C = 8 kN already supplying 6 kN laterally has only about 5.29 kN left longitudinally.同一个接触区同时传递纵向力与横向力。峰值能力为 C 时，可以用摩擦圆近似它们共用的预算。Kapania 的斯坦福博士论文用这一关系引入两者的耦合。例如，C = 8 kN 的轮胎已经提供 6 kN 横向力时，纵向只剩约 5.29 kN。

Real boundaries depend on slip and can be asymmetric. Brach and Brach explain why a circle or ellipse is not a complete tyre-response model. Here it is a screening constraint: a force vector outside it fails the assumed budget; a vector inside it still needs compatible slip states and vehicle dynamics.真实边界依赖滑移状态，也可能不对称。Brach 与 Brach 的分析说明了摩擦圆或椭圆为何不能替代完整轮胎响应模型。这里把它作为筛选约束：落在外面的力向量不满足假设预算，落在里面的向量仍需要相容滑移状态与整车动力学支持。

One explicit allocation for all four tyres给四条轮胎规定同一套分配规则

Return to the original −0.6g longitudinal, 1.0g lateral snapshot, with β = 0.5, λ = 0.6 and D = 4,905 N. Continue using the small-steering-angle approximation in body coordinates. Let b be the positive braking magnitude in units of g. Fix the front brake share γ at 0.6 and split each axle’s braking equally left/right. This removes the net yaw moment contributed by unequal left/right braking.回到开头的纵向 −0.6g、横向 1.0g 快照，取 β = 0.5、λ = 0.6、D = 4,905 N，继续在车身坐标中使用小转角近似。用 b 表示以 g 为单位的正制动强度。固定前轴制动力份额 γ = 0.6，每轴左右轮均分制动力，因此左右制动产生的横摆力矩相互抵消。

The centred mass and zero requested yaw moment then give each axle half of m aᵧ. Within an axle, split that lateral demand in proportion to the two capacities Cᵢ calculated from the current loads. The same rule applies to every state below. It is a constructed allocation for checking forces, not a demonstrated controller or a globally optimal distribution.质心居中且目标净横摆力矩为零，因此前后轴各承担 m aᵧ 的一半。轴内再按当前载荷对应的两轮能力 Cᵢ，成比例分配横向力。后续每个状态都使用这条规则。它是一套用于核算受力的构造分配，并不代表已经实现的控制器或全局最优分配。

For each wheel, compare the resultant with its capacity. Define utilisation uᵢ as below. The fraction 1 − uᵢ is the remaining radial force margin as a fraction of Cᵢ; it is not an extra braking allowance at fixed lateral force.逐轮比较合力需求与能力，定义利用率 uᵢ 如下。1 − uᵢ 表示相对于 Cᵢ 的合力径向余量；它不能直接当作保持横向力不变时，还能增加的制动力比例。

The original four wheel loads are paired with longitudinal and lateral force demands and utilisation. The inside rear is closest to its capacity. 在原来的四轮载荷旁列出纵横向力需求与利用率，内侧后轮最接近能力边界。
Figure 5 · The four contacts retain Figure 1’s positions and load values. Each uses its own C(Fz), not the independent 8 kN example. At the opening snapshot, all four utilisations are below one; the inner rear has the least margin.图 5 · 四个接触区沿用图 1 的位置与轮载，各自使用 C(Fz)，不套用前面独立的 8 kN 算例。开头工况的四轮利用率都小于 1，内侧后轮余量最少。

At b = 0.6 the front-outside, front-inside, rear-outside and rear-inside utilisations are about 0.54, 0.78, 0.69 and 0.84. The normal loads sum to 24.525 kN; the assigned forces sum to −11.772 kN longitudinally and +19.620 kN laterally. With contact positions (xᵢ, yᵢ), both the front/rear lateral moments and the left/right braking moments cancel, giving zero net yaw moment.b = 0.6 时，前外、前内、后外、后内四轮利用率依次约为 0.54、0.78、0.69、0.84。法向载荷合计 24.525 kN，所分配的纵向力合计 −11.772 kN、横向力合计 +19.620 kN。以 (xᵢ, yᵢ) 表示接触点位置，前后轴横向力产生的力矩相互抵消，左右制动力产生的力矩也相互抵消，净横摆力矩为零。

Keep lateral acceleration at 1g and increase b, recalculating loads, capacities and lateral allocation each time. At about b = 0.754, the inner rear reaches u = 1 first, while its load is still about 2.50 kN. The first constraint is the combined-force budget, not wheel lift. This is the boundary of the fixed 60% front-brake rule; changing brake balance can move it, so it is not the car’s maximum braking capability.保持横向 1g，逐步增大 b，每一步重新计算轮载、能力与横向力分配。约 b = 0.754 时，内侧后轮首先达到 u = 1，此时它仍承受约 2.50 kN 载荷。先触及的是合力能力限制，而非车轮离地。这只是前制动份额固定为 60% 的规则边界；改变制动分配会移动边界，不能把它称为整车最大制动能力。

The floor must keep its pressure distribution底板要守住的是压力分布

A fan-assisted floor creates a pressure difference over an area. The resulting downforce depends on the pressure integral; its front–rear balance depends on where that pressure acts. Equal total force does not imply equal pitch moment. McMurtry’s stationary inverted demonstration is evidence that fan-assisted downforce can exist without forward speed. It does not establish Roadster hardware or performance on an uneven road.风扇辅助底板在一定面积上建立压差。总下压力取决于压差积分，前后分配取决于压力作用的位置；总力相同，不代表俯仰力矩相同。McMurtry 的静止倒置演示说明风扇辅助下压力可以不依赖前进速度，但没有证明 Roadster 的硬件方案或起伏路面上的表现。

Braking pitches the body, cornering rolls it, and a bump changes the gap between the floor and road. A larger leak demands more airflow to maintain the same pressure; if the fan cannot supply it, downforce falls. Changes that are stronger at one end can shift β even before the total force loss looks severe. A rigid body with a perfectly sealed floor hides this failure mode.制动带来俯仰，转弯带来侧倾，路面凸起改变底板与路面的间隙。漏气增加后，维持同样压差需要更多流量；风扇供不上，便会损失下压力。如果一端变化更强，即使总力损失尚不明显，β 也可能已经偏移。刚性车身与完美密封的底板模型会掩盖这个问题。

The skirt is part of that tradeoff. McMurtry’s vehicle-skirt patent describes controlling seal contact to reduce leakage while managing road friction and drag. It is a design disclosure, not a verification of every road condition. The useful suspension target is enough control of ride height, pitch and roll to preserve the pressure field, while retaining compliance for tyre contact and avoiding bottoming.裙边也参与这项取舍。McMurtry 的车辆裙边专利描述了通过控制密封接触来减少泄漏，同时管理与路面的摩擦和阻力。这是设计披露，不是对各种路况的验证。悬架需要充分控制离地间隙、俯仰与侧倾以保持压力分布，同时保留轮胎跟随路面的顺从性，并避免触底。

Simply making every spring and anti-roll bar stiffer does not solve both jobs. It changes body motion and load-transfer distribution, and can make rough-road contact harder to maintain. A usable specification pairs the allowed ride-height and attitude range with the downforce and balance that the floor actually sustains throughout that range.把弹簧和防倾杆全部加硬，无法自动同时完成这两项任务：它会改变车身运动与载荷转移分配，也可能使起伏路面的接地更难维持。可用的设计指标需要把允许的离地间隙与姿态范围，同底板在整个范围内能够保持的下压力及其分配一起给出。

How much pressure loss can the budget tolerate?这套分配能容忍多少下压力损失？

Use the opening −0.6g / 1g snapshot again. Freeze γ = 0.6, geometry and tyre law. First reduce total downforce at β = 0.5; then hold D = 4,905 N and move β. At each candidate state, recompute loads and apply the same lateral allocation rule. Accept only uᵢ ≤ 1 at every wheel and a minimum load of 500 N, using the same illustrative reserve as above.仍用开头的 −0.6g 制动、1g 转弯快照，固定 γ = 0.6、几何参数与轮胎规律。先保持 β = 0.5，逐步减少总下压力；再保持 D = 4,905 N，移动 β。每个候选状态都重新计算轮载，并按同一规则分配横向力。只保留四轮都满足 uᵢ ≤ 1、最小轮载至少 500 N 的状态，沿用前面的示例余量约束。

Allowed downforce and front-balance combinations for the original braking-and-turning snapshot, screened against tyre-force and minimum-load constraints. 原始制动转弯工况下，总下压力与前轴比例的允许组合，以轮胎受力和最小轮载约束筛选。
Figure 6 · Both constraints are checked, but the 500 N floor is inactive throughout this plot: the visible boundaries come from tyre-force capacity. The region belongs to the stated allocation rule, not a controller-independent vehicle limit. A change of D and β is a comparison of recomputed states, not a simulation of the pressure-loss transient.图 6 · 两项约束都已检查，但本图范围内的最小轮载始终高于 500 N，图中边界均由轮胎受力能力决定。区域对应指定分配规则，并不是与控制分配无关的整车极限。D 与 β 的变化比较的是重新求解后的状态，没有模拟失吸瞬态。

A 20% downforce loss raises the largest utilisation from about 0.841 to 0.893. With half the downforce left it is about 0.987; with only 40% left it exceeds one, at about 1.024. At β = 0.5 the first force boundary is near D = 2.276 kN: about 46% of the original downforce remains, and the inner rear saturates. Thus even a small-looking reduction after the 50%-remaining point can exhaust this rule’s remaining budget.下压力损失 20% 时，最大利用率从约 0.841 升到 0.893。只剩一半下压力时，约为 0.987；只剩 40% 时，则升到约 1.024，已经越界。β = 0.5 时，第一条受力边界约为 D = 2.276 kN，即原下压力还剩约 46%，内侧后轮先饱和。因此，经过“只剩一半”这个位置后，再减掉少量下压力，就会耗尽这条分配规则的剩余预算。

Keeping the full downforce does not guarantee balance either. For this snapshot and allocation, the accepted front fraction is approximately 0.059 to 0.768; beyond the low end the inner front saturates, and beyond the high end the inner rear does. These values diagnose this assumed case, not a recommended aero range. This is braking at 1g lateral acceleration, whereas Figure 3 used pure cornering at 1.65g. A different brake split, tyre map or manoeuvre produces a different region.保住总下压力，也不等于保住前后平衡。在这个快照与分配规则下，允许的前轴比例约为 0.059 至 0.768；低于下界时内侧前轮饱和，高于上界时内侧后轮饱和。这些数值用于诊断当前假设工况，并不是推荐的空气动力学范围。这里是横向 1g 的弯中制动，图 3 则是横向 1.65g 的纯转弯。换一组制动分配、轮胎数据或车辆动作，区域就会改变。

The lateral forces are reallocated in every calculation. Real pressure, suspension motion and actuators take time to respond. This map therefore identifies where a requested force budget ceases to fit after recomputation; it does not establish recovery time, transient stability or a safe response to a sudden suction failure.每次计算都重新分配了横向力，而真实压力、悬架运动与执行器都需要时间响应。因此，这张图只指出重新核算之后，目标受力预算从哪里开始不再满足约束；它没有给出恢复时间，也没有证明瞬态稳定性或突发失吸时的安全响应。

Choose the next measurement that can overturn the result下一次测量，要能推翻当前判断

The calculations separate three decisions. A peak at β ≈ 0.56 does not select hardware without a wheel-load reserve and implementation costs. Four individually admissible loads do not establish a feasible braking manoeuvre until force and moment demands are assigned. A nominally feasible allocation also needs a pressure-loss boundary. The next work should test the assumptions that move those decisions most.这些计算把三项决策分开了：β 约为 0.56 的峰值，需要结合轮载余量与实现代价才能用于选硬件；四轮载荷各自合理，还需要分配合力并满足力矩约束，才能检查制动动作；名义工况满足预算之后，还要知道压力损失会从哪里使它失效。下一步应优先检验那些足以改变决策的假设。

Check先验证什么Keep comparable保持哪些条件一致What would change the decision?什么结果会改变判断？
Tyre force versus load轮胎能力随载荷的变化Tyre type, temperature, pressure, camber and surface; cover the inner/outer load range used here and measure combined-force response.轮胎型号、温度、胎压、外倾角与路面；覆盖算例中的内外轮载荷范围，并测复合受力响应。At matched loads, compare measured peak-force intervals, including measurement uncertainty, with the p-scenario predictions. Non-overlap rejects that prediction; replace C(Fz) and recompute the β interval.在匹配轮载下，对比含测量不确定度的实测峰值区间与 p 对照的预测范围；若不相交，就否定该项预测，替换 C(Fz) 后重算 β 区间。
A matched balance comparison同条件的分配对照At a controlled target below the predicted limits, compare two achievable pressure balances with measured total D, speed, tyre state and mass held comparable. Record loads and lateral response together.在低于预测边界的受控目标下，对比两个可实现的压力分配；保持实测总 D、车速、轮胎状态与质量可比，同步记录轮载与横向响应。If the assumed transfer split is wrong, or the axle capacities cannot be jointly realised, revise λ or the force-allocation model before attributing a gain to β.若假设的转移份额不成立，或两轴能力无法共同实现，应先修正 λ 或受力分配模型，不能直接把差异归因于 β。
Controlled pressure reduction受控降低压力Start on a controlled rig or in simulation with validated components. Prescribe D and balance changes and record pressure, ride height, wheel loads and requested/available forces on one clock.先在受控台架或经过部件验证的仿真中，规定 D 与分配的变化；在同一时间轴记录压力、离地间隙、轮载，以及需求与可用受力。If force demand exceeds measured capacity, load reserve disappears, or pressure balance leaves the assumed region, the nominal budget is invalid. Any lag requires a dynamic model before a vehicle-level conclusion.若需求超过实测能力、轮载余量消失，或压力分配离开所假设区域，名义预算就不成立。若存在响应滞后，还需建立动态模型，才能下整车结论。

Begin with the tyre data: every apparent gain and failure boundary above depends on it. The balance comparison should report minimum wheel load and force utilisation as well as lateral acceleration. The pressure test should report where the budget fails and what is uncertain, rather than only a peak downforce number. None of these tests has been performed for this article.先补轮胎数据，因为前面所有增益与失效边界都依赖它。分配对照除了横向加速度，还应报告最小轮载和受力利用率；压力试验应报告预算从哪里失效，以及测量仍有哪些不确定性，不能只留下一个峰值下压力。本文尚未执行这些试验。

What the four contacts change四个接触区，改变了什么判断

The useful output is a set of conditional choices. Compare changes at equal downforce and include wheel-load reserve. Close the requested braking-and-turning motion with four force vectors and a moment balance. Then identify where pressure loss or redistribution breaks that same budget. Each step makes a design claim more specific, and gives the next measurement a clear job.有用的结果是一组带条件的选择：在相同下压力下比较改动，同时检查轮载余量；用四个力向量与力矩平衡，算完目标制动转弯动作；再找出压力下降或重新分配，会从哪里打破同一份预算。每一步都把设计判断收窄，也让下一次测量有了明确任务。

The parameters and calculated cases include the reserve constraint, sensitivity scenarios, four-wheel allocations and failure boundaries. The calculations are reproducible; they remain illustrative force budgets. Measured tyre maps, compatible slip states, actuator limits and transient dynamics are still needed to predict a vehicle.参数与计算结果包含余量约束、敏感性对照、四轮受力分配与失效边界。计算可以复现，其结果仍属于示例受力预算。预测真实车辆，还需要实测轮胎图谱、相容滑移状态、执行器限制与瞬态动力学。

Suction still asks the tyres to transmit the final force. A jet provides another path by exchanging momentum with expelled gas. Part 4 will work backward from thrust and duration to gas mass, storage, valves and nozzles, then examine what remains as the stored pressure falls.吸附最终仍要通过轮胎传递力。喷流通过排出气体交换动量，提供另一条路径。第四期将从推力与持续时间反推气体质量、储罐、阀门与喷嘴，再检查储气压力下降时，这条受力路径还能维持多久。